300+ REAL TIME Network Theory Objective Questions & Answers 2023

250+ TOP MCQs on Tree and Co-Tree and Answers

Network Theory Multiple Choice Questions on “Tree and Co-Tree”.

1. 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
Clarification: 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.

2. 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
Clarification: 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.

3. 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
Clarification: 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.

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

Answer: D
Clarification: 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.

5. Number of twigs in a tree are? (where, n-number of nodes)
A. n
B. n+1
C. n-1
D. n-2

Answer: C
Clarification: 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.

6. Loops which contain only one link are independent are called?
A. open loops
B. closed loops
C. basic loops
D. none of the mentioned

Answer: C
Clarification: The addition of subsequent link forms one or more additional loops. Loops that contain only one link are independent are called basic loops.

7. 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

A.

Answer: B

Clarification: 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

considering the directions specified in the graph.

8. 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
Clarification: 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

9. If a graph consists of 5 nodes, then the number of twigs in the tree is?
A. 1
B. 2
C. 3
D. 4

Answer: D
Clarification: 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.

10. 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
Clarification: 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.

250+ TOP MCQs on Thevenin Theorem Involving Dependent and Independent Sources and Answers

Network Theory Multiple Choice Questions on “Thevenin Theorem Involving Dependent and Independent Sources”.

1. A circuit is given in the figure below. The Thevenin equivalent as viewed from terminals x and x’ is ___________

A. 8 V and 6 Ω
B. 5 V and 6 Ω
C. 5 V and 32 Ω
D. 8 V and 32 Ω

Answer: B
Clarification: We, Thevenized the left side of xx’ and source transformed right side of yy’.

V_xx’ = V_th = (displaystylefrac{frac{4}{8} + frac{8}{24}}{frac{1}{8} + frac{1}{24}}) = 5V
∴ R_th = 8 || (16 + 8)
= (frac{8×24}{8+24}) = 6 Ω.

2. For the circuit given in figure below, the Thevenin equivalent as viewed from terminals y and y’ is _________

A. 8 V and 32 Ω
B. 4 V and 32 Ω
C. 5 V and 6 Ω
D. 7 V and 6 Ω

Answer: D
Clarification: We, Thevenized the left side of xx’ and source transformed right side of yy’.

Thevenin equivalent as seen from terminal yy’ is
V_xx’ = V_th = (displaystylefrac{frac{4}{24} + frac{8}{8}}{frac{1}{24} + frac{1}{8}}) = 5V
= (frac{0.167+1}{0.04167+0.125}) = 7 V
∴ R_th = (8 + 16) || 8
= (frac{24×8}{24+8}) = 6 Ω.

3. In the following circuit, when R = 0 Ω, the current IR equals to 10 A. The value of R, for which maximum power is absorbed by it is ___________

A. 4 Ω
B. 3 Ω
C. 2 Ω
D. 1 Ω

Answer: C
Clarification: The Thevenin equivalent of the circuit is as shown below.

Therefore from the figure we can infer that R_th = 2 Ω

4. In the following circuit, when R = 0 Ω, the current I_R equals to 10 A. The maximum power will be?

A. 50 W
B. 100 W
C. 200 W
D. 400 W

Answer: A
Clarification: The Thevenin equivalent of the circuit is as shown below.
I = 10 A, R_th = 2 Ω
∴ P_max = ((frac{10}{2}))² × 2
= 5×5×2 = 50 W.

5. For the circuit given below, the Thevenin resistance across the terminals A and B is _____________

A. 5 Ω
B. 7 kΩ
C. 1.5 kΩ
D. 1.1 kΩ

Answer: B
Clarification: Let V_AB = 1 V
5 V_AB = 5
Or, 1 = 1 × I₁ or, I₁ = 1
Also, 1 = -5 + 1(I – I₁)
∴ I = 7
Hence, R = 0.2 kΩ.

6. For the circuit given below, the Thevenin voltage across the terminals A and B is ____________

A. 1.25 V
B. 0.25 V
C. 1 V
D. 0.5 V

Answer: D
Clarification: Current through 1 Ω = (frac{5}{2}) – I₁
Using source transformation to 5 V sources, V_OC = 1 × I₁
V_OC = -5 V_OC + ((frac{5}{2}) – I₁) × 1
Eliminating I1, we get, V_OC = 0.5 V.

7. In the following circuit, the value of open circuit voltage and the Thevenin resistance between terminals a and b are ___________

A. V_OC = 100 V, R_TH = 1800 Ω
B. V_OC = 0 V, R_TH = 270 Ω
C. V_OC = 100 V, R_TH = 90 Ω
D. V_OC = 0 V, R_TH = 90 Ω

Answer: D
Clarification: By writing loop equations for the circuit, we get,
V_S = V_X, I_S = I_X
V_S = 600(I₁ – I₂) + 300(I₁ – I₂) + 900 I₁
= (600+300+900) I₁ – 600I₂ – 300I₃
= 1800I₁ – 600I₂ – 300I₃
I₁ = I_S, I₂ = 0.3 V_S
I₃ = 3I_S + 0.2V_S
V_S = 1800IS – 600(0.01V_S) – 300(3I_S + 0.01V_S)
= 1800I_S – 6V_S – 900I_S – 3V_S
10V_S = 900I_S
For Thevenin equivalent, V_S = R_TH I_S + V_OC
So, Thevenin voltage V_OC = 0
Resistance R_TH = 90Ω.

8. In the circuit given below, it is given that V_AB = 4 V for R_L = 10 kΩ and V_AB = 1 V for R_L = 2kΩ. The values of the Thevenin resistance and voltage for the network N are ____________

A. 16 kΩ and 30 V
B. 30 kΩ and 16 V
C. 3 kΩ and 6 V
D. 50 kΩ and 30 V

Answer: B
Clarification: When R_L = 10 kΩ and V_AB = 4 V
Current in the circuit I = (frac{V_{AB}}{R_L} = frac{4}{10}) = 0.4 mA
Thevenin voltage is given by V_TH = I (R_TH + R_L)
= 0.4(R_TH + 10)
= 0.4R_TH + 4
Similarly, for R_L = 2 kΩ and V_AB = 1 V
So, I = (frac{1}{2}) = 0.5 mA
V_TH = 0.5(R_TH + 2)
= 0.5 R_TH + 1
∴ 0.1R_TH = 3
Or, R_TH = 30 kΩ
And V_TH = 12 + 4 = 16 V.

9. For the circuit shown in figure below, the value of the Thevenin resistance is _________

A. 100 Ω
B. 136.4 Ω
C. 200 Ω
D. 272.8 Ω

Answer: A
Clarification: I_X = 1 A, V_X = V_test
V_test = 100(1-2I_X) + 300(1-2I_X – 0.01V_S) + 800
Or, V_test = 1200 – 800I_X – 3V_test
Or, 4V_test = 1200 – 800 = 400
Or, V_test = 100V
∴ R_TH = (frac{V_{test}}{1}) = 100 Ω.

10. For the circuit shown in the figure below, the Thevenin voltage and resistance looking into X-Y are __________

A. (frac{4}{3}) V and 2 Ω
B. 4V and (frac{2}{3}) Ω
C. (frac{4}{3}) V and (frac{2}{3}) Ω
D. 4 V and 2 Ω

Answer: D
Clarification: (R_{TH} = frac{V_{OC}}{I_{SC}})
V_TH = V_OC
Applying KCL at node A, (frac{2I-V_{TH}}{1} + 2 = I + frac{V_{TH}}{2})
Or, I = (frac{V_{TH}}{1})
Putting, 2V_TH – V_TH + 2 = V_TH + (frac{V_{TH}}{2})
Or, V_TH = 4 V.
∴ R_TH = 4/2 = 2Ω.

11. In the figure given below, the value of the source voltage is ___________

A. 12 V
B. 24 V
C. 30 V
D. 44 V

Answer: C
Clarification: By applying KCL, (frac{V_P-E}{6} + frac{V_P}{6}) – 1 = 0
Or, 2 V_P – E = 6
Where, ((frac{-V_P+E}{6})) = 2
∴ E – V_P = 12
Or, V_P = 18 V
∴ E = 30V.

12. In the figure given below, the value of Resistance R by Thevenin Theorem is ___________

A. 10
B. 20
C. 30
D. 40

Answer: B
Clarification: (frac{V_P-100}{10} + frac{V_P}{10}) + 2 = 0
Or, 2V_P – 100 + 20 = 0
∴ V_P = 80/2 = 40V
∴ R = 20Ω.

13. In the figure given below, the Thevenin’s equivalent pair, as seen at the terminals P-Q, is given by __________

A. 2 V and 5 Ω
B. 2 V and 7.5 Ω
C. 4 V and 5 Ω
D. 4 V and 7.5 Ω

Answer: A
Clarification: For finding V_TH,

V_TH = (frac{4 ×10}{10+10}) = 2V
For finding R_TH,

R_TH = 10 || 10
= (frac{10×10}{10+10}) = 5 Ω.

14. The Thevenin equivalent impedance Z between the nodes P and Q in the following circuit is __________

A. 1
B. 1 + s + (frac{1}{s})
C. 2 + s + (frac{1}{s})
D. 3 + s + (frac{1}{s})

Answer: A
Clarification: To calculate the Thevenin resistance, all the current sources get open-circuited and voltage source short-circuited.
∴ R_TH = ((frac{1}{s}) + 1) || (1+s)
= (frac{left(frac{1}{s} + 1right)×(1+s)}{left(frac{1}{s} + 1right)+(1+s)})
= (frac{frac{1}{s}+1+1+s}{frac{1}{s}+1+1+s}) = 1
So, R_TH = 1.

15. While computing the Thevenin equivalent resistance and the Thevenin equivalent voltage, which of the following steps are undertaken?
A. Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
B. Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
C. The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched
D. The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched

Answer: C
Clarification: While computing the Thevenin equivalent voltage consisting of both dependent and independent sources, we first find the equivalent voltage called the Thevenin voltage by opening the two terminals. Then while computing the Thevenin equivalent resistance, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched and open-circuiting the dependent current sources keeping the independent current sources untouched.

250+ TOP MCQs on Series Resonance and Answers Quiz

Network Theory Multiple Choice Questions on “Series Resonance”.

1. The circuit is said to be in resonance if the current is ____ with the applied voltage.
A. in phase
B. out of phase
C. 45⁰ out of phase
D. 90⁰ out of phase

Answer: A
Clarification: The circuit is said to be in resonance if the current is in phase with the applied voltage and not if the current is out of phase with the applied voltage. The study of resonance is very useful particularly in the area of communications.

2. In a series resonance circuit, series resonance occurs when?
A. X_L = 1
B. X_C = 1
C. X_L = X_C
D. X_L = -X_C

Answer: C
Clarification: In a series resonance circuit, series resonance occurs when capacitive reactance is equal to the inductive reactance that is X_L = X_C.

3. As X_L = X_C in a series resonance circuit, the impedance is_________
A. purely capacitive
B. purely inductive
C. purely resistive
D. capacitive and inductive

Answer: C
Clarification: As X_L = X_C in a series resonance circuit, the impedance is purely resistive. In a series RLC circuit the current lags behind or leads the applied voltage depending on the values of X_L and X_C.

4. At resonant frequency, the voltage across capacitor is _______ the voltage across inductor.
A. greater than
B. less than
C. greater than or equal to
D. equal to

Answer: D
Clarification: At resonant frequency, the voltage across capacitor is equal to the voltage across inductor. If one of the parameters of the series RLC circuit is varied in such a way that the current in the circuit is in phase with the applied voltage, then the circuit is said to be in resonance.

5. In series RLC circuit, the voltage across capacitor and inductor are ______ with each other.
A. in phase
B. 180⁰ out of phase
C. 90⁰ out of phase
D. 45⁰ out of phase

Answer: B
Clarification: In series RLC circuit, the voltage across capacitor and inductor are 180⁰ out of phase with each other. The frequency at which the resonance occurs is called resonant frequency.

6. The voltage across the LC combination in a series RLC circuit is?
A. 0
B. 1
C. 2
D. 3

Answer: A
Clarification: Since the voltage across capacitor and inductor are 180⁰ out of phase with each other, the voltage across the LC combination in a series RLC circuit is 0V.

7. The expression of resonant frequency in a series resonant circuit is?
A. 1/(2π√C.
B. 1/( 2π√L)
C. 2π√LC
D. 1/(2π√LC.

Answer: D
Clarification: The expression of resonant frequency is resonant frequency = 1/(2π√LC.. In a series RLC circuit resonance may be produced by varying the frequency, keeping L and C constant.

250+ TOP MCQs on Voltage, Current and Power in a Star Connected System and Answers

Network Theory Multiple Choice Questions on “Voltage, Current and Power in a Star Connected System”.

1. In star connected system, V_RY is equal to?
A. V_YR
B. -V_YR
C. 2V_YR
D. 3V_YR

Answer: B
Clarification: The voltage available between any pair of terminals is called the line voltage. The double script notation is purposefully used to represent voltages and currents in poly phase circuits. In star connected system, V_RY = – V_YR.

2. In three phase system, the line voltage V_RY is equal to?
A. phasor sum of V_RN and V_NY
B. phasor difference of V_RN and V_NY
C. phasor sum of V_RN and V_NY
D. algebraic sum of V_RN and V_NY

Answer: A
Clarification: In three phase system, the line voltage V_RY is equal to the phasor sum of V_RN and V_NY which is also equal to the phasor difference of V_RN and V_YN.

3. The relation between the lengths of the phasors V_RN and – V_YN is?
A. |V_RN| > – |V_YN|
B. |V_RN| < – |V_YN|
C. |V_RN| = – |V_YN|
D. |V_RN| >= – |V_YN|

Answer: C
Clarification: The voltage V_RY is found by compounding V_RN and V_YN reversed. The relation between the lengths of the phasors V_RN and – V_YN is |V_RN| = – |V_YN|.

4. In a star connected system, the phasors V_RN, V_YN are ____ apart.
A. 15⁰
B. 30⁰
C. 45⁰
D. 60⁰

Answer: D
Clarification: In a star connected system, the phasors V_RN, V_YN are separated by θ = 60⁰. To subtract V_YN from V_RN, we reverse the phase V_YN and find its phasor sum with V_RN.

5. The relation between V_RY, Vph in a star connected system is?
A. V_RY = V_ph
B. V_RY = √3V_ph
C. V_RY = 3√3V_ph
D. V_RY = 3V_ph

Answer: B
Clarification: The two phasors V_YN and V_BN are equal in length and are 60⁰apart. The relation between V_RY, V_ph in a star connected system is V_RY = √3V_ph.

6. In a star connected system, the relation between V_YB, V_ph is?
A. V_YB = V_ph
B. V_YB = 3√3V_ph
C. V_YB = 3V_ph
D. V_YB = √3V_ph

Answer: D
Clarification: In a star connected system, the relation between V_YB, V_ph is V_YB = √3V_ph. The line voltage V_YB is equal to the phasor difference of V_YN and V_BN and is equal to √3V_ph.

7. The voltages, V_BR ,V_ph are related in star connected system is?
A. V_BR = 3V_ph
B. V_BR = 3√3V_ph
C. V_BR = √3V_ph
D. V_BR = V_ph

Answer: C
Clarification: The voltages, V_BR, V_ph in star connected system are related as V_BR = √3V_ph. The line voltage V_YB is equal to the phasor difference of V_BN and V_RN and is equal to √3V_ph.

8. A symmetrical star connected system has V_RN = 230∠0⁰. The phase sequence is RYB. Find V_RY.

A. 398.37∠30⁰
B. 398.37∠-30⁰
C. 398.37∠90⁰
D. 398.37∠-90⁰

Answer: A
Clarification: Since the system is a balanced system, all the phase voltages are equal in magnitude but displaced by 120⁰. V_RN = 230∠0⁰V. V_RY = √3×230∠(0^o+30^o)V=398.37∠30^oV.

9. A symmetrical star connected system has V_RN = 230∠0⁰. The phase sequence is RYB. Find V_YB.

A. 398.37∠-30⁰
B. 398.37∠210⁰
C. 398.37∠90⁰
D. 398.37∠-90⁰

Answer: D
Clarification: Corresponding line voltages are equal to √3 times the phase voltages and are 30⁰ ahead of the respective phase voltages. V_YN = 230∠-120⁰V. V_YB = √3×230∠(-120^o+30^o)V=398.37∠-90⁰V.

10. A symmetrical star connected system has V_RN = 230∠0⁰. The phase sequence is RYB. Find V_BR.

A. 398.37∠210⁰
B. 398.37∠-210⁰
C. 398.37∠120⁰
D. 398.37∠-120⁰

Answer: B
Clarification: All the line voltages are equal in magnitude and are displaced by 120⁰. V_BN = 230∠-240⁰V. V_BR = √3×230∠(-240^o+30^o)V=398.37∠-210^oV.

250+ TOP MCQs on The Impulse Function in Circuit Analysis and Answers

Network Theory Multiple Choice Questions on “The Impulse Function in Circuit Analysis”.

1. For the circuit shown below, find the voltage across the capacitor C₁ at the time the switch is closed.
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A. 0
B. V/4
C. V/2
D. V

Answer: D
Clarification: We use two different cicuits to illustrate how an impulse function can be created with a switching operation. The capacitor is charged to an initial voltage of V at the time the switch is closed.

2. The charge on capacitor C₂ in the circuit shown below is?
A. 0
B. 1
C. 2
D. 3

Answer: A
Clarification: In the circuit, the initial charge on C₂ is zero. So, charge on capacitor C₂ = zero. Capacitor circuit and series inductor circuit are two different cicuits to illustrate how an impulse function can be created with a switching operation.

3. The current in the circuit shown below is?

A. (V/R)/(s+1/C_e)
B. (V/R)/(s+1/RC_e)
C. (V/R)/(s-1/RC_e)
D. (V/R)/(s-1/C_e)

Answer: B
Clarification: The current flowing through the circuit is given by I = (V/s)/(R+1/sC₁+1/sC₂) = (V/R)/(s+1/RC_e) where the equivalent capacitance C₁C₂/(C₁+C₂) and is replaced by C_e.

4. By taking the inverse transform of the current (V/R)/(s+1/RC_e), the value of the current is?
A. V/R e^t/C_e
B. V/R e^t/RC_e
C. V/R e^-t/RC_e
D. V/R e^-t/C_e

Answer: C
Clarification: By taking the inverse transform of the current, we obtain i = V/R e^-t/RC_e which indicates that as R decreases, the initial current increases and the time constant decreases.

5. Consider the circuit shown below. On applying the Kirchhoff’s current law, the equation will be?
A. V/(2s-15)+(V-[(100/s)+30])/(3s+10)=0
B. V/(2s-15)+(V-[(100/s)+30])/(3s-10)=0
C. V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0
D. V/(2s+15)+(V-[(100/s)+30])/(3s-10)=0

Answer: C
Clarification: The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero. Applying Kirchhoff’s current law, we get V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0.

6. The value of the voltage V in the circuit shown below is?
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A. 40(s+7.5)/s(s+5) – 12(s+7.5)/(s-5)
B. 40(s+7.5)/s(s+5) – 12(s+7.5)/(s+5)
C. 40(s+7.5)/s(s+5) + 12(s+7.5)/(s-5)
D. 40(s+7.5)/s(s+5) + 12(s+7.5)/(s+5)

Answer: D
Clarification: Solving for V yields V=40(s+7.5)/s(s+5) + 12(s+7.5)/(s+5). The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero.

7. The value of the voltage V after taking the partial fractions in the equation V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0 is?
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A. 12 + 60/s + 10/(s+5)
B. 12 – 60/s + 10/(s+5)
C. 12 – 60/s – 10/(s+5)
D. 12 + 60/s – 10/(s+5)

Answer: A
Clarification: By taking the partial fractions we get V = 60/s-20/(s+5)+12+30/(s+5) and on solving the equation we get V = 12+60/s+10/(s+5).

8. Determine the voltage V after taking the inverse transform (12 + 60/s + 10/(s+5)).
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A. 12δ(t)-(60-10e^-5t)u(t)
B. 12δ(t)+(60+10e^-5t)u(t)
C. 12δ(t)-(60+10e^-5t)u(t)
D. 12δ(t)+(60-10e^-5t)u(t)

Answer: B
Clarification: By taking inverse transform of V = 12+60/s+10/(s+5) we have v=12δ(t)+(60+10e^-5t)u(t) volts and we have to derive the expression for the current when t > 0.

9. The current equation for the circuit shown below is?
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A. I=4/s-2/(s-5)
B. I=4/s-2/(s+5)
C. I=4/s+2/(s+5)
D. I=4/s+2/(s-5)

Answer: C
Clarification: After the switch has been opened, the current in L₁ is the same as the current in L₁. The current equation is I=(100/s+30)/(5s+25). On solving we get I = 4/s-2/(s+5).

10. The value of the current after taking the inverse transform of the current is?
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A. (4-2e^5t)u(t)
B. (4-2e^-5t)u(t)
C. (4+2e^5t)u(t)
D. (4-2e^-5t)u(t)

Answer: D
Clarification: By taking the inverse transform of I = 4/s-2/(s+5) gives i = (4-2e^-5t)u(t). Before the switch is opened,the current in L₁ is 10A and the current in L₂ is 0A.

250+ TOP MCQs on Advanced Problems Involving Parameters and Answers

Network Theory Multiple Choice Questions on “Advanced Problems Involving Parameters”.

1. For the circuit given below, the value of Transmission parameter A and C are ____________

A. A = -0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω
B. A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω
C. A = -0.7692 – j0.3461 Ω, C = -0.03461 + j0.023 Ω
D. A = 0.7692 – j0.3461 Ω, C = 0.023 + j0.03461 Ω

Answer: B
Clarification: V = [20 + (-j10) || (j15 − j20)] I₁
V₁ = (Big[20 + frac{(-j10)(-j5)}{-j15}Big]) I₁
= [20 – j(frac{10}{3})] I₁
I₀ = (left(frac{-j10}{-j10-j5}right)) I₁ = (frac{2}{3})I₁
V₂ = (-j20) I₀ + 20I’₀
= –(frac{j40}{3}I_1 + 20I_1 = (20 – frac{j40}{3}) I_1 )
∴ A = (frac{V_1}{V_2} = frac{(20-frac{j10}{3})I_1}{20-frac{j40}{3}) I_1}) = 0.7692 + j0.3461 Ω
∴ C = (frac{I_1}{V_2} = frac{1}{20-j40/3}) = 0.03461 + j0.023 Ω.

2. For the circuit given below, the value of the Transmission parameter B and D are __________

A. D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω
B. D = 0.6923 + j0.5385 Ω, B = 6.923 + j25.385 Ω
C. D = -0.6923 + j0.5385 Ω, B= 25.385 + j6.923 Ω
D. D = -0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

Answer: A
Clarification: Z₁ = (frac{(-j15)(-j10)}{-j15-j10-j20}) = j10
Z₂ = (frac{(-j10)(-j20)}{-j15}) = (-frac{j40}{3})
Z₃ = (frac{(j15)(-j20)}{-j15}) = j20
-I₂ = (frac{20-j40/3}{20-frac{j40}{3}+j20}I_1 = frac{3-j2}{3+j}) I₁
∴ D = (frac{-I_1}{I_2} = frac{3+j}{3-j2}) = 0.5385 + j0.6923 Ω
V₁ = [j10 + 2(9+j7)] I₁
= jI₁ (24 – j18)
So, B = –(frac{V_1}{I_2} = frac{-jI_1 (24-j18)}{-frac{3-j2}{3+j} I_1})
= (frac{6}{13})(-15+j55)
∴ B = -6.923 + j25.385 Ω.

3. For the circuit given below, the value of the Transmission parameters A and C are _________________

A. A = 0, C = 1
B. A = 1, C = 0
C. A = Z, C = 1
D. A = 1, C = Z

Answer: B
Clarification: V₁ = V₂
Or, A = (frac{V_1}{I_2}) = 1
I₁ = 0 or, C = (frac{I_1}{V_2}) = 0.

4. For the circuit given below, the value of the Transmission parameters B and D are _________________

A. B = Z, D = 1
B. B = 1, D = Z
C. B = Z, D = Z
D. B = 1, D = 1

Answer: A
Clarification: V₁ = ZI₁
And I₂ = -I₁
B = (frac{V_1}{I_2})
= (frac{-ZI_1}{-I_1}) = Z
D = (frac{-I_1}{I_2}) = 1.

5. For the circuit given below, the value of the Transmission parameters A and C are _______________

A. A = 1, C = 0
B. A = 0, C = 1
C. A = Y, C = 1
D. A = 1, C = Y

Answer: D
Clarification: V₁ = V₂
∴ A = (frac{V_1}{V_2}) = 1
And V₁ = ZI₁
∴ C = (frac{I_1}{V_2} = frac{1}{Z}) = Y.

6. For the circuit given below, the value of the Transmission parameters B and D are ________________

A. B = Y, D = 1
B. B = 1, D = 0
C. B = 0, D = 1
D. B = 0, D = Y

Answer: C
Clarification: V₁ = V₂ = 0
And I₂ = -I₁
∴ B = (frac{V_1}{I_2}) = 0
∴ D = (frac{-I_1}{I_2}) = 1.

7. For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio (frac{V_2}{V_1}) is ______________

A. 0.3299
B. 0.8942
C. 1.6
D. 0.2941

Answer: D
Clarification: Replacing the given 2-port circuit by an equivalent circuit and applying nodal analysis, we get,
V₂ = (20) (2I₁) = 40 I₁
Or, -10 + 20I₁ + 3V₂ = 0
Or, 10 = 20I₁ + (3) (40I₁) = 140I₁
∴ I₁ = (frac{1}{14}) and V₂ = (frac{40}{14})
So, V₁ = 16I₁ + 3V₂ = (frac{136}{14})
And I₂ = ((frac{100}{125})) (2I₁) = (frac{-8}{70})
∴ (frac{V_2}{V_1} = frac{40}{136}) = 0.2941.

8. For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio (frac{I_2}{I_1}) is ______________

A. 0.3299
B. 0.8942
C. -1.6
D. 0.2941

Answer: C
Clarification: Replacing the given 2-port circuit by an equivalent circuit and applying nodal analysis, we get,
V₂ = (20) (2I₁) = 40 I₁
Or, -10 + 20I₁ + 3V₂ = 0
Or, 10 = 20I₁ + (3) (40I₁) = 140I₁
∴ I₁ = (frac{1}{14}) and V₂ = (frac{40}{14})
So, V₁ = 16I₁ + 3V₂ = (frac{136}{14})
And I₂ = ((frac{100}{125})) (2I₁) = (frac{-8}{70})
∴ (frac{I_2}{I_1}) = -1.6.

9. If for a circuit the value of the h parameter is given as h = [8, 2/3; -2/3, 4/9]. Then the value of the voltage source V is _________________
A. 2.38 V
B. 1.19 V
C. 1.6 V
D. 3.2 V

Answer: B
Clarification: 8I₁ + (frac{2}{3V_2}) = 10
V₂ = (frac{2}{3})I₁ (5||(frac{9}{4}))
= (frac{2}{3})I₁ ((frac{45}{29})= frac{30}{29}I_1)
I₁ = (frac{29}{30})V₂
(8)((frac{29}{30})) V₂ + (frac{2}{3})V₂ = 10
V₂ = (frac{300}{252}) = 1.19 V.

10. For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10^-3]. Given that, Z_S = 2 kΩ and Z_L = 400 Ω. The value of the parameter Z_in is ______________
A. 250 Ω
B. 333.33 Ω
C. 650 Ω
D. 600 Ω

Answer: B
Clarification: Z_in = h_ie – (frac{h_{re} h_{fe} R_L}{1 + h_{oe} R_L})
= h₁₁ – (frac{h_{12} h_{21} R_L}{1+h_{22} R_L})
= 600 – (frac{0.04×30×400}{1+2×10^{-3}×400}) = 333.33 Ω.

11. For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10^-3]. Given that, Z_S = 2 kΩ and Z_L = 400 Ω. The value of the parameter Z_out is ______________
A. 650 Ω
B. 500 Ω
C. 250 Ω
D. 600 Ω

Answer: A
Clarification: Z_out = (frac{R_s+h_{ie}}{(R_s+h_{ie}) h_{oe}-h_{re} h_{fe}})
= (frac{R_s+h_{11}}{(R_s+h_{11}) h_{22}-h_{21} h_{12}})
= (frac{2000+600}{2600×2×10^{-3}-30×0.04}) = 650 Ω.

12. For the circuit given below, the value of the g₁₁ and g₂₁ are _________________

A. g₁₁ = –(frac{1}{R_1+R_2}), g₂₁ = (frac{R_2}{R_1+R_2})
B. g₁₁ = (frac{1}{R_1-R_2}), g₂₁ = –(frac{R_2}{R_1+R_2})
C. g₁₁ = (frac{1}{R_1+R_2}), g₂₁ = (frac{R_2}{R_1+R_2})
D. g₁₁ = (frac{1}{R_1-R_2}), g₂₁ = (frac{R_2}{R_1-R_2})

Answer: C
Clarification: I₁ = (frac{V_1}{R_1+R_2})
Or, g₁₁ = (frac{I_1}{V_1} = frac{1}{R_1+R_2})
By voltage division, V₂ = (frac{R_2}{R_1+R_2})V₁
Or, g₂₁ = (frac{V_2}{V_1} = frac{R_2}{R_1+R_2}).

13. For the circuit given below, the value of the g₁₂ and g₂₂ are _______________

A. g₁₂ = –(frac{R_2}{R_1+R_2}), g₂₂ = R₃ + (frac{R_1 R_2}{R_1+R_2})
B. g₁₂ = (frac{R_2}{R_1+R_2}), g₂₂ = -R₃ + (frac{R_1 R_2}{R_1+R_2})
C. g₁₂ = –(frac{R_2}{R_1+R_2}), g₂₂ = R₃ – (frac{R_1 R_2}{R_1+R_2})
D. g₁₂ = (frac{R_2}{R_1+R_2}), g₂₂ = -R₃ – (frac{R_1 R_2}{R_1+R_2})

Answer: A
Clarification: I₁ = –(frac{R_2}{R_1+R_2})I₂
Or, g₁₂ = (frac{I_1}{I_2} = -frac{R_2}{R_1+R_2})
Also, I₂ (R₃ + R₁ //R₂)
= I2 ((R_3 + frac{R_1 R_2}{R_1+R_2}))
∴ g₂₂ = (frac{V_2}{I_2} = R_3 + frac{R_1 R_2}{R_1+R_2}).

14. For the circuit given below, the value of g₁₁ and g₂₁ are _________________

A. g₁₁ = 0.0667 – j0.0333 Ω, g₂₁ = 0.8 + j0.4 Ω
B. g₁₁ = -0.0667 – j0.0333 Ω, g₂₁ = -0.8 – j0.4 Ω
C. g₁₁ = 0.0667 + j0.0333 Ω, g₂₁ = 0.8 + j0.4 Ω
D. g₁₁ = -0.0667 + j0.0333 Ω, g₂₁ = 0.8 – j0.4 Ω

Answer: C
Clarification: V₁ = (12-j6) I₁
Or, g₁₁ = (frac{I_1}{V_1} = frac{1}{12-j6}) = 0.0667 + j0.0333 Ω
g₂₁ = (frac{V_2}{V_1} = frac{12I_1}{(12-j6) I_1})
= (frac{2}{2-j}) = 0.8 + j0.4 Ω.

15. For the circuit given below, the value of g₁₂ and g₂₂ are ________________

A. g₁₂ = 0.8 + j0.4 Ω, g₂₂ = 2.4 + j5.2 Ω
B. g₁₂ = -0.8 + j0.4 Ω, g₂₂ = -2.4 – j5.2 Ω
C. g₁₂ = 0.8 – j0.4 Ω, g₂₂ = 2.4 – j5.2 Ω
D. g₁₂ = -0.8 – j0.4 Ω, g₂₂ = 2.4 + j5.2 Ω

Answer: D
Clarification: I₁ = (frac{-12}{12-j6})I₂
Or, g₁₂ = (frac{I_1}{I_2}) = -g₂₁ = -0.8 – j0.4 Ω
V₂ = (j10 + 12 || -j6) I₂
g₂₂ = (frac{V_2}{I_2}) = 2.4 + j5.2 Ω.