Category: Network Theory Objective Questions

Network Theory Multiple Choice Questions on “Phase Relation in a Pure Resistor”.

1. The phase difference between voltage and current in case of resistor is?
A. in phase
B. out of phase
C. 45⁰ out of phase
D. 90⁰ out of phase
Answer: A
Clarification: The phase difference between voltage and current in case of resistor is in phase. The amplitudes of the waveform may differ according to the value of resistance.

2. In the case of inductor, the voltage?
A. is in phase with the current
B. lags behind the current by 90⁰
C. leads the current by 90⁰
D. is out of phase with the current
Answer: C
Clarification: In the case of inductor, the voltage and current are out of phase. So, the voltage leads the current by 90⁰.

3. For inductor, what is the current?
A. is out of phase with the current
B. leads the current by 90⁰
C. is in phase with the current
D. lags behind the current by 90⁰
Answer: D
Clarification: For inductor, the voltage and current are out of phase. So, the current lags behind the current by 90⁰.

4. The value of inductance reactance is?
A. R
B. ωL
C. 1/ωL
D. ωC
Answer: B
Clarification: The value of inductance reactance is ωL. Hence a pure inductor has an impedance whose value is ωL.

5. The current in the pure capacitor?
A. lags behind the voltage by 90⁰
B. is in phase with the voltage
C. lags behind the voltage by 45⁰
D. leads the voltage by 90⁰
Answer: D
Clarification: In a capacitor, there exists a phase difference between current and voltage. The current in pure capacitor leads the voltage by 90⁰.

6. In case of pure capacitor, the voltage?
A. leads the voltage by 90⁰
B. lags behind the voltage by 45⁰
C. lags behind the voltage by 90⁰
D. is in phase with the voltage
Answer: C
Clarification: In a capacitor, there exists a phase difference between current and voltage. In case of pure capacitor, the voltage lags behind the voltage by 90⁰.

7. The impedance value of a pure capacitor is?
A. ωC
B. 1/ ωC
C. ωL
D. R
Answer: B
Clarification: The impedance value of a pure capacitor is 1/ ωC. And the impedance of the capacitor is called capacitive reactance.

8. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. How much is first sine wave shifted in phase from the 0⁰ reference?
A. leads reference angle by 10⁰
B. lags reference angle by 10⁰
C. leads reference angle by 15⁰
D. lags reference angle by 15⁰
Answer: C
Clarification: As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. First sine wave leads reference angle by 15⁰ from the 0⁰ reference.

9. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. How much is second sine wave shifted in phase from the 0⁰ reference?
A. lags reference angle by 15⁰
B. leads reference angle by 15⁰
C. lags reference angle by 10⁰
D. leads reference angle by 10⁰
Answer: A
Clarification: As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. Second sine wave lags reference angle by 15⁰ from the 0⁰ reference.

10. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. What is the phase angle between two sine waves?
A. 10⁰
B. 15⁰
C. 20⁰
D. 25⁰
Answer: D
Clarification: As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. The phase angle between two sine waves mentioned above is 25⁰.

Network Theory Multiple Choice Questions on “Dot Convention in Magnetically Coupled Circuits”.

1. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________
A. 25
B. 50
C. 100
D. 200

Answer: B
Clarification: Q = (frac{f_0}{BW})
And f₀ = 1/2π (LC.^0.5
BW = R/L
Or, Q = (frac{1}{R} (frac{L}{C})^{0.5})
When R, L and C are doubled, Q’ = 50.

2. In the circuit given below, the input impedance Z_IN of the circuit is _________

A. 0.52 – j4.30 Ω
B. 0.52 + j15.70 Ω
C. 64.73 + j17.77 Ω
D. 0.3 – j33.66 Ω

Answer: C
Clarification: Z_IN = (-6j) || (Z_A)
Z_A = j10 + (frac{12^2}{(j30+j6-j2+4)})
= 0.49 + j5.82
Z_IN = (frac{(-j6)(0.49+j5.82)}{(-j6+0.49+j5.82)})
= 64.73 + j17.77 Ω.

3. The switch in the circuit shown was on position X for a long time. The switch is then moved to position Y at time t=0. The current I(t) for t>0 is ________

A. 0.2e^-125tu(t) mA
B. 20e^-1250tu(t) mA
C. 0.2e^-1250tu(t) mA
D. 20e^-1000tu(t) mA

Answer: C
Clarification: C_EQ = (frac{0.8 × 0.2}{0.8+0.2}) = 0.16
V_C (t=0^–) = 100 V
At t≥0,
The discharging current I (t) = (frac{V_O}{R} e^{-frac{t}{RC}})
= (frac{100}{5000} e^{- frac{t}{5×10^3×0.16×10^{-6}}})
= 0.2e^-1250tu(t) mA.

4. In the circuit shown, the voltage source supplies power which is _____________

A. Zero
B. 5 W
C. 10 W
D. 100 W

Answer: A
Clarification: Let the current supplied by a voltage source.
Applying KVL in outer loop,
10 – (I+3) × (I+1) – (I+2) × 2 = 0
10 – 2(I+3) – 2(I-2) = 0
Or, I = 0
∴ Power VI = 0.

5. In the circuit shown below the current I(t) for t≥0⁺ (assuming zero initial conditions) is ___________

A. 0.5-0.125e^-1000t A
B. 1.5-0.125e^-1000t A
C. 0.5-0.5e^-1000t A
D. 0.375e^-1000t A

Answer: A
Clarification: I (t) = (frac{1.5}{3}) = 0.5
L_EQ = 15 mH
R_EQ = 5+10 = 15Ω
L = (frac{L_{EQ}}{R_{EQ}})
= (frac{15 × 10^3}{15} = frac{1}{1000})
I (t) A – (A – B. e^-t = 0.5 – (0.5-B. e^-1000t
= 0.5(0.5 – 0.375) e^-1000t
= 0.5 – 0.125 e^-1000t
I (t) = 0.5-0.125e^-1000t.

6. Initial voltage on capacitor VO as marked |V_O| = 5 V, V_S = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0⁺ is _____________

A. 1 V
B. -1 V
C. (frac{13}{3}) V
D. –(frac{13}{3}) V

Answer: C
Clarification: Applying voltage divider method, we get,
I = (frac{V}{R_{EQ}} )
= (frac{8}{1+1||1} )
= (frac{8}{1+frac{1}{2}} = frac{16}{3}) A
I₁ = (frac{16}{3} × frac{1}{2} = frac{8}{3}) A
And (I’_2 = frac{V}{R_{EQ}} = frac{5}{1+1||1})
= (frac{5}{1+frac{1}{2}} = frac{10}{3}) A
Now, (I’_1 = I’_2 × frac{1}{1+1} )
= (frac{10}{3} × frac{1}{2} = frac{5}{3}) A
Hence, the net current in 1Ω resistance = I1 + (I’_1)
= (frac{8}{3} + frac{5}{3} = frac{13}{3}) A
∴ Voltage drop across 1Ω = (frac{13}{3} × 1 = frac{13}{3}) V.

7. For a unit step signal u (t), the response is V₁ (t) = (1-e^-3t) for t>0. If a signal 3u (t) + δ(t) is applied, the response will be (considering zero initial conditions)?
A. (3-6e^-3t)u(t)
B. (3-3e^-3t)u(t)
C. 3u(t)
D. (3+3e^-3t)u(t)

Answer: C
Clarification: For u (t) = 1, t>0
V₁ (t) = (1-e^-3t)
Or, V₁ (s) = (left(frac{1}{s} + frac{1}{s+3}right) = frac{3}{s(s+3)})
And T(s) = (frac{V_1 (S)}{u(S)} = frac{3}{s+3})
Now, for R(s) = ((frac{3}{s}) + 1)
Response, H(s) = R(s) T(s) = ((frac{3+s}{s}) (frac{3}{s+3}) = frac{3}{s})
Or, h (t) = 3 u (t).

8. In the circuit given below, for time t<0, S₁ remained closed and S₂ open., S₁ is initially opened and S₂ is initially closed. If the voltage V₂ across the capacitor C₂ at t=0 is zero, the voltage across the capacitor combination at t=0⁺ will be ____________

A. 1 V
B. 2 V
C. 1.5 V
D. 3 V

Answer: D
Clarification: When S₁ is closed and S₂ is open,
V_C1 (0^–) = V_C1 (0⁺) = 3V
When S₁ is opened and S₂ is closed, V_C2 (0⁺) = V_C2 (0⁺) = 3V.

9. In the circuit given below, the switch S₁ is initially closed and S₂ is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through C during t=0⁺ is _____________

A. 55 A
B. 5.5 A
C. 45 A
D. 4.5 A

Answer: D
Clarification: By KCL, we get,
(frac{V_L}{10} – 10 + frac{V_L-10}{10}) = 0
Hence, 2 V_L = 110
∴ V_L = 55 V
Or, I_C = (frac{55-10}{10}) = 4.5 A.

10. In the circuit given below, the switch S₁ is initially closed and S₂ is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through L during t=0⁺ is _____________

A. 55 A
B. 5.5 A
C. 45 A
D. 4.5 A

Answer: A
Clarification: By KCL, we get,
(frac{V_L}{10} – 10 + frac{V_L-10}{10}) = 0
Hence, 2 V_L = 110
∴ V_L = 55 V.

11. An ideal capacitor is charged to a voltage V_O and connected at t=0 across an ideal inductor L. If ω = (frac{1}{sqrt{LC}}), the voltage across the capacitor at time t>0 is ____________
A. V_O
B. V_O cos(ωt)
C. V_O sin(ωt)
D. V_O e^-ωtcos(ωt)

Answer: B
Clarification: Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor.
Voltage across capacitor will be decreasing from V_O and periodic and is not decaying since both L and C is ideal.
∴ Voltage across the capacitor at time t>0 is V_O cos(ωt).

12. In the figure given below, what is the RMS value of the periodic waveform?

A. 2(sqrt{6}) A
B. 6(sqrt{2}) A
C. (sqrt{frac{4}{3}}) A
D. 1.5 A

Answer: A
Clarification: The rms value for any waveform is = (sqrt{frac{1}{T} int_0^T f^2 (t)dt)})
= ([frac{1}{T}(int_0^{frac{T}{2}}(mt)^2 dt + int_{frac{T}{2}}^T 6^2 dt]^{1/2})
= ([frac{1}{T}(frac{144}{T^2} × frac{T^3}{3×8} + 36 × frac{T}{2})]^{frac{1}{2}})
= ([6+18]^{frac{1}{2}} = sqrt{24} = 2sqrt{6}) A.

13. In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t=1 sec after the switch S is closed will be ___________

A. 14.7 A
B. 18.5 A
C. 40 A
D. 50 A

Answer: A
Clarification: Using KVL, 100 = R(frac{dq}{dt} + frac{q}{C} )
Or, 100 C = RC(frac{dq}{dt}) + q
Now, (int_{q_o}^q frac{dq}{100C-q} = frac{1}{RC} ∫_0^t dt)
Or, 100C – q = (100C – q_o) e^-t/RC
I = (frac{dq}{dt} = frac{(100C – q_0)}{RC} e^{-1/1})
= 40e^-1 = 14.7 A.

14. In the circuit given below, the switch is closed at time t=0. The voltage across the inductance just at t=0⁺ is ____________

A. 2 V
B. 4 V
C. -6 V
D. 8 V

Answer: B
Clarification: A t=0⁺,
I (0⁺) = (frac{10}{4||4+3} = frac{10}{5}) = 2A
∴ I_2(0⁺) = (frac{2}{2}) = 1A
V_L(0⁺) = 1 × 4 = 4 V.

15. A rectangular voltage wave of magnitude A and duration B is applied to a series combination of resistance R and capacitance C. The voltage developed across the capacitor is ____________
A. A[1 – exp(-(frac{B}{RC}))]
B. (frac{AB}{RC})
C. A
D. A exp(-(frac{B}{RC}))

Answer: A
Clarification: V_C = (frac{1}{C} ∫Idt)
= (frac{1}{C} ∫_0^B frac{A}{R} e^{-frac{t}{RC}}) dt
V_C = A[1 – exp(-(frac{B}{RC}))]
Hence, maximum voltage = V [1 – exp (-(frac{B}{RC}))].

Network Theory Multiple Choice Questions on “Sinusoidal Response of an R-L Circuit”.

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

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

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

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

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

Answer: B
Clarification: Since the inductor does not allow sudden changes in currents, at t = 0, i = 0. So, c = -V/√(R²+(ωL)²) cos⁡(θ-tan^-1(ωL/R)).

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

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

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complementary function of the solution of ‘i’ is?

A. i_c = ce^-100t
B. i_c = ce^100t
C. i_c = ce^-200t
D. i_c = ce^200t

Answer: C
Clarification: By applying Kirchhoff’s voltage law to the circuit, we have 20i+0.1di/dt=100cos⁡(10³ t+π/2) => (D+200)i=1000cos⁡(1000t+π/2). The complementary function is i_c = ce^-200t.

6. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The particular integral of the solution of ‘i_p’ is?

A. i_p = 0.98cos⁡(1000t+π/2-78.6^o)
B. i_p = 0.98cos⁡(1000t-π/2-78.6^o)
C. i_p = 0.98cos⁡(1000t-π/2+78.6^o)
D. i_p = 0.98cos⁡(1000t+π/2+78.6^o)

Answer: A
Clarification: Assuming particular integral as i_p = A cos (ωt + θ) + B sin(ωt + θ). We get i_p = V/√(R²+(ωL)²) cos⁡(ωt+θ-tan^-1(ωL/R)) where ω = 1000 rad/sec, V = 100V, θ = π/2, L = 0.1H, R = 20Ω. On substituting, we get i_p = 0.98cos⁡(1000t+π/2-78.6^o).

7. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complete solution of ‘i’ is?

A. i = ce^-200t + 0.98cos⁡(1000t-π/2-78.6^o)
B. i = ce^-200t + 0.98cos⁡(1000t+π/2-78.6^o)
C. i = ce^-200t + 0.98cos⁡(1000t+π/2+78.6^o)
D. i = ce^-200t + 0.98cos⁡(1000t-π/2+78.6^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 = ce^-200t + 0.98cos⁡(1000t+π/2-78.6^o).

8. The current flowing through the circuit at t = 0 in the circuit shown below is?

A. 1
B. 2
C. 3
D. 0

Answer: D
Clarification: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

9. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The value of c in the complementary function of ‘i’ is?

A. c = -0.98cos⁡(π/2-78.6^o)
B. c = -0.98cos⁡(π/2+78.6^o)
C. c = 0.98cos⁡(π/2+78.6^o)
D. c = 0.98cos⁡(π/2-78.6^o)

Answer: A
Clarification: At t = 0, the current flowing through the circuit is zero. Placing i = 0 in the current equation we get c = -0.98cos⁡(π/2-78.6^o).

10. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complete solution of ‘i’ is?

A. i = [-0.98 cos⁡(π/2-78.6^o)] exp⁡(-200t)+0.98cos⁡(1000t+π/2-78.6^o)
B. i = [-0.98 cos⁡(π/2-78.6^o)] exp⁡(-200t)-0.98cos⁡(1000t+π/2-78.6^o)
C. i = [0.98 cos⁡(π/2-78.6^o)] exp⁡(-200t)-0.98cos⁡(1000t+π/2-78.6^o)
D. i = [0.98 cos⁡(π/2-78.6^o)] exp⁡(-200t)+0.98cos⁡(1000t+π/2-78.6^o)

Answer: A
Clarification: The complete solution for the current is the sum of the complementary function and the particular integral.
So, i = [-0.98 cos⁡(π/2-78.6^o)] exp⁡(-200t)+0.98cos⁡(1000t+π/2-78.6^o).

Basic Network Theory Questions on “Poles and Zeros of Network Functions”.

1. 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
Clarification: 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.

2. The scale factor is denoted by the letter?
A. G
B. H
C. I
D. J

Answer: B
Clarification: The scale factor is denoted by the letter ‘H’ and its value is equal to the ratio of a_o to b_o.

3. The zeros in the transfer function are denoted by?
A. 3
B. 2
C. 1
D. 0

Answer: D
Clarification: The roots of the equation P (S) = 0 are zeros of the transfer function. The zeros in the transfer function are denoted by ‘o’.

4. The poles in the transfer function are denoted by?
A. x
B. y
C. z
D. w

Answer: A
Clarification: The roots of the equation Q (S) = 0 are poles of the transfer function. The poles in the transfer function are denoted by ‘x’.

5. The network function N (S) becomes _________ when s is equal to anyone of the zeros.
A. 1
B. 2
C. 0
D. ∞

Answer: C
Clarification: 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.

6. The N (S) becomes ________ when s is equal to any of the poles.
A. ∞
B. 0
C. 1
D. 2

Answer: A
Clarification: 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.

7. 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
Clarification: 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.

8. 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
Clarification: 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.

9. 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
Clarification: 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.

10. 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
Clarification: 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.

Network Theory Multiple Choice Questions on “Basic Network Concepts”.

1. Energy per unit charge is ____________
A. Power
B. Voltage
C. Current
D. Capacitance
Answer: B
Clarification: The work or energy per unit charge utilised in the process of separation of charges is known as Voltage or Potential difference. The phenomenon of transfer of charge from one point to another is termed Current. The rate at which the work is done is called Power. Charge per unit voltage is Capacitance.

2. A conductor is said to have resistance of one ohm if a potential difference of one volt across its terminals causes a current of X ampere to flow through it. What will be the value of X?
A. 4
B. 2
C. 3
D. 1
Answer: D
Clarification: Ohm’s law states that the potential difference (voltage) across a conductor is proportional to the current through it. The constant of proportionality is called the Resistance(R).
According to Ohm’s law, V = IR (where V is the potential difference between two points which include a resistance R).
–> I = V/R = 1V/1Ω = 1A.

3. Resistance depends on the temperature of the conductor.
A. True
B. False
Answer: A
Clarification: Resistance is directly proportional to its length, inversely proportional to the area of cross section of the conductor, depends on the nature of the material and on the temperature of the conductor.

4. A 25 Ω resistor has a voltage of 150 sin377 t. Find the corresponding power.
A. 900 sin² 337 t
B. 90 sin² 337 t
C. 900 sin² 377 t
D. 9 sin² 337 t
Answer: C
Clarification: Given R = 25 Ω and v = 150 sin 377 t
i = (frac{v}{R} = frac{150 sin⁡ 377 t}{25}) = 6 sin 377 t
p = vi = (150 sin 377 t)(6 sin 377 t) = 900 sin² 377 t.

5. Unit of inductance is ________
A. Weber
B. Henry
C. Farad
D. Tesla
Answer: B
Clarification: The unit of inductance is Henry. Weber is the unit of magnetic flux. Tesla is the unit of flux density. Farad is the unit of capacitance.

6. Inductance of an inductor is inversely proportional to its ___________
A. Number of turns
B. Area of cross section
C. Absolute permeability
D. Length
Answer: D
Clarification: Inductance of an inductor, L = µN²A/l
From the above equation, Inductance of an inductor is inversely proportional to its length.

7. Energy stored in an inductor is ________
A. LI
B. LI²
C. LI/2
D. LI²/2
Answer: D
Clarification: V = L (frac{di}{dt})
dE = Vidt = L (frac{di}{dt} idt) = Lidt
E = (int_0^I dE = int_0^I Lidt = frac{1}{2} LI^2).

8. An inductor of 3mH has a current i = 5(1 – e^-5000t). Find the corresponding maximum energy stored.
A. 37.5 mJ
B. 375 J
C. 37.5 kJ
D. 3.75 mJ
Answer: A
Clarification: Given L = 3 mH, i = 5(1 – e^-5000t)
V = L (frac{di}{dt} = 3 × 10^{-3} frac{d}{dt}[5(1-e^{-5000t})] = 75 e^{-5000t})
I = i(∞) = 5(1 – e^-∞) = 5 A
E = (frac{1}{2}) LI² = 0.5 × 3 × 10^-3 × 5² = 37.5 mJ.

9. The capacitance of a capacitor does not depend on the absolute permittivity of the medium between the plates.
A. True
B. False
Answer: B
Clarification: C = Ɛ (frac{A}{d})
Where d is the distance between the plates, A is the cross-sectional area of the plates and Ɛ is absolute permittivity of the medium between the plates.
Hence, the capacitance of a capacitor depends on the absolute permittivity of the medium between the plates.

10. Which of the following is not the energy stored in a capacitor?
A. (frac{CV^2}{2})
B. (frac{QV}{2})
C. (frac{Q^2}{2C})
D. (frac{QC}{2})
Answer: D
Clarification: Energy stored in a capacitor, E = (frac{CV^2}{2})
Since C = Q/V
E = (frac{CV^2}{2} = frac{QV}{2} = frac{QC}{2}).

11. A voltage is defined by (
v(t)=begin{cases}
0\
2t\
4e^{-(t-2)}\
end{cases} ) for (
begin{cases}
t < 0 \
0t>2s \
end{cases} ) respectivelyand is applied to the 10 µF capacitor. Which of the following is incorrect?
A. i = 0 for t < 0
B. i = 20µA for 0 < t < 2s
C. i = 40e^t-2µA for t > 2s
D. i = -40e^t-2µA for t > 2s
Answer: C
Clarification: Using i = C (frac{dv}{dt})
For t < 0, i = 0
For 0 < t < 2s, v = 2t; i = 10 × 10^-6 (frac{d}{dt})(2t) = 20 µA
For t > 2s, v = 4e^-(t-2)
i = 10 × 10^-6 (frac{d}{dt})[4e^-(t-2)] = 10 × 10^-6[-4e^-(t-2)] = -40e^-(t-2)µA.

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

1. Consider the circuit shown below. Find the equivalent Thevenin’s voltage between nodes A and B.

A. 8
B. 8.5
C. 9
D. 9.5

Answer: B
Clarification: The thevenin’s voltage is equal to the open circuit voltage across the terminals AB that is across 12Ω resistor. V_th = 10×12/14 = 8.57V.

2. Consider the circuit shown below. Find the thevenin’s resistance between terminals A and B.

A. 1
B. 2
C. 1.7
D. 2.7

Answer: C
Clarification: The resistance into the open circuit terminals is equal to the thevenin’s resistance => R_th = (12×2)/14 = 1.71Ω.

3. Consider the circuit shown below. Find the current flowing through 24Ω resistor.

A. 0.33
B. 0.66
C. 0
D. 0.99

4. Determine the equivalent thevenin’s voltage between terminals A and B in the circuit shown below.

A. 0.333
B. 3.33
C. 33.3
D. 333

Answer: C
Clarification: Let us find the voltage drop across terminals A and B. 50-25=10I+5I => I=1.67A. Voltage drop across 10Ω resistor = 10×1.67=16.7V. So, V_th=V_AB=50-V=50-16.7=33.3V.

5. Find the equivalent thevenin’s resistance between terminals A and B in the circuit shown below.

A. 333
B. 33.3
C. 3.33
D. 0.333

Answer: C
Clarification: To find R_th, two voltage sources are removed and replaced with short circuit. The resistance at terminals AB then is the parallel combination of the 10Ω resistor and 5Ω resistor => R_th=(10×5)/15=3.33Ω.

6. Determine the equivalent thevenin’s voltage between terminals A and B in the circuit shown below.

A. 5
B. 15
C. 25
D. 35

Answer: C
Clarification: Current through 3Ω resistor is 0A. The current through 6Ω resistor = (50-10)/(10+6)=2.5A. The voltage drop across 6Ω resistor = 25×6=15V. So the voltage across terminals A and B = 0+15+10 = 25V.

7. Find the equivalent thevenin’s resistance between terminals A and B in the following circuit.

A. 6
B. 6.25
C. 6.5
D. 6.75

Answer: D
Clarification: To find R_th, two voltage sources are removed and replaced with short circuit => R_th=(10×6)/(10+6)+3=6.75Ω.

8. Determine the equivalent thevenin’s voltage between terminals ‘a’ and ‘b’ in the circuit shown below.

A. 0.7
B. 1.7
C. 2.7
D. 3.7

Answer: C
Clarification: The voltage at terminal a is V_a=(100×6)/16=37.5V, The voltage at terminal b is V_b=(100×8)/23=34.7V. So the voltage across the terminals ab is V_ab=V_a-V_b=37.5-34.7=2.7V.

9. Find the equivalent thevenin’s resistance between terminals A and B in the circuit shown below.

A. 6
B. 7
C. 8
D. 9

Answer: D
Clarification: To find R_th, two voltage sources are removed and replaced with short circuit => R_ab=(6×10)/(6+10)+(8×15)/(8+15)=8.96≅9V.

10. Find the current through 5Ω resistor in the following circuit.

A. 0.1
B. 0.2
C. 0.3
D. 0.4