1) According to the graph theory of loop analysis, how many equilibrium equations are required at a minimum level in terms of number of branches (b) and number of nodes (n) in the graph?
a. n-1
b. b+(n-1)
c. b-(n-1)
d. b/ n-1
ANSWER: b-(n-1)
Explanation:
No explanation is available for this question!

2) What would be an order of branch impedance matrix for the below stated KVL equilibrium equation on the basis of loop or mesh analysis?
E = B (Vs – Zb Is)
a. b x 1
b. b x b
c. b-n+1) x 1
d. (b-n+1) x b
ANSWER: b x b
Explanation:
No explanation is available for this question!

3) Consider the assertions given below. Which among them do/does not specify/ies the property of ‘Complete Incidence Matrix’?
a. Determinant of a loop of a complete incidence matrix is always zero
b. Addition of all entries in any column should never be equal to zero
c. Rank of connected or oriented graph is always ‘n-1’
d. All of the above
ANSWER: Addition of all entries in any column should never be equal to zero
Explanation:
No explanation is available for this question!

4) Which parameter should be essentially equal to the number of nodes in the network in accordance to the principle of duality?
a. Total impedance
b. Total admittance
c. Number of meshes
d. Number of voltage sources
ANSWER: Number of meshes
Explanation:
No explanation is available for this question!

5) How many fundamental cutsets will be generated for a graph with ‘n’ number of nodes?
a. n+1
b. n-1
c. n2(n-1)
d. n/ n-1
ANSWER: n-1
Explanation:
No explanation is available for this question!

6) What will be the number of trees, if the graph exhibits reduction in the form of reduced incident matrix given below?
a. 16
b. 24
c. 26
d. 28
ANSWER: 24
Explanation:
No explanation is available for this question!

7) According to the linear graph theory, the number of possible trees is always equal to the determinant of product of ______
a. Only complete incidence matrix
b. Reduced incidence matrix & its transpose
c. Cut-Set matrix
d. Tie-set matrix
ANSWER: Reduced incidence matrix & its transpose
Explanation:
No explanation is available for this question!

8) What will be the value of a rectangular (complete incidence) matrix, if an associated branch is oriented towards the node?
a. 1
b. -1
c. 0
d. Not defined (∞)
ANSWER: -1
Explanation:
No explanation is available for this question!

9) How many number of minimum end nodes or terminal nodes are involved in a tree, according to its properties?
a. Only one
b. Two
c. Four
d. Infinite
ANSWER: Two
Explanation:
No explanation is available for this question!

10) Which among the below specified assertions are precisely related to the conditions applicable for a path to be an improper subgraph?
A. Incidence of a single branch at a terminating node
B. Incidence of two branches at the remaining nodes
a. A is true & B is false
b. A is false & B is true
c. Both A & B are true
d. Both A & B are false
ANSWER: Both A & B are true

11. The direction of the cut-set is?
a) same as the direction of the branch current
b) opposite to the direction of the link current
c) same as the direction of the link current
d) opposite to the direction of the branch current
Answer: a
Explanation: A cut-set is a minimal set of branches of a connected graph such that the removal of these branches causes the graph to be cut into exactly two parts. The direction of the cut-set is same as the direction of the branch current.

12. Consider the graph shown below. The direction of the cut-set of node ‘a’ is?

a) right
b) left
c) upwards
d) downwards
Answer: c
Explanation: The direction of the cut set at node ‘a’ will be the direction of the branch current at node ‘a’. So the direction of the current will be upwards.

13. Consider the graph shown above in question 2. The direction of the cut-set at node ‘b’ will be?
a) upwards
b) right
c) downwards
d) left
Answer: b
Explanation: The direction of the current will be towards right. The direction of the cut set at node ‘b’ will be the direction of the branch current at node ‘b’. So the direction of the current will be towards right.

14. In the graph shown above in the question 2, the direction of the cut-set at node ‘c’ is?
a) downwards
b) upwards
c) left
d) right
Answer: b
Explanation: The direction of the cut set at node ‘c’ will be the direction of the branch current at node ‘c’. So the direction of the current will be upwards.

15. In the graph shown in the question 2, the direction of the cut-set at node ‘d’ will be?
a) left
b) downwards
c) right
d) upwards
Answer: c
Explanation: The direction of the cut set at node ‘d’ will be the direction of the branch current at node ‘d’. So the direction of the current will be upwards.

16. The row formed at node ‘a’ in the cut set matrix in the figure shown in question 2 is?
I₁ I₂ I₃ I₄ I₅ I₆ I₇ I₈
a) +1 +1 +1 +1 0 0 0 0
b) +1 0 0 0 +1 0 0 +1
c) -1 0 0 0 -1 0 0 -1
d) -1 -1 0 0 -1 -1 0 0
Answer: b
Explanation: The direction of the cut set at node ‘a’ is towards node ‘a’. So the current direction of I₁ is same as cut set direction. So it is +1. Similarly for all other currents.

17. The row formed at node ‘c’ in the cut set matrix in the figure shown question 2 is?
a) -1 -1 0 0 +1 -1 0 0
b) 0 0 +1 0 0 -1 -1 0
c) +1 0 0 0 +1 0 0 +1
d) -1 0 0 0 -1 0 0 -1
Answer: b
Explanation: The direction of the cut set at node ‘c’ is away from node ‘c’. So the current direction of I₃ is same as cut set direction. So it is +1. Similarly for all other currents.

18. The number of cut set matrices formed from a graph is?
a) N^N-1
b) N^N
c) N^N-2
d) N^N+1
Answer: c
Explanation: For every tree, there will be a unique cut set matrix. So there will be N^N-2 cut set matrices.

19. For every tree there will be _____ number of cut set matrices.
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: For every tree, there will a unique cut set matrix. So, number of cut-set matrices for every tree = 1.

20. If a row of the cut set matrix formed by the branch currents of the graph is shown below. Then which of the following is true?
I₁ I₂ I₃ I₄ I₅ I₆ I₇ I₈
-1 -1 0 0 +1 -1 0 0
a) -V₁-V₂+V₅-V₆=0
b) -I₁-I₂+I₅-I₆=0
c) -V₁+V₂+V₅-V₆=0
d) -I₁+I₂+I₅-I₆=0
Answer: b
Explanation: KCL equations are derived from cut set matrix and these include currents not voltages. So, -I₁-I₂+I₅-I₆=0.

21. A graph is said to be a directed graph if ________ of the graph has direction.
a) 1 branch
b) 2 branches
c) 3 branches
d) every branch
Answer: d
Explanation: If every branch of the graph has direction, then the graph is said to be a directed graph. If the graph does not have any direction then that graph is called undirected graph.

22. The number of branches incident at the node of a graph is called?
a) degree of the node
b) order of the node
c) status of the node
d) number of the node
Answer: a
Explanation: Nodes can be incident to one or more elements. The number of branches incident at the node of a graph is called degree of the node.

23. If no two branches of the graph cross each other, then the graph is called?
a) directed graph
b) undirected graph
c) planar graph
d) non-planar graph
Answer: c
Explanation: If a graph can be drawn on a plane surface such that no two branches of the graph cross each other, then the graph is called planar graph .

24. Consider the graph given below. Which of the following is a not a tree to the graph?


Answer: d
Explanation: Tree is sub graph which consists of all node of original graph but no closed paths. So, ‘d’ is not a tree to the graph.

25. Number of twigs in a tree are? n- number of nodes
a) n
b) n+1
c) n-1
d) n-2
Answer: c
Explanation: Twig is a branch in a tree. Number of twigs in a tree are n-1. If there are 4 nodes in a tree then number of possible twigs are 3.

26. Loops which contain only one link are independent are called?
a) open loops
b) closed loops
c) basic loops
d) none of the above
Answer: c
Explanation: The addition of subsequent link forms one or more addition al loops. Loops which contain only one link are independent are called basic loops.

27. If the incident matrix of a graph is given below. The corresponding graph is?

     a    b   c  d   e    f
     1   +1   0  +1   0   0   +1
     2   -1  -1   0  +1   0    0
     3    0  +1   0   0   +1  -1
     4    0   0  -1  -1   -1   0

Answer: b
Explanation: For the given incidence matrix,

     a    b   c  d   e    f
     1   +1   0  +1   0   0   +1
     2   -1  -1   0  +1   0    0
     3    0  +1   0   0   +1  -1
     4    0   0  -1  -1   -1   0

the corresponding graph is ‘b’ considering the directions specified in the graph.

28. If A represents incidence matrix, I represents branch current vectors, then?
a) AI = 1
b) AI = 0
c) AI = 2
d) AI= 3
Answer: b
Explanation: If A represents incidence matrix, I represents branch current vectors, then the relation is AI= 0 that is its characteristic equation must be equated to zero

29. If a graph consists of 5 nodes, then the number of twigs in the tree are?
a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: Number of twigs= n-1. As given number of nodes are 5 then n = 5. On substituting in the equation, number of twigs =5 -1 = 4.

30. If there are 4 branches, 3 nodes then number of links in a co-tree are?
a) 2
b) 4
c) 6
d) 8
Answer: a
Explanation: Number of links = b-n+1. Given number of branches = 4 and number of nodes = 3. On substituting in the equation, number of links in a co-tree = 4 – 3 + 1 = 2.

31. The current in a closed path in a loop is called?
a) loop current
b) branch current
c) link current
d) twig current
Answer: c
Explanation: In a loop there exists a closed path and a circulating current which is called link current. The current in any branch can be found by using link currents.

32. Tie-set is also called?
a) f loop
b) g loop
c) d loop
d) e loop
Answer: a
Explanation: The fundamental loop formed by one link has a unique path in the tree joining the two nodes of the link. This loop is also called f-loop.

33. Consider the graph shown below. If a tree of the graph has branches 4, 5, 6, then one of the twigs will be?

a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: Branches of the tree are called twigs. So 4, 5, 6 are the twigs of the tree. The current in any branch of a graph can be found by using link currents.

34. Consider the graph shown in the question 3 above. If a tree of the graph has branches 4, 5, 6, then one of the links will be?
a) 3
b) 4
c) 5
d) 6
Answer: a
Explanation: The branches of the co-tree are called links. So the links will be 1, 2, 3. For a given tree of a graph addition of each link between any two nodes form a loop called fundamental loop.

35. The loop current direction of the basic loop formed from the tree of the graph is?
a) same as the direction of the branch current
b) opposite to the direction of the link current
c) same as the direction of the link current
d) opposite to the direction of the branch current
Answer: c
Explanation: The loop current direction of the basic loop formed from the tree of the graph is same as the direction of the link current.

36. Consider the graph shown below. The direction of the loop currents will be? (ACW – Anticlockwise, CW – Clockwise).

a) I₁ ACW
b) I₂ ACW
c) I₃ CW
d) I₄ ACW
Answer: a
Explanation: The direction of the loop current will be along the direction of the link current in a basic loop. So I₁ – ACW, I₂ – CW, I₃ – ACW, I₄ – CW.

37. For Tie-set matrix, if the direction of current is same as loop current, then we place ___ in the matrix.
a) +1
b) -1
c) 0
d) +1 or -1
Answer: a
Explanation: For Tie-set matrix, if the direction of current is same as loop current, then we place +1 in the matrix.

38. If a row of the tie set matrix is as given below, then its corresponding equation will be?
1 2 3 4 5 6 7 8
I₁ -1 +1 0 0 +1 0 0 0
a) -V₁+V₂+V₃=0
b) -I₁+I₂+I₃=0
c) -V₁+V₂-V₃=0
d) -I₁+I₂-I₃=0
Answer: a
Explanation: KVL equations are derived from tie set matrix and these include voltages not currents. So, -V₁+V₂+V₃=0.

39. The matrix formed by link branches of a tie set matrix is?
a) Row matrix
b) Column matrix
c) Diagonal matrix
d) Identity matrix
Answer: d
Explanation: As the direction of the basic loops of the tree are taken along the direction of the link currents, then the matrix formed by the link currents will be a identity matrix.

40. The number of tie set matrices formed from a graph are?
a) N^N-1
b) N^N
c) N^N-2
d) N^N+1
Answer: c
Explanation: For every tree, there will be a unique tie set matrix. So there will be N^N-2 tie set matrices.

1. The expression of current in R- L circuit is?
a) i=(V/R)(1+exp⁡((R/L)t))
b) i=-(V/R)(1-exp⁡((R/L)t))
c) i=-(V/R)(1+exp⁡((R/L)t))
d) i=(V/R)(1-exp⁡((R/L)t))
Answer: d
Explanation: The expression of current in R- L circuit is i = (V/R)-(V/R)exp⁡((R/L)t). On solving we get i = (V/R)(1-exp⁡((R/L)t) ).

2. The steady state part in the expression of current in the R-L circuit is?
a) (V/R)(exp⁡((R/L)t))
b) (V/R)(-exp⁡((R/L)t))
c) V/R
d) R/V
Answer: c
Explanation: The steady state part in the expression of current in the R-L circuit is steady state part = V/R. When the switch S is closed, the response reaches a steady state value after a time interval.

3. In the expression of current in the R-L circuit the transient part is?
a) R/V
b) (V/R)(-exp⁡((R/L)t))
c) (V/R)(exp⁡((R/L)t))
d) V/R
Answer: b
Explanation: The expression of current in the R-L circuit has the transient part as
(V/R)(-exp⁡((R/L)t) ). The transition period is defined as the time taken for the current to reach its final or steady state value from its initial value.

4. The value of the time constant in the R-L circuit is?
a) L/R
b) R/L
c) R
d) L
Answer: a
Explanation: The time constant of a function (V/R)e^-(R/L)t is the time at which the exponent of e is unity where e is the base of the natural logarithms. The term L / R is called the time constant and is denoted by ‘τ’.

5. After how many time constants, the transient part reaches more than 99 percent of its final value?
a) 2
b) 3
c) 4
d) 5
Answer: d
Explanation: After five time constants, the transient part of the response reaches more than 99 percent of its final value.

6. A series R-L circuit with R = 30Ω and L = 15H has a constant voltage V = 60V applied at t = 0 as shown in the figure. Determine the current (A) in the circuit at t = 0+.

a) 1
b) 2
c) 3
d) 0
Answer: d
Explanation: Since the inductor never allows sudden changes in currents. At t = 0+ that just after the initial state the current in the circuit is zero.

7. The expression of current obtained from the circuit in terms of differentiation from the circuit shown in the question 6?
a) di/dt+i=4
b) di/dt+2i=0
c) di/dt+2i=4
d) di/dt-2i=4
Answer: c
Explanation: Let the i be the current flowing through the circuit. By applying Kirchhoff’s voltage law, we get 15 di/dt+30i=60 => di/dt+2i=4.

8. The expression of current from the circuit shown in the question 6 is?
a) i=2(1-e^-2t)A
b) i=2(1+e^-2t)A
c) i=2(1+e^2t)A
d) i=2(1+e^2t)A
Answer: a
Explanation: At t = 0+ the current in the circuit is zero. Therefore at t = 0+, i = 0 => 0 = c + 2 =>c = -2. Substituting the value of ‘c’ in the current equation, we have i = 2(1-e^-2t)A.

9. The expression of voltage across resistor in the circuit shown in the question 6 is?
a) V_R =60(1+e^2t)V
b) V_R =60(1-e^-2t)V
c) V_R =60(1-e^2t)V
d) V_R =60(1+e^-2t)V
Answer: b
Explanation: Voltage across the resistor V_R = iR. On substituting the expression of current we get voltage across resistor = (2(1-e^-2t) )×30=60(1-e^-2t)V.

10. Determine the voltage across the inductor in the circuit shown in the question 6 is?
a) V_L =60(-e^-2t)V
b) V_L =60(e^2t)V
c) V_L =60(e^-2t)V
d) V_L =60(-e^2t)V
Answer: c
Explanation: Voltage across the inductor V_L = Ldi/dt. On substituting the expression of current we get voltage across the inductor = 15×(d/dt)(2(1-e^-2t)))=60(e^-2t)V.

11. The current in the R-L circuit at a time t = 0+ is?
a) V/R
b) R/V
c) V
d) R
Answer: a
Explanation: The capacitor never allows sudden changes in voltage, it will act as a short circuit at t = 0+. So the current in the circuit at t = 0+ is V/R.

12. The expression of current in R- C circuit is?
a) i=(V/R)exp⁡(t/RC )
b) i=(V/R)exp⁡(-t/RC )
c) i=(V/R)-exp(⁡t/RC )
d) i=(V/R)-exp⁡(-t/RC )
Answer: b
Explanation: The particular solution of the current equation is zero. So the expression of current in R- C circuit is i=(V/R)exp⁡(-t/RC ).

13. In an R-C circuit, when the switch is closed, the response ____________
a) do not vary with time
b) decays with time
c) rises with time
d) first increases and then decreases
Answer: b
Explanation: In a R-C circuit, when the switch is closed, the response decays with time that is the response V/R decreases with increase in time.

14. The time constant of an R-C circuit is?
a) RC
b) R/C
c) R
d) C
Answer: a
Explanation: The time constant of an R-C circuit is RC and it is denoted by τ and the value of τ in dc response of R-C circuit is RC sec.

15. After how many time constants, the transient part reaches more than 99 percent of its final value?
a) 2
b) 3
c) 4
d) 5
Answer: d
Explanation: After five time constants, the transient part of the response reaches more than 99 percent of its final value.

16.A series R-C circuit consists of resistor of 10 and capacitor of 0.1F as shown in the figure. A constant voltage of 20V is applied to the circuit at t = 0. What is the current in the circuit at t = 0?

a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A.

17. The expression of current obtained from the circuit in terms of differentiation from the circuit shown in the question 6?
a) di/dt+i=1
b) di/dt+i=2
c) di/dt+i=3
d) di/dt+i=0
Answer: d
Explanation: By applying Kirchhoff’s law, we get

Differentiating with respect to t, we get 10 di/dt+i/0.1=0 => di/dt+i=0.

18. The current equation in the circuit shown in the question 6 is?
a) i=2(e^-2t)A
b) i=2(e^2t)A
c) i=2(-e^-2t)A
d) i=2(-e^2t)A
Answer: a
Explanation: At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A. The current equation is i=2(e^-2t)A.

19. The expression of voltage across resistor in the circuit shown in the question 6 is?
a) V_R =20(e^t)V
b) V_R =20(-e^-t)V
c) V_R =20(-e^t)V
d) V_R =20(e^-t)V
Answer: d
Explanation: The expression of voltage across resistor in the circuit is V_R = iR =(2(e^-t ) )×10=20(e^-t )V.

20. Determine the voltage across the capacitor in the circuit shown in the question 6 is?
a) V_C =60(1-e^-t )V
b) V_C =60(1+e^t )V
c) V_C =60(1-e^t )V
d) V_C =60(1+e^-t )V
Answer: a
Explanation: The expression of voltage across capacitor in the circuit V_C = V(1-e^-t/RC) =20(1-e^-t)V.

21. For an R-L-C circuit, we get [D – (K₁ + K₂)][D – (K₁ – K₂)] i = 0. If K₂ is positive, then the curve will be?
a) damped
b) over damped
c) under damped
d) critically damped
Answer: b
Explanation: For an R-L-C circuit, we get [D – (K₁ + K₂)][D – (K₁ – K₂)] i = 0. If K₂ is positive, then the curve will be over damped response.

22. If the roots of an equation are real and unequal, then the response will be?
a) critically damped
b) under damped
c) over damped
d) damped
Answer: c
Explanation: If the roots of an equation are real and unequal, then the response will be over damped response. Over damped response of a system is defined as the system returns (exponentially decays) to equilibrium without oscillating.

23. If the roots of an equation are complex conjugate, then the response will be?
a) over damped
b) critically damped
c) damped
d) under damped
Answer: d
Explanation: If the roots of an equation are complex conjugate, then the response will be under damped response. Damping is an influence within or upon an oscillatory system that has the effect of reducing, restricting or preventing its oscillations.

24. If the roots of an equation are real and equal, then the response will be?
a) over damped
b) damped
c) critically damped
d) under damped
Answer: c
Explanation: If the roots of an equation are real and equal, then the response will be critically damped response. For a critically damped system, the system returns to equilibrium as quickly as possible without oscillating.

25. The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the switch is closed at t = 0. Find the equation obtained from the circuit in terms of current.

Answer: a
Explanation: At t = 0, switch S is closed when the 100V source is applied to the circuit and results in the following differential equation.

26. Replacing the differentiation with D₁, D₂ in the equation obtained from the question 5. Find the values of D₁, D₂.
a) 200±j979.8
b) -200±j979.8
c) 100±j979.8
d) -100±j979.8
Answer: b
Explanation: Let the roots of the characteristic equation are denoted by D₁, D₂. So on differentiating the equation obtained in the question 5, we get D₁ = -200+j979.8, D₂ = -200-j979.8.

27. The expression of current from the circuit shown in the question 5.
a) i=e^-200t [c₁ cos979.8t+c₂ 979.8t]A
b) i=e^200t [c₁ cos979.8t-c₂ 979.8t]A
c) i=e^-200t [c₁ cos979.8t-c₂ 979.8t]A
d) i=e^200t [c₁ cos979.8t+c₂ 979.8t]A
Answer: a
Explanation: The expression of current from the circuit will be i = e^K₁t[c₁cosK₁t + c₂sinK₂t]. So, i=e^-200t [c₁ cos979.8t+c₂ 979.8t]A.

28. At time t = 0, the value of current in the circuit shown in the question 5.
a) 1
b) 2
c) 3
d) 0
Answer: d
Explanation: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

29. The voltage across the inductor at t = 0 in the circuit shown in the question 5.
a) 50
b) 100
c) 150
d) 200
Answer: b
Explanation: At t = 0, that is initially the voltage across the inductor is 100V. => V = 100V. So we can write Ldi/dt = 100.

30. The current equation obtained from the circuit shown in the question 5.
a) i=e^-200t (1.04 sin979.8t)A
b) i=e^-200t (2.04 sin979.8t)A
c) i=e^-200t (3.04 sin979.8t)A
d) i=e^-200t (4.04 sin979.8t)A
Answer: b
Explanation: On solving the values of c₁, c₂ are obtained as c₁ = 0, c₂ = 2.04. So, the current equation is i=e^-200t (2.04 sin979.8t)A.

31. The resistance element __________ while going from the time domain to frequency domain.
a) does not change
b) increases
c) decreases
d) increases exponentially
Answer: a
Explanation: The s-domain equivalent circuit of a resistor is simply resistance of R ohms that carries a current I ampere seconds and has a terminal voltage V volts-seconds. The resistance element does not change while going from the time domain to the frequency domain.

32. The relation between current and voltage in case of inductor is?
a) v=Ldt/di
b) v=Ldi/dt
c) v=dt/di
d) v=di/dt
Answer: b
Explanation: Consider an inductor with an initial current I_o. The time domain relation between current and voltage is v=Ldi/dt.

33. The s-domain equivalent of the inductor reduces to an inductor with impedance?
a) L
b) sL
c) s²L
d) s³L
Answer: b
Explanation: If the initial energy stored in the inductor is zero, the equivalent circuit of the inductor reduces to an inductor with impedance sL ohms.

34. The voltage and current in a capacitor are related as?
a) i=Cdt/dv
b) v=Cdv/dt
c) i=Cdv/dt
d) v=Cdt/dv
Answer: c
Explanation: Consider an initially charged capacitor and the initial voltage on the capacitor is V_o. The voltage current relation in the time domain is i=Cdv/dt.

35. The s-domain equivalent of the capacitor reduces to an capacitor with impedance?
a) sC
b) C
c) 1/C
d) 1/sC
Answer: d
Explanation: The s-domain equivalent of the capacitor can be derived for the charged capacitor and it reduces to an capacitor with impedance 1/sC.

36. From the circuit shown below, find the value of current in the loop.

a) (V/R)/(s+1/RC)
b) (V/C)/(s+1/R)
c) (V/C)/(s+1/RC)
d) (V/R)/(s+1/R)
Answer: a
Explanation: Applying Kirchhoff’s law around the loop, we have V/s=1/sC I+RI. Solving above equation yields I=CV/(RCS+1)=(V/R)/(s+1/RC).

37. After taking the inverse transform of current in the circuit shown in question 6, the value of current is?
a) i=(V/C)e^-t/R
b) i=(V/C)e^-t/RC
c) i=(V/R)e^-t/RC
d) i=(V/R)e^-t/R
Answer: c
Explanation: We had assumed the capacitor is initially charged to V_o volts. By taking the inverse transform of the current, we get i=(V/R) e^-t/RC.

38. The voltage across the resistor in the circuit shown in question 6 is?
a) Ve^t/R
b) Ve^-t/RC
c) Ve^-t/R
d) Ve^t/RC
Answer: b
Explanation: We can determine the voltage v by simply applying the ohm’s law from the circuit. And applying the Ohm’s law from the circuit v = Ri = Ve^-t/RC.

39. The voltage across the resistor in the parallel circuit shown is?

a) V/(s-1/R)
b) V/(s-1/RC)
c) V/(s+1/RC)
d) V/(s+1/C)
Answer: c
Explanation: The given circuit is converted to parallel equivalent circuit. By taking the node equation, we get v/R+sCv=CV. Solving the above equation, v=V/(s+1/RC).

40. Taking the inverse transform of the voltage across the resistor in the circuit shown in question 9.
a) Ve^-t/τ
b) Ve^t/τ
c) Ve^tτ
d) Ve^-tτ
Answer: a
Explanation: By taking the inverse transform, we get v=Ve^-t/RC=Ve^-t/τ, where τis the time constant and τ = RC. And v is the voltage across the resistor.

41. What is the steady state value of F (t), if it is known that F(s) = 2s(S+1)(s+2)(s+3)?
a) 12
b) 13
c) 14
d) Cannot be determined
Answer: b
Explanation: From the equation of F(s), we can infer that, a simple pole is at origin and all other poles are having negative real part.
∴ F(∞) = lim_s→0 s F(s)
= lim_s→0 2ss(S+1)(s+2)(s+3)
= 2(s+1)(s+2)(s+3)
= 26=13.

42. What is the steady state value of F (t), if it is known that F(s) = 1(s−1)(s+2)?
a) 1
b) –12
c) 12
d) Cannot be determined
Answer: d
Explanation: The steady state value of this Laplace transform is cannot be determined since; F(s) has a pole s = 1. Hence the answer is that it cannot be determined.

43. What is the steady state value of F (t), if it is known that F(s) = 1(s+2)2(s+4)?
a) 116
b) Cannot be determined
c) 0
d) 18
Answer: c
Explanation: The steady state value of F(s) exists since all poles of the given Laplace transform have negative real part.
∴F(∞) = lim_s→0 s F(s)
= lim_s→0 s(s+2)2(s+4)
= 0.

44. What is the steady state value of F (t), if it is known that F(s) = 10(s+1)(s2+1)?
a) -5
b) 5
c) 10
d) Cannot be determined
Answer: d
Explanation: The steady state value of this Laplace transform is cannot be determined since; F(s) is having two poles on the imaginary axis (j and –j). Hence the answer is that it cannot be determined.

45. What is the steady state value of F (t), if it is known that F(s) =b/(s(s+1)(s+a)), where a>0?
a) ba
b) ab
c) 1
d) Cannot be determined
Answer: a
Explanation: F (∞) = lim_s→0 s F(s)
= lim_s→0 sbs(s+1)(s+a)
= lim_s→0 b(s+1)(s+a)
= ba.

46. The inverse Laplace transform of F(s) = 2s2+3s+2 is ______________
a) -2e^-2t + 2e^-t
b) 2e^-2t + 2e^-t
c) -2e^-2t – 2e^-t
d) 2e^-t + e^-2t
Answer: a
Explanation: s² + 3s + 2 = (s+2) (s+1)
Now, F(s) = A(s+2)+B(s+1)
Hence, A = (s+2) F(s) |_s=-2
= 2s+1|_s=-2 = -2
And, B = (s+1) F(s) |_s=-1
= 2s+2|_s=-1 = 2
∴ F(s) = −2(s+2)+2(s+1)
∴ F (t) = L^-1{F(s)}
= -2e^-2t + 2e^-t for t≥0.

47. The inverse Laplace transform of F(s) = 2s+ce−bs is _____________
a) 2e^{-k (t+b)} u (t+b)
b) 2e^{-k (t-b)} u (t-b)
c) 2e^{k (t-b)} u (t-b)
d) 2e^{k (t-b)} u (t+b)
Answer: b
Explanation: Let G(s) = 2s+c
Or, G (t) = L^-1{G(s)} = 2e^-ct
∴ F (t) = L^-1{G(s) e^-bs}
= 2e^{-k (t-b)} u (t-b).

48. The inverse Laplace transform of F(s) = 3s+5s2+7 is _____________
a) 3 sin (7–√t) – 57√ cos (7–√ t)
b) 3 sin (7–√t) + 57√ cos (7–√ t)
c) 3 cos (7–√t) + 57√ sin (7–√ t)
d) 3 cos (7–√t) – 57√ sin (7–√ t)
Answer: c
Explanation: F (t) = L^-1{F(s)}
= L^-1{3ss2+7+5s2+7}
= L^-1{3ss2+7√2+57√7√s2+7√2}
The inverse Laplace transform is = 3 cos (7–√t) + 57√ sin (7–√ t).

49. The inverse Laplace transform of F(s) = 5s2−9 is ______________
a) 16e3t+56e−3t
b) 16e3t–56e−3t
c) 56e3t+56e−3t
d) 56e3t–56e−3t
Answer: d
Explanation: F (t) = L^-1{F(s)}
= L^-1{5s2−9}
= L^-1As−3+Bs+3
Hence, A = (s-3) 5(s−3)(s+3)|s=3=56
And, B = (s+3) 5(s−3)(s+3)|s=−3=−56
The inverse Laplace transform is 56e3t–56e−3t.

50. The inverse Laplace transform of F(s) = e−3ss(s2+3s+2) is ______________
a) {0.5 + 0.5e^-2(t+3) – e^-(t+3)} u (t+3)
b) {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3)
c) {0.5 – 0.5e^-2(t-3) – e^-(t-3)} u (t-3)
d) 0.5 + 0.5e^-2t – e^-t)
Answer: b
Explanation: Let G(s) = 1s(s2+3s+2)
Or, F(s) = G(s) e^-3s
G (t) = L^-1{G(s)}
= L^-1As+Bs+2+Cs+1
Solving we get, A = 0.5, B = 0.5, C = -1
So, G (t) = 0.5 + 0.5e^-2t-e^-t
The inverse Laplace transform is F (t) = {0.5 + 0.5e^-2(t+3) – e^-(t-3)} u (t-3).

51. Given a sinusoidal voltage that has a peak to peak value of 100 V. The RMS value of the sinusoidal voltage is ___________
a) 50 V
b) 70.7 V
c) 35.35 V
d) 141.41 V
Answer: c
Explanation: Given that,
Peak value = 50 V
We know that, RMS value = PeakValue2√
= 502√ = 35.35 V.

52. The Laplace transform of F(t) = sin(2t)cos(2t) is ______________
a) 42(s2+16)
b) 1s+4–2s+2+1s
c) 2s3
d) 12s+s2(s2+36)
Answer: a
Explanation: Using the Trigonometric Identity,
We get, sin (2t) cos (2t) = 12 (sin (4t))
∴L {sin (2t) cos (2t)} = 42(s2+16).

53. Convolution of step signal 100 times that is 100 convolution operations. The Laplace transform is ______________
a) 1s100
b) 1s50
c) 1
d) s¹⁰⁰
Answer: a
Explanation: n times = u (t) * u (t) * …… * u (t)
Laplace transform of the above function = 1sn, where n is number of convolutions.
∴ Laplace transform for 100 convolutions = 1s100.

54. For the circuit given below, the Time-constant is __________

a) 1.5
b) 1.25
c) 2.5
d) 2.25
Answer: b
Explanation: We know that,
Time constant is given by R_eq.C
The equivalent resistance is given by,
R_eq = R || R
= R∗RR+R
= R2
= 5 Ω
So, Time-constant = 5X0.52
= 1.25.

55. In a dual slop integrating type digital voltmeter, the firs integrating is carried out for 50 periods of the supply frequency of 50 Hz. If the reference voltage used is 10 V, the total conversion time for an input of 40 V is?
a) 3 s
b) 2 s
c) 4 s
d) 1 s
Answer: c
Explanation: In a dual slope integrating digital voltmeter,
(t1t2) V_in = V_ref
Where, t₁ = first integration time = 50 × 150 = 1
But V_in = 40 V and V_ref = 10 V
∴ t₂ = Vint1Vref = 4 s.

56. A coil of inductance 10 H, resistance 40 Ω is connected as shown in the figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. If, at t=0⁺, the voltage across the coil is 120 V, the value of the resistance R is __________

a) Zero
b) 20 Ω
c) 40 Ω
d) 60 Ω
Answer: a
Explanation: I_L at 0^– = 12060 = 2A
120 = 2(R + 40 + 20)
∴ R = 0.

57. A coil of inductance 10 H, resistance 40 Ω is connected as shown in figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. For the value of R obtained, the time taken for 95 % of the stored energy dissipated is ________

a) 0.10 s
b) 0.15 s
c) 0.50 s
d) 1.0 s
Answer: c
Explanation: For source free circuit,
I (t) = I_oe−RTt
∴ I (t) = 0.05 = 2 × e−6010t
Or, t = 0.61 ≈ 0.5 s.

58. In the circuit shown below steady state is obtained before the switch closes at t = 0. The switch is closed for 1.5 s and is then opened. At t = 1 s, V (t) will be?

a) – 3.24 V
b) 1.97 V
c) 5.03 V
d) 13.24 V
Answer: c
Explanation: V (0^–) = 10 V = V (0⁺)
For 0<t≤1.5 s, τ = 4 × 0.05 = 0.2 s
V_OC = 5 V
V (t) = 5 + (10 – 5)e−t0.2 = 5 + 5e^-5t
V (1s) = 5.03 V.

59. The circuit shown below is at steady state before the switch closes at t=0. The switch is closed for 1.5 s and is then opened. At t = 2 s, V (t) will be?

a) 5.12 V
b) 6.43 V
c) 8.57 V
d) 9.88 V
Answer: c
Explanation: V (1.5s) = 5.002 V
For t > 1.5 s, τ = 8 × 0.05 = 0.4
V (t) = 10+ (5-10)e−(t−1.5)0.4 = 10 – 5e^-2.5(t-1.5)
For t≥1.5 s, V (2 s) = 8.57 V.

60. In the circuit given below, the voltage across capacitor when switch is closed at t = ∞ is ____________

a) 50 V
b) 20 V
c) 30 V
d) 7.5 V
Answer: a
Explanation: From the figure, we can infer that,
Voltage across capacitor = voltage across 10 Ω resistance.
Now, voltage across the 10 Ω resistance = 10010+10 X 10
= 10020.10
= 50 V.

61. In the circuit given below, the current source is 1 A, voltage source is 5 V, R₁ = R₂ = R₃ = 1 Ω, L₁ = L₂ = L₃ = 1 H, C₁ = C₂ = 1 F. The current through R₃ is _________

a) 1 A
b) 5 A
c) 6 A
d) 8 A
Answer: b
Explanation: At steady state, the circuit becomes,

∴ The current through R₃ = 51 = 5 A.

62. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________

a) 14.7 A
b) 18.5 A
c) 40.0 A
d) 50.0 A
Answer: a
Explanation: Using KVL, 100 = Rdqdt+qC
100 C = RCdqdt + q
Or, ∫qqodq100C−q=1RC∫t0dt
100C – q = (100C – q_o)e^-t/RC
I = dqdt=(100C–qo)RCe−1/1
∴ e^-t/RC = 40e^-1 = 14.7 A.

63. In the circuit given below, the current source is 1 A, voltage source is 5 V, R₁ = R₂ = R₃ = 1 Ω, L₁ = L₂ = L₃ = 1 H, C₁ = C₂ = 1 F. The current through the voltage source V is _________

a) 1 A
b) 3 A
c) 2 A
d) 4 A
Answer: d
Explanation: At steady state, the circuit becomes,

∴ The current through the voltage source V = 5 – 1 = 4 A.

64. In the circuit given below, the voltage across R by mesh analysis is _________

a) 1.59 V
b) 1 V
c) 2.54 V
d) 1.54 V
Answer: d
Explanation: In loop 1, by KVL, (10 + 2) I₁ + (-2) I₂ + 0I₃ = 5
Or, 12 I₁ – 2I₂ = 5
In loop 2, -2I₁ + (10+2+20+2) I₂ + (-2) I₃ = 0
Or, -2I₁ + 34 I₂ – 2I₃ = 0
In loop 3, 0I₁ + – 2I₂ + (2+10) I₃ = 10
Or, 0I₁ – 2I₂ + 12I₃ = 10
Now, the voltage across R is V_R = (I₂ – I₃) R
Now, I₂ = 3604800=340
Now, I₃ = 40604800=203240 A
Therefore, V_R = (I₂ – I₃) R = [340–203240]2 = 1.54 V.

65. In a dual slop integrating type digital voltmeter, the firs integrating is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V, the total conversion time for an input of 1 V is _________
a) 0.01 s
b) 0.05 s
c) 0.1 s
d) 1 s
Answer: c
Explanation: In a dual slope integrating digital voltmeter,
(t1t2) V_in = V_ref
Where, t₁ = first integration time = 10 × 150 = 0.25
But V_in = 1 V and V_ref = 2 V
∴ t₂ = Vint1Vref = 0.1 s.

66. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. The internal resistance and a full- scale deflection produced by a DC current are 100 Ω and 1 mA respectively. A voltage of 100 V (rms) is applied to the input terminals. The value of R required is _________
a) 63.56 Ω
b) 89.83 Ω
c) 89.83 kΩ
d) 141.3 kΩ
Answer: c
Explanation: V_OAverage = 0.636 × 2–√V_rms = 0.8993 V_rms
The deflection with AC is 0.8993 times that with DC for the same value of voltage V
S_AC = 0.8993 S_DC
S_DC of a rectifier type instrument is 1Ifs where I_fs is the current required to produce full scale deflection, I_fs = 1 mA; R_m = 100 Ω; S_DC = 10³ Ω/V
S_AC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R_S = S_AC V – R_m – 2R_d, where R_d is the resistance of diode, for ideal diode R_d = 0
∴ R_S = 899.3 × 100 – 100 = 89.83 kΩ.

67. A 1 μF capacitor is connected across a 50 V battery. The battery is kept closed for a long time. The circuit current and voltage across capacitor is __________
a) 0.5 A and 0 V
b) 0 A and 50 V
c) 20 A and 5 V
d) 0.05 A and 5 V
Answer: b
Explanation: We know that, when the capacitor is fully charged, it acts as an open circuit.
That is when the capacitor is fully charged.
So, the circuit current and voltage across capacitor are 0 A and 50 V respectively.

68. In the circuit given below, the switch is closed at t = 0. At t = 0⁺ the current through C is ___________

a) 2 A
b) 3 A
c) 4 A
d) 5 A
Answer: d
Explanation: We know that,
When the capacitor is fully charged, it acts as a short circuit.
The equivalent resistance R_eq = (10 + 10) Ω
= 20 Ω
Given voltage = 100 V
So, current through C = 10020 A = 5 A.

69. A digital Multimeter reads 10 V when fed with a triangular wave, which is symmetric about the time-axis. If the input is same, the rms reading meter will read?
a) 203√
b) –103√
c) –203√
d) 103√
Answer: d
Explanation: For triangular wave Average value = Vm3, rms value = Vm3√
∴ Vm3 = 10 V or, V_m = 30 V.

70. A resistance R is measured using the connection shown in the below figure.

The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω. The value of R is _________
a) 12.75 Ω
b) 12.0 Ω
c) 12.25 Ω
d) 12.5 Ω
Answer: c
Explanation: Percentage error in ammeter = ± 110×100 × 100 = ± 0.1%
Percentage error in voltmeter= ±110×150 × 100 = ± 0.067%
So, δI = ± 0.3 ± 0.1 = ± 0.4%
δV = ± 0.4 ± 0.067 = ± 0.467%
R = VI
So, error = ± δV ± δI = ± 0.867
Measured value of resistance = R_m = 12510 = 12.5
∴ True value = R_m(1-RaRm) = 12.25 Ω.

1. The ratio of voltage transform at first port to the voltage transform at the second port is called?
a) Voltage transfer ratio
b) Current transfer ratio
c) Transfer impedance
d) Transfer admittance
Answer: a
Explanation: Voltage transfer ratio is the ratio of voltage transform at first port to the voltage transform at the second port and is denoted by G(s). G₂₁ = V₂(s)/V₁(s) G₁₂ = V₁(s)/V₂(s).

2. The ratio of the current transform at one port to current transform at other port is called?
a) Transfer admittance
b) Transfer impedance
c) Current transfer ratio
d) Voltage transfer ratio
Answer: c
Explanation: Current transfer ratio is the ratio of the current transform at one port to current transform at other port and is denoted by α(s). α₁₂(s) = I₁(s)/I₂(s) α₂₁(s) = I₂(s)/I₁(s).

3. The ratio of voltage transform at first port to the current transform at the second port is called?
a) Voltage transfer ratio
b) Transfer admittance
c) Current transfer ratio
d) Transfer impedance
Answer: d
Explanation: Transfer impedance is the ratio of voltage transform at first port to the current transform at the second port and is denoted by Z(s). Z₂₁(s) = V₂(s)/I₁(s) Z₁₂(s) = V₁(s)/I₂(s).

4. For the network shown in the figure, find the driving point impedance.

a) (s²-2s+1)/s
b) (s²+2s+1)/s
c) (s²-2s-1)/s
d) (s²+2s-1)/s
Answer: b
Explanation: Applying Kirchoff’s law at port 1, Z(S)=V(S)/I(S), where V(s) is applied at port 1 and I(s) is current flowinmg through the network. Then Z(S)=V(S)/I(S) = 2+S+1/S = (s²+2s+1)/s.

5. Obtain the transfer function G₂₁ (S) in the circuit shown below.

a) (s+1)/s
b) s+1
c) s
d) s/(s+1)
Answer: d
Explanation: Applying Kirchhoff’s law V₁ (S) = 2 I₁ (S) + 2 sI₁ (S) V₂ (S) = I₁ (S) X 2s Hence G₂₁ (S) = V₂(s)/V₁(s) =2 s/(2+2 s)=s/(s+1).

6. Determine the transfer function Z₂₁ (S) in the circuit shown in question 5.
a) s
b) 2 s
c) 3 s
d) 4 s
Answer: b
Explanation: The transfer function Z₂₁ (S) is Z₂₁ (S) = V₂(S)/I₁(S). V₂ (S) = I₁ (S) X 2s. V₂(S)/I₁(S)=2s. On substituting Z₂₁ (S) = 2s.

7. Find the driving point impedance Z₁₁ (S) in the circuit shown in question 5.
a) 2(s+2)
b) (s+2)
c) 2(s+1)
d) (s+1)
Answer: c
Explanation: The driving point impedance Z₁₁ (S) is Z₁₁ (S)=V₁(S)/I₁(S). V₁ (S) = 2 I₁ (S) + 2 sI₁ (S) => V₁(S) = (2+2s)I₁(S) => V₁(S)/I₁(S) = 2(s+1). On substituting Z₁₁ (S) = 2(S+1).

8. Obtain the transfer function G₂₁ (s) in the circuit shown below.

a) (8 S+2)/(8 S+1)
b) (8 S+2)/(8 S+2)
c) (8 S+2)/(8 S+3)
d) (8 S+2)/(8 S+4)
Answer: d
Explanation: From the circuit, the parallel combination of resistance and capacitance can be combined into equivalent in impedance. Z_eq(S) = 1/(2 S+1/2)=2/(4 S+1). Applying Kirchhoff’s laws, we have V₂ (S) = 2 I₁(S) => V₁ (S) = I₁ (S)[2/(4 S+1)+2] = I₁ (S)[(8 S+4)/(4 S+1)] The transfer function G₂₁ (s) = V₂(s)/V₁(s) =2 I₁(S)/((8 S+4)/(4 S+1))I₁(S) =(8 S+2)/(8 S+4).

9. Obtain the transfer function Z₂₁(s) in the circuit shown in question 8.
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: The transfer function Z₂₁(s) is Z₂₁ (S) = V₂(S)/I₁(S). V₂ (S) = 2 I₁(S) => V₂ (S)/I₁ =2. On substituting Z₂₁(s) = 2.

10. Determine the driving point impedance Z₁₁(S) in the circuit shown in question 8.
a) (8 S+4)/(4 S+4)
b) (8 S+4)/(4 S+3)
c) (8 S+4)/(4 S+2)
d) (8 S+4)/(4 S+1)
Answer: d
Explanation: The driving point impedance Z₁₁(S) is Z₁₁(S) = V₁(s)/I₁(s). V₁(s) = I₁(s)((2/(4s+1))+2) = I₁(s)((8s+4)/(4s+1)) => V₁(s)/I₁(s) = ((8s+4)/(4s+1)). On substituting we get Z₁₁(S) = (8S+4)/(4S+1).

11. The coefficients of the polynomials P (S) and Q (S) in the network function N (S) are ________ for passive network.
a) real and positive
b) real and negative
c) complex and positive
d) complex and negative
Answer: a
Explanation: The coefficients of the polynomials P (S) and Q (S) in the network function N (S) are real and positive for passive network. On factorising the network function we obtain the poles and zeros.

12. The scale factor is denoted by the letter?
a) G
b) H
c) I
d) J
Answer: b
Explanation: The scale factor is denoted by the letter ‘H’ and its value is equal to the ratio of a_o to b_o.

13. The zeros in the transfer function are denoted by?
a) 3
b) 2
c) 1
d) 0
Answer: d
Explanation: The roots of the equation P (S) = 0 are zeros of the transfer function. The zeros in the transfer function are denoted by ‘o’.

14. The poles in the transfer function are denoted by?
a) x
b) y
c) z
d) w
Answer: a
Explanation: The roots of the equation Q (S) = 0 are poles of the transfer function. The poles in the transfer function are denoted by ‘x’.

15. The network function N (S) becomes _________ when s is equal to anyone of the zeros.
a) 1
b) 2
c) 0
d) ∞
Answer: c
Explanation: The network function N (S) becomes zero when s in the transfer function is equal to anyone of the zeros as the network function is completely defined by its poles and zeros.

16. The N (S) becomes ________ when s is equal to any of the poles.
a) ∞
b) 0
c) 1
d) 2
Answer: a
Explanation: The network function is completely defined by its poles and zeros and the network function N (S) becomes infinite when s in the transfer function is equal to anyone of the poles.

17. If the poles or zeros are not repeated, then the function is said to be having __________ poles or ________ zeros.
a) simple, multiple
b) multiple, simple
c) simple, simple
d) multiple, multiple
Answer: c
Explanation: If the poles or zeros are not repeated, then the function is said to be having simple poles or simple zeros and the network function is said to be stable when the real parts of the poles and zeros are negative.

18. If the poles or zeros are repeated, then the function is said to be having __________ poles or ________ zeros.
a) multiple, multiple
b) simple, simple
c) multiple, simple
d) simple, multiple
Answer: a
Explanation: If there are repeated poles or zeros, then function is said to be having multiple poles or multiple zeros and the network function is stable if the poles and zeros lie within the left half of the s-plane.

19. If the number of zeros (n) are greater than the number of poles (m), then there will be _________ number of zeros at s = ∞.
a) n
b) m
c) n-m
d) n+m
Answer: c
Explanation: If the number of zeros (n) are greater than the number of poles (m), then there will be (n-m) number of zeros at s = ∞ and to obtain (n-m) zeros at s = ∞ the condition is n>m.

20. If the number of poles (m)are greater than the number of zeros (n), then there will be _________ number of zeros at s = ∞.
a) m+n
b) m-n
c) m
d) n
Answer: b
Explanation: If the number of poles (m)are greater than the number of zeros (n), then there will be (m-n) number of zeros at s = ∞ and to obtain (m-n) poles at s = ∞ the condition is m>n.

21. The driving point function is the ratio of polynomials in s. Polynomials are obtained from the __________ of the elements and their combinations.
a) transform voltage
b) transform current
c) transform impedance
d) transform admittance
Answer: c
Explanation: The driving point function is the ratio of polynomials in s. Polynomials are obtained from the transform impedance of the elements and their combinations and if the zeros and poles are not repeated then the poles or zeros are said to be distinct or simple.

22. The pole is that finite value of S for which N (S) becomes __________
a) 0
b) 1
c) 2
d) ∞
Answer: d
Explanation: The quantities P₁, P₂ … P_m are called poles of N (S) if N (S) = ∞ at those points. The pole is that finite value of S for which N (S) becomes infinity.

23. A function N (S) is said to have a pole (or zero) at infinity, if the function N (1/S) has a pole (or zero) at S = ?
a) ∞
b) 2
c) 0
d) 1
Answer: c
Explanation: A function N (S) is said to have a pole (or zero) at infinity, if the function N (1/S) has a pole (or zero) at S = infinity. A zero or pole is said to be of multiplicity ‘r’ if (S-Z)^r or(S-P)^r is a factor of P(s) or Q(s).

24. The number of zeros including zeros at infinity is __________ the number of poles including poles at infinity.
a) greater than
b) equal to
c) less than
d) greater than or equal to
Answer: b
Explanation: The number of zeros including zeros at infinity is equal to the number of poles including poles at infinity and it cannot be greater than or less than the number of poles including poles at infinity.

25. The poles of driving point impedance are those frequencies corresponding to ___________ conditions?
a) short circuit
b) voltage source
c) open circuit
d) current source
Answer: c
Explanation: A zero of N(s) is a zero of V(s),it signifies a short circuit. Similarly a pole of Z(s) is a zero of I(s). The poles of driving point impedance are those frequencies corresponding to open circuit conditions.

26. The zeros of driving point impedance are those frequencies corresponding to ___________ conditions?
a) current source
b) open circuit
c) voltage source
d) short circuit
Answer: d
Explanation: The zeros of driving point impedance are those frequencies corresponding to short circuit conditions as pole of Z(s) is a zero of I(s) and zero of N(s) is a zero of V(s),it signifies a short circuit.

27. In the driving point admittance function, a zero of Y (s) means a _______of I (S).
a) 1
b) 2
c) 3
d) zero
Answer: d
Explanation: In the driving point admittance function, a zero of Y (s) means a zero of I (S) i.e., the open circuit condition as the driving point admittance function is the ratio of I(s) to V(s).

28. In the driving point admittance function, a pole of Y (s) means a _______ of V (S).
a) zero
b) 1
c) 2
d) 3
Answer: a
Explanation: The driving point admittance function Y(s) = I(s)/V(s). In the driving point admittance function, a pole of Y (s) means a zero of V (S) i.e., the short circuit condition.

29. The real part of all zeros and poles must be?
a) positive or zero
b) negative or zero
c) positive
d) negative
Answer: b
Explanation: The real part of all zeros and poles must be negative or zero. But the poles or zeros should not be positive because if they are positive, then they will lie in the right-half of the s-plane.

30. Poles or zeros lying on the jω axis must be?
a) complex
b) at least one complex pole
c) at least one complex zero
d) simple
Answer: d
Explanation: Poles or zeros lying on the jω axis must be simple because on jω axis the imaginary part of poles or zeros will be zero.

31. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?
a) all odd terms are missing
b) all even terms are missing
c) all even or odd terms are missing
d) all even and odd terms are missing
Answer: c
Explanation: All the quotients in the polynomial P(s) are positive. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing.For example P(s) = s³+3s is Hurwitz because all quotient terms are positive and all even terms are missing.

32. The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on ____________
a) right half of s plane
b) left half of s-plane
c) on jω axis
d) on σ axis
Answer: c
Explanation: The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on jω axis not on right half of s plane or on left half of s-plane.

33. If the polynomial P (s) is either even or odd, then the roots of P (s) lie on __________
a) on σ axis
b) on jω axis
c) left half of s-plane
d) right half of s plane
Answer: b
Explanation: If the polynomial P (s) is either even or odd, then the roots of P (s) lie on jω axis not on right half of s plane or on left half of s-plane.

34. If the ratio of the polynomial P (s) and its derivative gives a continued fraction expansion with ________ coefficients, then the polynomial P (s) is Hurwitz.
a) all negative
b) all positive
c) positive or negative
d) positive and negative
Answer: b
Explanation: If the ratio of the polynomial P (s) and its derivative P^‘(s) gives a continued fraction expansion with all positive coefficients, then the polynomial P (s) is Hurwitz. If all the quotients in the continued fraction expansion are positive, the polynomial P(s) is positive.

35. Consider the polynomial P(s)=s⁴+3s²+2. The given polynomial P (s) is Hurwitz.
a) True
b) False
Answer: a
Explanation: P(s)=s⁴+3s²+2 => P^‘ (s)=4s³+6s
After doing the continued fraction expansion, we get all the quotients as positive. So, the polynomial P (s) is Hurwitz.

36. When s is real, the driving point impedance function is _________ function and the driving point admittance function is _________ function.
a) real, complex
b) real, real
c) complex, real
d) complex, complex
Answer: b
Explanation: When s is real, the driving point impedance function is real function and the driving point admittance function is real function because the quotients of the polynomials P(s) and Q(s) are real. When Z(s) is determined from the impedances of the individual branches, the quotients are obtained by adding together, multiplying or dividing the branch parameters which are real.

37. The poles and zeros of driving point impedance function and driving point admittance function lie on?
a) left half of s-plane only
b) right half of s-plane only
c) left half of s-plane or on imaginary axis
d) right half of s-plane or on imaginary axis
Answer: c
Explanation: The poles and zeros of driving point impedance function and driving point admittance function lie on left half of s-plane or on imaginary axis of the s-plane..

38. For real roots of s_k, all the quotients of s in s²+ω_k² of the polynomial P (s) are __________
a) negative
b) non-negative
c) positive
d) non-positive
Answer: b
Explanation: For real roots of s_k, all the quotients of s in s²+ω_k² of the polynomial P (s) are non-negative. So by multiplying all factors in P(s) we find that all quotients are positive.

39. The real parts of the driving point function Z (s) and Y (s) are?
a) positive and zero
b) positive
c) zero
d) positive or zero
Answer: d
Explanation: The real parts of the driving point impedance function Z (s) and driving point admittance function Y (s) are positive or zero.

40. For the complex zeros to appear in conjugate pairs the poles of the network function are ____ and zeros of the network function are ____________
a) complex, complex
b) complex, real
c) real, real
d) real, complex
Answer: c
Explanation: P(s) and Q(s) are real when s is real. So the poles of the network function are real and zeros of the network function are real, the complex zeros to appear in conjugate pairs.

41. The coefficients of numerator polynomial and the denominator polynomial in a transfer function must be?
a) real
b) complex
c) at least one real coefficient
d) at least one complex coefficient
Answer: a
Explanation: The coefficients of P(s), the numerator polynomial and of Q(s), the denominator polynomial in a transfer function must be real. Therefore all poles and zeros if complex must occur in conjugate pairs.

42. In a transfer function, the degree of numerator polynomial is ___________ than the degree of the denominator polynomial.
a) greater than
b) less than
c) equal to
d) less than or equal to
Answer: d
Explanation: In a transfer function, the degree of numerator polynomial is less than or equal to than the degree of the denominator polynomial. And the degree of the numerator polynomial of Z₂₁(s) or Y₂₁(s) is less than or equal to the degree of the denominator polynomial plus one.

43. The real parts of all poles and zeros in a driving point function must be?
a) zero
b) negative
c) zero or negative
d) positive
Answer: c
Explanation: The real parts of all poles and zeros in a driving point function must be zero or negative but should not be positive and the complex or imaginary poles and zeros must occur in conjugate pairs.

44. If the real part of driving point function is zero, then the pole and zero must be?
a) complex
b) simple
c) one complex pole
d) one complex zero
Answer: b
Explanation: If the real part of driving point function is zero, then the pole and zero must be simple but should not contain any complex pole or complex zero.

45. The degree of the numerator polynomial and denominator polynomial in a driving point function may differ by?
a) 0
b) 1
c) 0 or 1
d) 2
Answer: c
Explanation: The degree of numerator polynomial and denominator polynomial in a driving point function may differ by zero or one. And the polynomials P(s) and Q(s) may not have any missing terms between the highest and lowest degrees unless all even or odd terms are missing.

46. The lowest degree in numerator polynomial and denominator polynomial in a driving point function may differ by at most __________
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: The lowest degree in numerator polynomial and denominator polynomial in a driving point function may differ by at most one and the coefficients in the polynomials P(s) and Q(s) of network function must be real and positive.

47. The coefficients in the denominator polynomial of the transfer function must be?
a) positive
b) negative
c) positive or zero
d) negative or zero
Answer: a
Explanation: The coefficients in the denominator polynomial of the transfer function must be positive but should not be negative and the coefficients in the polynomials P(s) and Q(s) of transfer function must be real.

48. The coefficients in the numerator polynomial of the transfer function may be?
a) must be negative
b) must be positive
c) may be positive
d) may be negative
Answer: d
Explanation: The coefficients in the numerator polynomial of the transfer function may be negative and the complex or imaginary poles and zeros must occur in conjugate pairs.

49. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?
a) all odd terms are missing
b) all even terms are missing
c) all even or odd terms are missing
d) all even and odd terms are missing
Answer: c
Explanation: The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing and the polynomial P(s) may have missing terms between the lowest and the highest degree.

50. The degree of numerator polynomial in a transfer function may be as small as _________ independent of the degree of the denominator polynomial.
a) 1
b) 2
c) 0
d) 3
Answer: c
Explanation: The degree of numerator polynomial in a transfer function may be as small as zero, independent of the degree of the denominator polynomial and for the voltage transfer ratio and the current transfer ratio, the maximum degree of P(s) must be equal to the degree of Q(s).

51. In the circuit shown below, switch K is moved from position to position 2 at time t = 0. At time t = 0-, the current in the inductor is I₀ and the voltage at the capacitor is V₀. The inductor is represented by a transform impedance _________ in series with a voltage source __________

a) Ls, L V₀
b) Ls, LI₀
c) 1/Ls, LI₀
d) 1/Ls, L V₀
Answer: a
Explanation: The inductor has an initial current I₀. It is represented by a transform impedance Ls in series with a voltage source L V₀.

52. In the circuit shown in question 1, the capacitor is replaced by a transform impedance of __________ with an initial voltage ___________
a) 1/Cs, V₀/S
b) 1/Cs, I₀/S
c) Cs, I₀/S
d) Cs, V₀/S
Answer: a
Explanation: The capacitor has an initial voltage V₀ across it. It is represented by a transform impedance of 1/Cs with an initial voltage V₀/S.

53. The value of the total voltage after replacing the inductor and capacitor in question 1 is?
a) V₁(S)-LI₀-V₀/S
b) V₁(S)+LI₀-V₀/S
c) V₁(S)+LI₀+V₀/S
d) V₁(S)-LI₀+V₀/S
Answer: b
Explanation: The current I(s) is given as the total transform voltage in the circuit divided by the total transform impedance. The value of the total voltage after replacing the inductor and capacitor is V (s) = V₁(S)+LI₀-V₀/S.

54. The value of the total impedance after replacing the inductor and capacitor in question 1 is?
a) R-LS-1/CS
b) R-LS+1/CS
c) R+LS+1/CS
d) R+LS-1/CS
Answer: c
Explanation: The value of the total impedance after replacing the inductor and capacitor is
Z (s) = R+LS+1/CS. By knowing the V(s) and Z(s) we can calculate I(s) as I(s) is given as the total transform voltage in the circuit divided by the total transform impedance.

55. The current flowing in the circuit in question 1 is?
a) (V₁(S)-LI₀-V₀/S)/( R+LS+1/CS)
b) (V₁(S)-LI₀+V₀/S)/( R+LS+1/CS)
c) (V₁(S)+LI₀+V₀/S)/( R+LS+1/CS)
d) (V₁(S)+LI₀-V₀/S)/( R+LS+1/CS)
Answer: d
Explanation: The current I(s) is given as the total transform voltage in the circuit divided by the total transform impedance. The current flowing in the circuit is I (s) = V(s)/I(s) =(V₁(S)+LI₀-V₀/S)/( R+LS+1/CS).

56. Obtain the admittance of the last two elements in the parallel combination after transformation in the circuit shown below.

a) 1+s
b) 2+s
c) 3+s
d) 4+s
Answer: d
Explanation: The term admittance is defined as the inverse of impedance. The admittance of capacitor is 1/s and the admittance of resistor is 1/4 mho. So the admittance of the last two elements in the parallel combination is Y₁(s) = 4 + s.

57. The impedance of the last two elements in the parallel combination after transformation in the circuit shown in question 6 is?
a) 1/(s+4)
b) 1/(s+3)
c) 1/(s+2)
d) 1/(s+1)
Answer: a
Explanation: The impedance of resistor is 4Ω and the impedance of capacitor is s. So the impedance of the last two elements in the parallel combination is Z₁(s) = 1/(s+4).

58. The series combination of the last elements after replacing 1/s and 1/4Ω with 1/(S+4) in the question 6 is?
a) (3s+4)/2s(s-4)
b) (3s-4)/2s(s-4)
c) (3s+4)/2s(s+4)
d) (3s-4)/2s(s+4)
Answer: c
Explanation: We got the impedance of last two elements in parallel combination as Z₁(s) = 1/(s+4) and now the impedance of capacitor is 1/2s. So the series combination of the last elements is Z₂(s) =1/2s+1/(s+4)=(3s+4)/2s(s+4).

59. Determine the admittance parallel combination of the last elements after replacing with (3s+4)/2s(s+4) in the question 6 is?
a) (4s²-19s+4)/(6s-8)
b) (4s²+19s-4)/(6s+8)
c) (4s²+19s-4)/(6s-8)
d) (4s²+19s+4)/(6s+8)
Answer: d
Explanation: The term admittance is defined as the inverse of the term impedance. As the impedance is Z₂(s) =1/2s+1/(s+4)=(3s+4)/2s(s+4) , the admittance parallel combination of the last elements is Y₂(s) = 1/2+2s(s+4)/( 3s+4)=(4s²+19s+4)/(6s+8).

60. Obtain the transform impedance of the network shown in question 6.
a) (6s-8)/( 4s²+19s-4)
b) (6s+8)/( 4s²+19s+4)
c) (6s+8)/( 4s²-19s+4)
d) (6s-8)/( 4s²+19s+4)
Answer: b
Explanation: The term impedance is the inverse of the term admittance. We got admittance as Y₂(s) = (4s²+19s+4)/(6s+8). So the transform impedance of the network is
Z (s) = 1 / Y₂(s) = (6s+8)/( 4s²+19s+4).

1. Two ports containing no sources in their branches are called?
a) active ports
b) passive ports
c) one port
d) three port
Answer: b
Explanation: Two ports containing no sources in their branches are called passive ports; among them are power transmission lines and transformers.

2. Two ports containing sources in their branches are called?
a) three port
b) one port
c) passive ports
d) active ports
Answer: d
Explanation: Two ports containing sources in their branches are called active ports. A voltage and current is assigned to each of the two ports.

3. In determining open circuit impedance parameters, among V₁, V₂, I₁, I₂, which of the following are dependent variables?
a) V₁ and V₂
b) I₁ and I₂
c) V₁ and I₂
d) I₁ and V₂
Answer: a
Explanation: In determining open circuit impedance parameters, among V₁, V₂, I₁, I₂; V₁ and V₂ are dependent variables and I₁, I₂ are independent variables i.e., dependent variables depend on independent variables.

4. In determining open circuit impedance parameters, among V₁, V2, I₁, I2, which of the following are independent variables?
a) I₁ and V₂
b) V₁ and I₂
c) I₁ and I₂
d) V₁ and V₂
Answer: c
Explanation: In determining open circuit impedance parameters, among V₁, V₂, I₁, I₂; I₁ and I₂ are independent variables and V₁, V₂ are dependent variables. Independent variables are the variables that do not depend on any other variable.

5. Which of the following expression is true in case of open circuit parameters?
a) V₁ = Z₁₁ V₁ + Z₁₂ I₂
b) V₁ = Z₁₁I₁ + Z₁₂ V₂
c) V₁ = Z₁₁I₁ + Z₁₂ I₂
d) V₂ = Z₁₁I₁ + Z₁₂ I₂
Answer: c
Explanation: The expression relating the open circuit parameters Z₁₁, Z₁₂ and currents I₁, I₂ and voltage V₁ is V₁ = Z₁₁I₁ + Z₁₂ I₂.

6. Which of the following expression is true in case of open circuit parameters?
a) V₂ = Z₂₁I₂ + Z₂₂ I₂
b) V₂ = Z₂₁I₁ + Z₂₂ I₂
c) V₁ = Z₂₁I₂ + Z₂₂ I₂
d) V₁ = Z₂₁I₁ + Z₂₂ I₂
Answer: b
Explanation: The expression relating the currents I₁, I₂ and voltage V₁ and open circuit parameters Z₂₁, Z₂₂ is V₂ = Z₂₁I₁ + Z₂₂ I₂.

7. Find the Z – parameter Z₁₁ in the circuit shown below.

a) 1
b) 1.5
c) 2
d) 2.5
Answer: d
Explanation: For determining Z₁₁, the current I₂ is equal to zero. Now we obtain Z_eq as 1+ (6×2)/(6+2)=2.5Ω. So, Z₁₁ = 2.5Ω.

8. The value of Z₂₁ in the circuit shown in the question 7 is?
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: V₂ is the voltage across the 4Ω impedance. The current through 4Ω impedance is I₁/4. And V₂ = (I₁/4) x 4 = I₁. So, Z₂₁ = 1Ω.

9. Find the value of Z₁₂ in the circuit shown in the question 7.
a) 3
b) 2
c) 1
d) 0
Answer: c
Explanation: The current through vertical 2Ω resistor is = I₂/2. So, V₁ = 2 x (I₂/2). On solving and substituting we get Z₁₂ = 1Ω.

10. Determine the value of Z₂₂ in the circuit shown in the question 7.
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: Open circuiting port 1, we get V₂ = I₂((2+2)||4) => V₂ = I₂ x 2 =>V₂/I₂ = 2. Therefore the value of Z₂₂ is 2Ω.

11. In determining short circuit impedance parameters, among V₁, V₂, I₁, I₂, which of the following are dependent variables?
a) V₁ and V₂
b) I₁ and I₂
c) V₁ and I₂
d) I₁ and V₂
Answer: b
Explanation: In determining short circuit impedance parameters, among V₁, V₂, I₁, I₂; I₁ and I₂ are dependent variables and V₁, V₂ are independent variables i.e., dependent variables depend on independent variables.

12. In determining short circuit impedance parameters, among V₁, V₂, I₁, I₂, which of the following are independent variables?
a) I₁ and V₂
b) V₁ and I₂
c) I₁ and I₂
d) V₁ and V₂
Answer: d
Explanation: In determining short circuit impedance parameters, among V₁, V₂, I₁, I₂; V₁ and V₂ are independent variables and I₁, I₂ are dependent variables. Independent variables are the variables that do not depend on any other variable.

13. Which of the following expression is true in case of short circuit parameters?
a) I₁ = Y₁₁ V₁ + Y₁₂ V₂
b) I₁ = Y₁₁ I₁ + Y₁₂ V₂
c) V₁ = Y₁₁ I₁ + Y₁₂ V₂
d) V₁ = Y₁₁ V₁ + Y₁₂ V₂
Answer: a
Explanation: The expression relating the short circuit parameters Y₁₁, Y₁₂ and voltages V₁, V₂ and current is I₁, is I₁ = Y₁₁ V₁ + Y₁₂ V₂.

14. Which of the following expression is true in case of short circuit parameters?
a) I₂ = Y₂₁I₁ + Y₂₂ I₂
b) V₂ = Y₂₁I₁ + Y₂₂ V₂
c) I₂ = Y₂₁V₁ + Y₂₂ V₂
d) I₂ = Y₂₁V₁ + Y₂₂ I₂
Answer: c
Explanation: The expression relating the voltages V₁, V₂ and current is I₂ and short circuit parameters Y₁₁, Y₁₂ is I₂ = Y₂₁V₁ + Y₂₂V₂.

15. The parameters Y₁₁, Y₁₂, Y₂₁, Y₂₂ are called?
a) Open circuit impedance parameters
b) Short circuit admittance parameters
c) Inverse transmission parameters
d) Transmission parameters
Answer: b
Explanation: The parameters Y₁₁, Y₁₂, Y₂₁, Y₂₂ are called short circuit admittance parameters also called network functions as they are obtained by short circuiting port 1 or port 2.

16. Find the Y – parameter Y₁₁ in the circuit shown below.

a) 2
b) 3/2
c) 1
d) 1/2
Answer: d
Explanation: After short circuiting b-b’, the equation will be V₁ = (I₁) x 2. We know Y₁₁ = I₁/V₁. From the equation we get I₁/V₁ = 2. On substituting we get Y₁₁ = 2 mho.

17. Find the Y – parameter Y₂₁ in the circuit shown in question 6.
a) -1/4
b) 1/4
c) 1/2
d) -1/2
Answer: a
Explanation: After short circuiting b-b’, the equation will be -I₂=I₁ × 2/4=I₁/2 and -I₂= V₁/4 and on solving and substituting we get Y₂₁ =I₂/V₁=-1/4 mho.

18. Find the Y – parameter Y₂₂ in the circuit shown in question 6.
a) 3/8
b) 5/8
c) 7/8
d) 9/8
Answer: b
Explanation: On short circuiting a-a’,we get Z_eq = 8/5 Ω. V₂=I₂× 8/5. We know Y₂₂ = I₂/V₂. We got I₂/V₁ = 5/8. ON substituting we get Y₂₂ = 5/8 mho.

19. Find the Y – parameter Y₁₂ in the circuit shown in question 6.
a) 1/2
b) -1/2
c) -1/4
d) 1/4
Answer: c
Explanation: Short circuiting a-a’, -I₁= 2/5 I₂ and I₂= 5 V₂/8. On solving -I₁= 2/5×5/8 V₂= V₂/4. We know
Y₁₂ = I₁/V₂. We got I₁/V₂ = -1/4. So the value of Y₁₂ will be -1/4 mho.

20. Which of the following equation is true in the circuit shown in question 6?
a) I₁=0.5(V₁)+0.25(V₂)
b) I₁=0.25(V₁)+0.625(V₂)
c) I₁=-0.25(V₁)+0.625(V₂)
d) I₁=0.5(V₁)-0.25(V₂)
Answer: d
Explanation: We got the admittance parameters as Y₁₁ = 0.5, Y₁₂ = -0.25, Y₂₁ = -0.25, Y₂₂ = 0.625. So the equations in terms of admittance parameters is
I₁=0.5(V₁)-0.25(V₂) and I₂=-0.25(V₁)+0.625(V₂).

21.In the circuit shown below, find the transmission parameter A.

a) 6/5
b) 5/6
c) 3/4
d) 4/3
Answer: a
Explanation: Open circuiting b-b^‘, V₁ = 6 I₁, V₂ = 5I₁. On solving V₁/V₂ = 6/5. On substituting we get A = V₁/V₂=6/5.

22. In the circuit shown above, find the transmission parameter C.
a) 4/5
b) 3/5
c) 2/5
d) 1/5
Answer: d
Explanation: C = I₁/V₂ |I₂=0. By open circuiting b-b^‘ we get V₂ = 5 I₁ =>I₁/V₂ = 1/5. On substituting we get C = I₁/V₂=1/5 Ω.

23. In the circuit shown above, find the transmission parameter B.
a) 15/5
b) 17/5
c) 19/5
d) 21/5
Answer: b
Explanation: The transmission parameter B is given by B = -V₁/I₂ |V₂=0. Short circuiting b-b^‘, -I₂= 5/17 V₁ => -V₁/I₂ = 17/5. On substituting we get B=17/5 Ω.

24. In the circuit shown above, find the transmission parameter D.
a) 1/5
b) 3/5
c) 7/5
d) 9/5
Answer: c
Explanation: D is a transmission parameter and is given by D = -I₁/I₂ |V₂=0. Short circuiting b-b^‘, I₁= 7/17 V₁ and-I₂= 5/17 V₁. So we get I₁/I₂ = 7/5. So D=7/5.

25. The hybrid parameter h₁₁ is called?
a) short circuit input impedance
b) short circuit forward current gain
c) open circuit reverse voltage gain
d) open circuit output admittance
Answer: a
Explanation: h₁₁=V₁/I₁ |V₂=0. So the hybrid parameter h₁₁ is called short circuit input impedance.

26. The hybrid parameter h₂₁ is called?
a) open circuit output admittance
b) open circuit reverse voltage gain
c) short circuit forward current gain
d) short circuit input impedance
Answer: c
Explanation: h₂₁=I₂/I₁ |V₂=0. So the hybrid parameter h₂₁ is called short circuit forward current gain.

27. In the circuit shown below, find the h-parameter h₁₁.

a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: h₁₁=V₁/I₁ |V₂=0. So short circuiting b-b^‘, V₁ = I₁((2||2)+1) = 2I₁ => V₁/I₁= 2. On substituting we get h₁₁ = V₁/I₁= 2Ω.

28. In the circuit shown in question 7, find the h-parameter h₂₁.
a) 1
b) -1
c) 1/2
d) -1/2
Answer: d
Explanation: Short circuiting b-b^‘, h₂₁ = I₂/I₁ when V₂=0 and -I₂= I₁/2 => I₂/I₁ = -1/2. So h₂₁ = -1/2.

29. In the circuit shown in question 7, find the h-parameter h₁₂.
a) 1/2
b) -1/2
c) 1
d) -1
Answer: a
Explanation: Open circuiting a-a^‘ we get V₁=I_y×2 and
I_y=I₂/2 and V₂=I_x×4 and I_x=I₂/2. On solving and substituting, we get h₁₂ =V₁/V₂=1/2.

30. In the circuit shown in question 7, find the h-parameter h₂₂.
a) 1
b) 2
c) 1/2
d) 3/2
Answer: c
Explanation: Open circuiting a-a^‘ we get V₁=I_y×2 and I_y=I₂/2 and V₂=I_x×4 and I_x=I₂/2. On solving and substituting, we get h₂₂ =I₂/V₂=1/2 Ω.

31. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?
a) all odd terms are missing
b) all even terms are missing
c) all even or odd terms are missing
d) all even and odd terms are missing
Answer: c
Explanation: All the quotients in the polynomial P(s) are positive. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing.For example P(s) = s³+3s is Hurwitz because all quotient terms are positive and all even terms are missing.

32. The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on ____________
a) right half of s plane
b) left half of s-plane
c) on jω axis
d) on σ axis
Answer: c
Explanation: The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on jω axis not on right half of s plane or on left half of s-plane.

33. If the polynomial P (s) is either even or odd, then the roots of P (s) lie on __________
a) on σ axis
b) on jω axis
c) left half of s-plane
d) right half of s plane
Answer: b
Explanation: If the polynomial P (s) is either even or odd, then the roots of P (s) lie on jω axis not on right half of s plane or on left half of s-plane.

34. If the ratio of the polynomial P (s) and its derivative gives a continued fraction expansion with ________ coefficients, then the polynomial P (s) is Hurwitz.
a) all negative
b) all positive
c) positive or negative
d) positive and negative
Answer: b
Explanation: If the ratio of the polynomial P (s) and its derivative P^‘(s) gives a continued fraction expansion with all positive coefficients, then the polynomial P (s) is Hurwitz. If all the quotients in the continued fraction expansion are positive, the polynomial P(s) is positive.

35. Consider the polynomial P(s)=s⁴+3s²+2. The given polynomial P (s) is Hurwitz.
a) True
b) False
Answer: a
Explanation: P(s)=s⁴+3s²+2 => P^‘ (s)=4s³+6s
After doing the continued fraction expansion, we get all the quotients as positive. So, the polynomial P (s) is Hurwitz.

36. When s is real, the driving point impedance function is _________ function and the driving point admittance function is _________ function.
a) real, complex
b) real, real
c) complex, real
d) complex, complex
Answer: b
Explanation: When s is real, the driving point impedance function is real function and the driving point admittance function is real function because the quotients of the polynomials P(s) and Q(s) are real. When Z(s) is determined from the impedances of the individual branches, the quotients are obtained by adding together, multiplying or dividing the branch parameters which are real.

37. The poles and zeros of driving point impedance function and driving point admittance function lie on?
a) left half of s-plane only
b) right half of s-plane only
c) left half of s-plane or on imaginary axis
d) right half of s-plane or on imaginary axis
Answer: c
Explanation: The poles and zeros of driving point impedance function and driving point admittance function lie on left half of s-plane or on imaginary axis of the s-plane..

38. For real roots of s_k, all the quotients of s in s²+ω_k² of the polynomial P (s) are __________
a) negative
b) non-negative
c) positive
d) non-positive
Answer: b
Explanation: For real roots of s_k, all the quotients of s in s²+ω_k² of the polynomial P (s) are non-negative. So by multiplying all factors in P(s) we find that all quotients are positive.

39. The real parts of the driving point function Z (s) and Y (s) are?
a) positive and zero
b) positive
c) zero
d) positive or zero
Answer: d
Explanation: The real parts of the driving point impedance function Z (s) and driving point admittance function Y (s) are positive or zero.

40. For the complex zeros to appear in conjugate pairs the poles of the network function are ____ and zeros of the network function are ____________
a) complex, complex
b) complex, real
c) real, real
d) real, complex
Answer: c
Explanation: P(s) and Q(s) are real when s is real. So the poles of the network function are real and zeros of the network function are real, the complex zeros to appear in conjugate pairs.

41.In the circuit shown below, find the Z-parameter Z₁₁.

a) 1
b) 2
c) 3
d) 4
Answer: c
Explanation: The Z –parameter Z₁₁ is V₁/I₁, port 2 is open circuited. V₁ = (1+2)I₁ => V₁/I₁ = 3 and on substituting, we get Z₁₁ = 3Ω.

42. In the circuit shown in question 1, find the Z-parameter Z₁₂.
a) 4
b) 3
c) 2
d) 1
Answer: c
Explanation: The Z-parameter Z₁₂ is V₂/I₁ |I₂=0. On open circuiting port 2 we obtain the equation, V₁ = (2) I₂ => V₁/I₁ = 2. On substituting we get Z₁₂ = 2Ω.

43. In the circuit shown in question 1, find the Z-parameter Z₂₁.
a) 2
b) 4
c) 1
d) 3
Answer: a
Explanation: The Z-parameter Z₂₁ is V₂/I₁ |I₂=0. On open circuiting port 2, we get V₂ = (2)I₁ => V₂/I₁ = 2. On substituting we get Z₂₁ = 2Ω.

44. In the circuit shown in question 1, find the Z-parameter Z₂₂.
a) 3
b)2
c) 4
d) 1
Answer: a
Explanation: The Z-parameter Z₂₁ is V₂/I₂ |I₁=0. This parameter is obtained by open circuiting port 1. So we get V₂ = (2 + 1)I₂ => V₂ = 3(I₂) => V₂/I₂ = 3. On substituting Z₂₁ = 3Ω.

45. In the circuit shown below, find the Z-parameter Z₁₁.

a) 10
b) 15
c) 20
d) 25
Answer: b
Explanation: The Z –parameter Z₁₁ is V₁/I₁, port 2 is open circuited. V₁ = (10 + 5)I₁ => V₁/I₁ = 15 and on substituting, we get Z₁₁ = 2.5Ω.

46. In the circuit shown in question 5, find the Z-parameter Z₁₂.
a) 15
b) 10
c) 5
d) 1
Answer: c
Explanation: The Z-parameter Z₁₂ is V₂/I₁ |I₂=0. On open circuiting port 2 we obtain the equation, V₁ = (5) I₂ => V₁/I₁ = 5. On substituting we get Z₁₂ = 5Ω.

47. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z₁₁.
a) 8
b) 18
c) 28
d) 38
Answer: b
Explanation: The Z-parameter Z₁₁ is Z₁₁ = Z_11x + Z_11y and Z_11x = 3, Z_11y = 15. On substituting we get Z₁₁ = 3 +15 = 18Ω.

48. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z₁₂.
a) 4
b) 5
c) 6
d) 7
Answer: d
Explanation: The Z-parameter Z₁₂ is Z₁₂ = Z_12x + Z_12y and we have Z_12x = 2, Z_12y. On substituting we get Z₁₂ = 2 + 5 = 7Ω.

49. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z₂₁.
a) 7
b) 6
c) 5
d) 4
Answer: a
Explanation: The Z-parameter Z₂₁ is Z₂₁ = Z_21x + Z_21y and we have Z_21x = 2, Z_21y = 5. On substituting we get Z₂₁ = 2 + 5 = 7Ω.

50. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z₂₂.
a) 38
b) 28
c) 18
d) 8
Answer: b
Explanation: The Z-parameter Z₂₂ is Z₂₂ = Z_22x + Z_22y and we have Z_22x = 3, Z_22y = 25. On substituting we get Z₂₂ = 3 +25 = 28Ω.

51. A periodic voltage v (t) = 1 + 4 sin ωt + 2 cos ωt is applied across a 1Ω resistance. The power dissipated is ____________
a) 1 W
b) 11 W
c) 21 W
d) 24.5 W
Answer: b
Explanation: Given that, v (t) = 1 + 4 sin ωt + 2 cos ωt
So, Power is given by,
Power, P = 121+422√1+222√1
= 11 W.

52. A constant k high pass p section has a characteristic impedance of 300 Ω at f = ∞. At f = f_c, the characteristic impedance will be?
a) 0
b) ∞
c) 300 Ω
d) More than 300 Ω
Answer: b
Explanation: For constant k high pass p section is given by,
Z = R1–(fdf)2√
At f = f_d, denominator term is 0.
So, Z = infinite.

53. In the circuit given below, the current through R is 2 sin 8t. The value of R is ___________

a) (0.18 + j0.72)
b) (0.46 + j1.90)
c) – (0.18 + j1.90)
d) (0.23 – 0.35 j)
Answer: d
Explanation: Here, Inductor is not given, hence ignoring the inductance. Let I₁ and I₂ are currents in the loop then,
I₁ = 2sin8t3
= 0.66 sin 8t
Again, I₂ = −jX4X0.75I13.92−2.56j
= (0.23 – 0.35j) sin 8t
So, R = (0.23 – 0.35 j).

54. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₂₁ in terms of Hybrid parameters can be expressed as ________
a) y₂₁ = (−h21h12h11+h22)
b) y₂₁ = h21h11
c) y₂₁ = –h12h11
d) y₂₁ = 1h11
Answer: b
Explanation: We know that, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (1)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
I₁ = V1h11–h12V2h11 ………. (5)
And I₂ = h21V1h11+(−h21h12h11+h22)V2 ………. (6)
∴ Comparing (1), (2) and (5), (6), we get,
y₁₁ = 1h11
y₁₂ = –h12h11
y₂₁ = h21h11
y₂₂ = (−h21h12h11+h22).

55. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₂₁ in terms of Hybrid parameters can be expressed as ________
a) z₂₁ = (h11–h21h12h22)
b) z₂₁ = – h21h22
c) z₂₁ = h12h22
d) z₂₁ = 1h22
Answer: b
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.

56. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₁ in terms of Hybrid parameters can be expressed as ________
a) z₁₁ = (h11–h21h12h22)
b) z₁₁ = – h21h22
c) z₁₁ = h12h22
d) z₁₁ = 1h22
Answer: a
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.

57. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₂ in terms of Hybrid parameters can be expressed as ________
a) z₁₂ = (h11–h21h12h22)
b) z₁₂ = – h21h22
c) z₁₂ = h12h22
d) z₁₂ = 1h22
Answer: c
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.

58. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₂₂ in terms of Hybrid parameters can be expressed as ________
a) z₂₂ = (h11–h21h12h22)
b) z₂₂ = – h21h22
c) z₂₂ = h12h22
d) z₂₂ = 1h22
Answer: d
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, V₁ = h₁₁ I₁ + h₁₂ V₂ ………. (3)
I₂ = h₂₁ I₁ + h₂₂ V₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = (h11–h21h12h22)I1+h12h22I2 ………… (5)
And V₂ = I2h22–h21I1h22 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = (h11–h21h12h22)
z₁₂ = h12h22
z₂₁ = – h21h22
z₂₂ = 1h22.

59. Permeability is analogous to _____________
a) Conductivity
b) Resistivity
c) Retentivity
d) Coercivity
Answer: a
Explanation: We know that resistance and reluctance are given by,
R = ρLA
And Reluctance = LμA
So, Permeability is analogous to conductivity.

60. A resistance and an inductance are connected in parallel and fed from 50 Hz ac mains. Each branch takes a current of 5 A. The current supplied by source is ____________
a) 10 A
b) 7.07 A
c) 5 A
d) 0 A
Answer: b
Explanation: The current is given by,
|5 – j5| = 52+52−−−−−−√
= 50−−√=52–√ = 7.07 A.

61. A triangular Pulse of 50 V peak is applied to a capacitor of 0.1 F. The change of the capacitor and its waveform shape is ___________
a) 10 rectangular
b) 5 rectangular
c) 5 triangular
d) 10 triangular
Answer: c
Explanation: We know that,
Q = CV
Or, 0.1 X 50 = 5
And it is a triangular pulse.

62. A 10 μF capacitor is charged from a 5 volt source through a resistance of 10 kΩ. The charging current offer 35 m sec. If the initial voltage on C is – 3 V is ___________
a) 0.56 mA
b) 5.6 mA
c) 6 mA
d) 5 μA
Answer: a
Explanation: Initial current immediately after charging is given by,
VR=5+310000
= 0.8 mA
Now, i = i₀e^-t/RC
= 0.8 mA x et10kX10X10−6
= 0.8 X 10^-3 X e35X10−310−1
= 0.56 mA.

63. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₁ in terms of Inverse Hybrid parameters can be expressed as ________
a) z₁₂ = 1g11
b) z₁₂ = – g12g11
c) z₁₂ = – g21g11
d) z₁₂ = (g22–g21g12g11)
Answer: a
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = I1g11–g12g11I2 ………… (5)
And V₂ = (g22–g21g12g11)I2–g21I1g11 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = 1g11
z₁₂ = – g12g11
z₂₁ = – g21g11
z₂₂ = (g22–g21g12g11).

64. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₁₂ in terms of Inverse Hybrid parameters can be expressed as ________
a) z₁₂ = 1g11
b) z₁₂ = – g12g11
c) z₁₂ = – g21g11
d) z₁₂ = (g22–g21g12g11)
Answer: b
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = I1g11–g12g11I2 ………… (5)
And V₂ = (g22–g21g12g11)I2–g21I1g11 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = 1g11
z₁₂ = – g12g11
z₂₁ = – g21g11
z₂₂ = (g22–g21g12g11).

65. For a T-network if the Open circuit impedance parameters are given as z₁₁, z₂₁, z₁₂, z₂₂, then z₂₂ in terms of Inverse Hybrid parameters can be expressed as ________
a) z₂₂ = 1g11
b) z₂₂ = – g12g11
c) z₂₂ = – g21g11
d) z₂₂ = (g22–g21g12g11)
Answer: d
Explanation: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ ……… (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ………. (2)
And, I₁ = g₁₁ V₁ + g₁₂ I₂ ………. (3)
V₂ = g₂₁ V₁ + g₂₂ I₂ ……….. (4)
Now, (3) and (4) can be rewritten as,
V₁ = I1g11–g12g11I2 ………… (5)
And V₂ = (g22–g21g12g11)I2–g21I1g11 ……….. (6)
∴ Comparing (1), (2) and (5), (6), we get,
z₁₁ = 1g11
z₁₂ = – g12g11
z₂₁ = – g21g11
z₂₂ = (g22–g21g12g11).

66. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Transmission parameters can be expressed as ________
a) y₁₁ = DB
b) y₁₁ = C−AB
c) y₁₁ = – 1B
d) y₁₁ = AB
Answer: a
Explanation: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = ABV2–1BV1 …………. (5)
And I₁ = CV₂ – D (ABV2–1BV1)=DBV1+(C−AB)V2 …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = DB
y₁₂ = C−AB
y₂₁ = – 1B
y₂₂ = AB.

67. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Transmission parameters can be expressed as ________
a) y₁₂ = DB
b) y₁₂ = C−AB
c) y₁₂ = – 1B
d) y₁₂ = AB
Answer: b
Explanation: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = ABV2–1BV1 …………. (5)
And I₁ = CV₂ – D (ABV2–1BV1)=DBV1+(C−AB)V2 …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = DB
y₁₂ = C−AB
y₂₁ = – 1B
y₂₂ = AB.