300+ REAL TIME Network Theory Objective Questions & Answers 2023

250+ TOP MCQs on Impedence Diagram and Answers

Network Theory Multiple Choice Questions on “Impedence Diagram”.

1. Impedance is a complex quantity having the real part as _______ and the imaginary part as ______
A. resistance, resistance
B. resistance, reactance
C. reactance, resistance
D. reactance, reactance

Answer: B
Clarification: Almost all electric circuits offer impedance to the flow of current. Impedance is a complex quantity having the real part as resistance and the imaginary part as reactance.

2. The voltage function v(t) in the circuit shown below is?

A. v(t) = V_m e^-tjω
B. v(t) = V_me^tjω
C. v(t) = e^tjω
D. v(t) = e^-tjω

Answer: B
Clarification: The voltage function v(t) in the circuit is a complex function and is given by v(t) = V_m e^tjω=Vm(cosωt+jsinωt).

3. The current i(t) in the circuit shown below is?

A. i(t)=(V_m/(R-jωL))e^tjω
B. i(t)=(V_m(R+jωL)) e^tjω
C. i(t)=(V_m(R-jωL)) e^tjω
D. i(t)=(V_m/(R+jωL)) e^tjω

Answer: D
Clarification: i(t)=I_m e^tjω. V_m e^tjω=RI_m e^tjω+L d/dt (I_m e^tjω). V_me^tjω=RI_m e^tjω+L I_m(jω)e^tjω. I_m = V_m/(R+jωL). i(t)=(V_m/(R+jωL)) e^tjω.

4. The impedance of the circuit shown below is?

A. R + jωL
B. R – jωL
C. R + 1/jωL
D. R – 1/jωL

Answer: A
Clarification: Impedance is defined as he ratio of the voltage to current function. The impedance of the circuit Z= V_me^tjω/((V_m/(R+jωL)) e^tjω)=R+jωL.

5. What is the magnitude of the impedance of the following circuit?

A. √(R+ωL)
B. √(R-ωL)
C.√(R²+(ωL)²)
D. √(R²-(ωL)²)

Answer: C
Clarification: Complex impedance is the total opposition offered by the circuit elements to ac current and can be displayed on the complex plane. The magnitude of the impedance of the circuit is √(R²+(ωL)²).

6. The phase angle between current and voltage in the circuit shown below is?

A. tan^-1⁡ωL/R
B. tan^-1⁡ωR/
C. tan^-1⁡R/ωL
D. tan^-1⁡L/ωR

Answer: A
Clarification: The angle between impedance and reactance is the phase angle between the current and voltage applied to the circuit. θ=tan^-1⁡ωL/R

7. The voltage function v(t) in the circuit shown below is?

A. v(t) = e^-tjω
B. v(t) = e^tjω
C. v(t) = V_me^tjω
D. v(t) = V_me^-tjω

Answer: C
Clarification: If we consider the RC series circuit and apply the complex function v(t) to the circuit then voltage function is given by v(t) = V_me^tjω.

8. The impedance of the circuit shown below is?

A. R + jωC
B. R – jωC
C. R + 1/jωC
D. R – 1/jωC

Answer: C
Clarification: The impedance of the circuit shown above is R + 1/jωC. Here the impedance Z consists resistance which is real part and capacitive reactance which is imaginary part of the impedance.

9. The magnitude of the impedance of the circuit shown below is?

A. √(R+1/ωC.
B. √(R-1/ωC.
C. √(R²+(1/ωC.²)
D. √(R²-(1/ωC.²)

Answer: C
Clarification: The magnitude of the impedance of the circuit shown above is √(R²+(1/ωC.²). Here the impedance is the vector sum of the resistance and the capacitive reactance.

10. The angle between resistance and impedance in the circuit shown below.

A. tan^-11/ωRC
B. tan^-1⁡C/ωR
C. tan^-1⁡R/ωC
D. tan^-1⁡ωRC

Answer: A
Clarification: The angle between resistance and impedance in the circuit shown above tan^-1⁡1/ωRC. The angle between resistance and impedance is the phase angle between the applied voltage and current in the circuit.

250+ TOP MCQs on Problems Involving Dot Conventions and Answers

Network Theory Multiple Choice Questions on “Problems Involving Dot Conventions”.

1. The current through an electrical conductor is 1A when the temperature of the conductor 0°C and 0.7 A when the temperature is 100°C. The current when the temperature of the conductor is 1200°C is ___________
A. 0.08 A
B. 0.16 A
C. 0.32 A
D. 0.64 A

Answer: B
Clarification: (frac{1}{0.7} = frac{R_O (1+αt)}{R_O})
= 1 + α100
∴α = 0.0043 per °C
Current at 1200 °C is given by, (frac{1}{I} = frac{R_O (1+α1200)}{R_O})
= 1 + α1200
= 1 + 0.0043 × 1200 = 6.16
∴ I = (frac{1}{6.16}) = 0.16 A.

2. In the circuit given below, the equivalent capacitance is ____________

A. 1.625 F
B. 1.583 F
C. 0.583 F
D. 0.615 F

Answer: D
Clarification: C_CB = (left(frac{C_2 C_3}{C_2+ C_3}right)) + C₅ = 1.5 F
Now, C_AB =(left(frac{C_1 C_{CB}}{C_1+ C_{CB}}right)) + C₆ = 1.6 F
C_XY = (frac{C_{AB} × C_4}{C_{AB} + C_4}) = 0.615 F.

3. In the circuit given below, the equivalent capacitance is ______________

A. 3.5 μF
B. 1.2 μF
C. 2.4 μF
D. 4.05 μF

Answer: D
Clarification: The 2.5 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 1.5 μF.
Hence, C₁ = (frac{1.5(2.5+1)}{1.5+2.5+1})
= (frac{5.25}{5}) = 1.05
Now, C₁ is in parallel with the 3 μF capacitor.
∴ C_EQ = 1.05 + 3 = 4.05 μF.

4. In the circuit given below, the voltage across A and B is?

A. 13.04 V
B. 17.84 V
C. 12 V
D. 10.96 V

Answer: B
Clarification: Loop current I₁ = (frac{6}{10}) = 0.6 A
I₂ = (frac{12}{14}) = 0.86 A
V_AB = (0.6) (4) + 12 + (0.86) (4)
= 2.4 + 12 + 3.44
= 17.84 V.

5. In the figure given below, the voltage source provides the circuit with a voltage V. The number of non-planar graph of independent loop equations is ______________

A. 8
B. 12
C. 7
D. 5

Answer: D
Clarification: The total number of independent loop equations are given by L = B – N + 1 where,
L = number of loop equations
B = number of branches = 12
N = number of nodes = 8
∴ L = 12 – 8 + 1 = 5.

6. When a DC voltage is applied to an inductor, the current through it is found to build up in accordance with I = 20(1-e^-50t). After the lapse of 0.02 s, the voltage is equal to 2 V. What is the value of inductance?
A. 2 mH
B. 5.43 mH
C. 1.54 mH
D. 0.74 mH

Answer: B
Clarification: V_L = L(frac{dI}{dt})
Where, I = 20(1-e^-50t)
Therefore, V_L = L(frac{d 20(1-e^{-50t})}{dt})
= L × 20 × 50e^-50t
At t = 0.02 s, V_L = 2 V
∴ L = (frac{2}{20 × 50 × e^{-50×0.02}})
= 5.43 μH.

7. An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V. It is disconnected and then immersed in oil with a relative permittivity of 2.5. The energy after immersion is?
A. 275 μJ
B. 250 μJ
C. 225 μJ
D. 625 μJ

Answer: B
Clarification: E = (frac{1}{2}) CV²
Or, C = (frac{ε_0 ε_r A}{d} = ϵ_r (frac{ε_0 A}{d}))
= 2.5 × 0.005 × 10^-6
∴ C_NEW = 12.5 × 10^-9 F
Now, q = CV = 0.005 × 10^-6 × 500 = 2.5 × 10^-6
V_NEW = (frac{q} {C_{NEW}})
= (frac{2.5 × 10^{-6}}{12.5 × 10^{-9}}) V_NEW = 200
E = (frac{1}{2}) CV²
= (frac{1}{2}) × 12.5 × 10^-9 × (200)²
= 250 μJ.

8. The resistance of copper motor winding at t=20°C is 3.42 Ω. After extended operation at full load, the motor windings measures 4.22 Ω. If the temperature coefficient is 0.0426, what is the rise in temperature?
A. 60°C
B. 45.2°C
C. 72.9°C
D. 10.16°C

Answer: D
Clarification: Given that, R₁ = 3.42 Ω
T₁ = 20° and α = 0.0426
R₂ = 4.22 Ω
Now, (frac{R_1}{1 + αT_1} = frac{R_2}{1 + αT_2})
Or, (frac{3.42}{1 + 0.0426 × 20} = frac{4.22}{1 + 0.0426 T_2})
∴ Rise in temperature = T₂ – T₁
= 30.16 – 20
= 10.16°C.

9. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 50 μF is?
A. 1.57 A
B. 1.87 A
C. 1.67 A
D. 2.83 A

Answer: C
Clarification: As the capacitors are in parallel, then the voltage V is given by,
V = (frac{1}{C_1} int I_1 ,dt )
∴ I₁ = C₁ (frac{dV}{dt})
That is, (frac{I_1}{I_2} = frac{C_1}{C_2} = frac{50}{100} = frac{1}{2}) ………………….. (1)
Also, I₁ + I₂ = 5 A ………………………….. (2)
Solving (1) and (2), we get, I₁ = 1.67 A.

10. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 100 μF is?
A. 2.33 A
B. 3.33 A
C. 1.33 A
D. 4.33 A

Answer: B
Clarification: As the capacitors are in parallel, then the voltage V is given by,
V = (frac{1}{C_2} int I_2 ,dt )
∴ I₂ = C₂ (frac{dV}{dt})
That is, (frac{I_1}{I_2} = frac{C_1}{C_2} = frac{50}{100} = frac{1}{2}) ………………….. (1)
Also, I₁ + I₂ = 5 A ………………………….. (2)
Solving (1) and (2), we get, I₂ = 3.33 A.

11. In the circuit given below, the resonant frequency is _______________

A. (frac{1}{2sqrt{2} π}) Hz
B. (frac{1}{2π}) Hz
C. (frac{1}{4π}) Hz
D. (frac{1}{sqrt{2}2π}) Hz

Answer: C
Clarification: I_EQ = L₁ + L₂ + 2M
L_EQ = 1 + 2 + 2 × (frac{1}{2}) = 4 H
∴ F_O = (frac{1}{2πsqrt{LC}} )
= (frac{1}{2πsqrt{4 × 1}} )
= (frac{1}{4π}) Hz.

12. In a series resonant circuit, V_C = 300 V, V_L = 300 V and V_R = 100 V. What is the value of the source voltage?
A. Zero
B. 100 V
C. 350 V
D. 200 V

Answer: B
Clarification: As V_C and V_L are equal, then X_C is equal to X_L and both the voltages are then cancelled out.
That is V_S = V_R
∴ V_S = 100 V.

13. For the circuit given below, what is the value of the Q factor for the inductor?

A. 4.74
B. 4.472
C. 4.358
D. 4.853

Answer: C
Clarification: Q_IN = (sqrt{frac{L}{CR^2} – 1})
= (sqrt{frac{1}{2 × 5^2 × 10^{-3} – 1}})
= (sqrt{frac{1}{50 × 10^{-3} – 1}})
= (sqrt{19}) = 4.358.

14. In the circuit given below, the value of the voltage source E is _______________

A. -65 V
B. 40 V
C. -60 V
D. 65 V

Answer: A

15. In the circuit given below, bulb X uses 48 W when lit, bulb Y uses 22 W when lit and bulb Z uses 14.4 W when lit. The number of additional bulbs in parallel to this circuit, that would be required to below the fuse is _______________

A. 4
B. 5
C. 6
D. 7

Answer: A
Clarification: I_X = (frac{48}{12}) = 4 A
I_Y = (frac{22}{12}) = 1.8 A
I_Z = (frac{14.4}{12}) = 1.2 A
Current required to below the fuse = 20 A
∴ Additional bulbs must draw current = 20 – (4 + 18 + 1.2)
= 20 – 7 = 13
∴ Number of additional bulbs required = (frac{13}{3}) = 4.33
So, 4 additional bulbs are required.

250+ TOP MCQs on Sinusoidal Response of an R-C Circuit and Answers

Network Theory online test on “Sinusoidal Response of an R-C Circuit”.

1. In the sinusoidal response of R-C circuit, the complementary function of the solution of i is?
A. i_c = ce^-t/RC
B. i_c = ce^t/RC
C. i_c = ce^-t/RC
D. i_c = ce^t/RC

Answer: A
Clarification: From the R-c circuit, we get the characteristic equation as (D+1/RC.i=-Vω/R sin⁡(ωt+θ). The complementary function of the solution i is i_c = ce^-t/RC.

2. The particular current obtained from the solution of i in the sinusoidal response of R-C circuit is?
A. i_p = V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.)
B. i_p = -V/√(R²+(1/ωC.²) cos⁡(ωt+θ-tan^-1(1/ωRC.)
C. i_p = V/√(R²+(1/ωC.²) cos⁡(ωt+θ-tan^-1(1/ωRC.)
D. i_p = -V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.)

Answer: A
Clarification: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i_p = V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.).

3. The value of ‘c’ in complementary function of ‘i’ is?
A. c = V/R cosθ+V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)
B. c = V/R cosθ+V/√(R²+(1/(ωC.)²) cos⁡(θ-tan^-1(1/ωRC.)
C. c = V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ-tan^-1(1/ωRC.)
D. c = V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)

Answer: D
Clarification: Since the capacitor does not allow sudden changes in voltages, at t = 0, i = V/R cosθ. So, c = V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.).

4. The complete solution of the current in the sinusoidal response of R-C circuit is?
A. i = e^-t/RC[V/R cosθ+V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)+V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]
B. i = e^-t/RC[V/R cosθ-V/√(R²+(1/ωC.²) cos⁡(θ+tan^-1(1/ωRC.)-V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]
C. i = e^-t/RC[V/R cosθ+V/√(R²+(1/ωC.²) cos⁡(θ+tan^-1(1/ωRC.)-V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]
D. i = e^-t/RC[V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)+V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]

Answer: D
Clarification: The complete solution for the current becomes i = e^-t/RC[V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)+V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complementary function of the solution of ‘i’ is?

A. i_c = c exp (-t/10^-10)
B. i_c = c exp(-t/10¹⁰)
C. i_c = c exp (-t/10^-5)
D. i_c = c exp (-t/10⁵)

Answer: C

6. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The particular integral of the solution of ‘i_p’ is?

A. i_p = (4.99×10^-3) cos⁡(100t+π/4-89.94^o)
B. i_p = (4.99×10^-3) cos⁡(100t-π/4-89.94^o)
C. i_p = (4.99×10^-3) cos⁡(100t-π/4+89.94^o)
D. i_p = (4.99×10^-3) cos⁡(100t+π/4+89.94^o)

Answer: D
Clarification: Assuming particular integral as i_p = A cos (ωt + θ) + B sin (ωt + θ)
we get i_p = V/√(R²+(1/ωC.²) cos⁡(ωt+θ-tan^-1(1/ωRC.)
where ω = 1000 rad/sec, θ = π/4, C = 1µF, R = 10Ω. On substituting, we get i_p = (4.99×10^-3) cos⁡(100t+π/4+89.94^o).

7. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The current flowing in the circuit at t = 0 is?

A. 1.53
B. 2.53
C. 3.53
D. 4.53

Answer: C
Clarification: At t = 0 that is initially current flowing through the circuit is i = V/R cosθ = (50/10)cos(π/4) = 3.53A.

8. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complete solution of ‘i’ is?

A. i = c exp (-t/10^-5) – (4.99×10^-3) cos⁡(100t+π/2+89.94^o)
B. i = c exp (-t/10^-5) + (4.99×10^-3) cos⁡(100t+π/2+89.94^o)
C. i = -c exp(-t/10^-5) + (4.99×10^-3) cos⁡(100t+π/2+89.94^o)
D. i = -c exp(-t/10^-5) – (4.99×10^-3) cos⁡(100t+π/2+89.94^o)

Answer: B
Clarification: The complete solution for the current is the sum of the complementary function and the particular integral. The complete solution for the current becomes i = c exp (-t/10^-5) + (4.99×10^-3) cos⁡(100t+π/2+89.94^o).

9. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The value of c in the complementary function of ‘i’ is?

A. c = (3.53-4.99×10^-3) cos⁡(π/4+89.94^o)
B. c = (3.53+4.99×10^-3) cos⁡(π/4+89.94^o)
C. c = (3.53+4.99×10^-3) cos⁡(π/4-89.94^o)
D. c = (3.53-4.99×10^-3) cos⁡(π/4-89.94^o)

Answer: A
Clarification: At t = 0, the current flowing through the circuit is 3.53A. So, c = (3.53-4.99×10^-3) cos⁡(π/4+89.94^o).

10. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complete solution of ‘i’ is?

A. i = [(3.53-4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o)
B. i = [(3.53+4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o)
C. i = [(3.53+4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)-4.99×10^-3) cos⁡(100t+π/2+89.94^o)
D. i = [(3.53-4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)-4.99×10^-3) cos⁡(100t+π/2+89.94^o)

Answer: A
Clarification: The complete solution for the current is the sum of the complementary function and the particular integral. So, i = [(3.53-4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o).

250+ TOP MCQs on Properties of Driving Point Functions and Answers

Network Theory Multiple Choice Questions on “Properties of Driving Point Functions”.

1. 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
Clarification: 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.

2. The pole is that finite value of S for which N (S) becomes __________
A. 0
B. 1
C. 2
D. ∞
Answer: D
Clarification: 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.

3. 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
Clarification: 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).

4. 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
Clarification: 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.

5. 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
Clarification: 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.

6. 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
Clarification: The zeros of driving point impedance are those frequencies corresponding to short circuit conditions as a 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.

7. 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
Clarification: 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).

8. 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
Clarification: 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.

9. The real part of all zeros and poles must be?
A. positive or zero
B. negative or zero
C. positive
D. negative
Answer: B
Clarification: 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.

10. 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
Clarification: 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.

250+ TOP MCQs on Sources and Some Definitions and Answers

Network Theory Multiple Choice Questions on “Sources and Some Definitions”.

1. Source is a basic network element which supplies power to the networks.
A. True
B. False
Answer: B
Clarification: The basic network element which supplies energy to the networks is Source. Hence, it is true.

2. The dependent sources are of _____________ kinds.
A. 5
B. 2
C. 3
D. 4
Answer: D
Clarification: The dependent sources are of four kinds, depending on whether the control variable is voltage or current and the controlled source is a voltage source or current source. They are VCVS, VCCS, CCVS and CCCS.

3. The constant g_m has dimension of ___________
A. Ampere per volt
B. Ampere
C. Volt
D. Volt per ampere
Answer: A
Clarification: Transconductance is the ratio of current to voltage. Hence, the constant g_m has a dimension of ampere per volt or siemens (S).

4. In CCVS, voltage depends on the control current and the constant called __________
A. Transconductance
B. Transresistance
C. Current Gain
D. Voltage Gain
Answer: B
Clarification: In CCVS, voltage is directly proportional to the control current. The constant of proportionality is called Transresistance(r).
V = ri.

5. Every circuit is a network, but all networks are not circuits.
A. True
B. False
Answer: A
Clarification: The interconnection of two or more circuit elements is called a Network. If the network contains at least one closed path, it is called a Circuit.

6. Which of the following is not an example of a linear element?
A. Resistor
B. Thermistor
C. Inductor
D. Capacitor
Answer: B
Clarification: If the resistance, inductance or capacitance offered by an element does not change linearly with the change in applied voltage or circuit current, the element is termed as a linear element. Resistor, Inductor and Capacitor are examples of linear elements. Thermistor is an example of Non-Linear element.

7. Find the odd one out.
A. Resistor
B. Voltage-dependent resistor(VDR)
C. Temperature-dependent resistor(Thermistor)
D. Light-dependent resistor(LDR)
Answer: A
Clarification: A non-linear circuit element is one in which the current does not change linearly with the change in applied voltage. Examples of non-linear elements are VDR, thermistor and LDR. Hence resistor is the odd one.

8. Which of the following is an Active element?
A. Resistor
B. Inductor
C. Capacitor
D. OP-AMP
Answer: D
Clarification: OP-AMP is an active element because it can be used for the amplification or generation of signals. All the other circuit elements are passive elements.

9. A semiconductor diode is an ____________ element.
A. Bilateral
B. Unilateral
C. Active
D. Passive
Answer: B
Clarification: In semiconductor diode, current flows through the diode only in one direction. Hence, it is a unilateral element.

10. Example of distributed element is ___________
A. Resistor
B. Thermistor
C. Semiconductor diode
D. Transmission lines
Answer: D
Clarification: Distributed elements are those which are not separable for analysis purposes. Examples of distributed elements are transmission lines in which the resistance, inductance and capacitance are distributed along its length.

250+ TOP MCQs on Norton’s Theorem and Answers

Network Theory Multiple Choice Questions on “Norton’s Theorem”.

1. Find the current flowing between terminals A and B of the circuit shown below.

A. 1
B. 2
C. 3
D. 4

Answer: D
Clarification: The magnitude of the current in Norton’s equivalent circuit is equal to the current passing through the short circuited terminals that are I=20/5=4A.

2. Find the equivalent resistance between terminals A and B of the circuit shown below.

A. 0.33
B. 3.33
C. 33.3
D. 333

Answer: B
Clarification: Norton’s resistance is equal to the parallel combination of both the 5Ω and 10Ω resistors that is R = (5×10)/15 = 3.33Ω.

3. Find the current through 6Ω resistor in the circuit shown below.

A. 1
B. 1.43
C. 2
D. 2.43

Answer: B
Clarification: The current passing through the 6Ω resistor and the voltage across it due to Norton’s equivalent circuit is I = 4×3.33/(6+3.33) = 1.43A.

4. Find the voltage drop across 6Ω resistor in the circuit shown below.

A. 6.58
B. 7.58
C. 8.58
D. 9.58

Answer: C
Clarification: The voltage across the 6Ω resistor is V = 1.43×6 = 8.58V. So the current and voltage have the same values both in the original circuit and Norton’s equivalent circuit.

5. Find the current flowing between terminals A and B in the following circuit.

A. 1
B. 2
C. 3
D. 4

Answer: D
Clarification: Short circuiting terminals A and B, 20-10(I₁)=0, I₁=2A. 10-5(I₂), I₂=2A. Current flowing through terminals A and B = 2+2 = 4A.

6. Find the equivalent resistance between terminals A and B in the following circuit.

A. 3
B. 3.03
C. 3.33
D. 3.63

Answer: C
Clarification: The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 5Ω resistor => R = ((10×5))/(10+5) = 3.33Ω.

7. Find the current flowing between terminals A and B obtained in the equivalent Nortan’s circuit.

A. 8
B. 9
C. 10
D. 11

Answer: D
Clarification: To solve for Norton’s current we have to find the current passing through the terminals A and B. Short circuiting the terminals a and b, I=100/((6×10)/(6+10)+(15×8)/(15+8))=11.16 ≅ 11A.

8. Find the equivalent resistance between terminals A and B obtained in the equivalent Nortan’s circuit.

A. 8
B. 9
C. 10
D. 11

Answer: B
Clarification: The resistance at terminals AB is the parallel combination of the 10Ω resistor and the 6Ω resistor and parallel combination of the 15Ω resistor and the 8Ω resistor => R=(10×6)/(10+6)+(15×8)/(15+8)=8.96≅9Ω.

9. Find the current through 5Ω resistor in the circuit shown below.

A. 7
B. 8
C. 9
D. 10

Answer: A
Clarification: To solve for Norton’s current we have to find the current passing through the terminals A and B. Short circuiting the terminals a and b I=11.16×8.96/(5+8.96) = 7.16A.

10. Find the voltage drop across 5Ω resistor in the circuit shown below.

A. 33
B. 34
C. 35
D. 36

Answer: D
Clarification: The voltage drop across 5Ω resistor in the circuit is the product of current and resistance => V = 5×7.16 = 35.8 ≅ 36V.