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Network Theory and Complex Circuit Diagram MCQs ( Network Theory ) MCQs – Network Theory MCQs

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Advanced Problems on Network Theory – 1

1. Branch current and loop current relation is expressed in matrix form shown below, where Ij represents branch current and Ik represents loop current.
[I₁; I₂; I₃; I₄; I₅; I₆; I₇; I₈] = [0 0 1 0; -1 -1 -1 0; 0 1 0 0; 1 0 0 0; 0 0 -1 -1; 1 1 0 -1; 1 0 0 0; 0 0 0 1] [I₁; I₂; I₃; I₄] The rank of the incidence matrix is?
a) 4
b) 5
c) 6
d) 8
Answer: a
Explanation: Number of branches b = 8
Number of links l = 4
Number of twigs t = b – l = 4
Rank of matrix = n – 1 = t = 4.

2. A capacitor, used for power factor correction in a single phase circuit decreases which of the following?
a) Power factor
b) Line current
c) Both Line current and Power factor
d) Neither Line current nor Power factor
Answer: b
Explanation: We know that a capacitor is used to increase the Power factor. However, with decrease in line current the power factor is increased. Hence line current decreases.

3. D is the distance between the plates of a parallel plate capacitor. The dielectric constants are ∈₁ and ∈₂ respectively. The total capacitance is proportional to ____________
a) ∈1∈2∈1+∈2
b) ∈₁ – ∈₂
c) ∈1∈2
d) ∈₁ ∈₂
Answer: a
Explanation: The combination is equal to two capacitors in series.
So, C = [∈0∈1(A0.5d)][∈0∈2(A0.5d)]∈0∈1A0.5d+∈0∈1A0.5d
Hence, C is proportional to ∈1∈2∈1+∈2.

4. A two branch circuit has a coil of resistance R₁, inductance L₁ in one branch and capacitance C₂ in the second branch. If R is increased, the dynamic resistance is going to ___________
a) Increase
b) Decrease
c) Remains constant
d) May increase or decrease
Answer: b
Explanation: We know that,
Dynamic resistance = L1R1C2
So, if R₁ is increased, keeping Inductance and Capacitance same, so The Dynamic resistance will decrease, as the denomination is increasing.

5. A 1 μF capacitor is connected to 12 V batteries. The energy stored in the capacitor is _____________
a) 12 x 10^-6 J
b) 24 x 10^-6 J
c) 60 x 10^-6 J
d) 72 x 10^-6 J
Answer: d
Explanation: We know that,
Energy, E = 0.5 CV²
= 0.5 X 1 X 10^-6 X 144
= 72 x 10^-6 J.

6. For the two circuits shown below, the relation between I_A and I_B is ________

a) I_B = I_A + 6
b) I_B = I_A + 2
c) I_B = 1.5I_A
d) I_B = I_A
Answer: c
Explanation: In the circuit of figure (I_B), transforming 3A source into 18 V source, all sources are 1.5 times of that in circuit (I_A). Hence, I_B = 1.5I_A.

7. For the circuit given below, the current I in the circuit is ________

a) –j1 A
b) J1 A
c) Zero
d) 20 A
Answer: a
Explanation: X_EQ = sL + R×1/sCR+1/sC=sL+R1+sRC
I_O = VXEQ
∴ I = XCXC+R I_O
= 1/sC1sC+R×VsL(1+sRC)+R(1+sRC)
= 11+sRC×VsL(1+sRC)+R(1+sRC)
= VsL(1+sRC)+R
= Vj×103×20×10−3(1+j×103×50×10−6+1)
= V20j(1+j50×10−3)+1
= V20j−1+1=2020j = -j1 A.

8. An AC source of RMS voltage 20 V with internal impedance Z_S = (1+2j) Ω feeds a load of impedance Z_L = (7+4j) Ω in the circuit given below. The reactive power is _________

a) 8 VAR
b) 16 VAR
c) 28 VAR
d) 32 VAR
Answer: b
Explanation: Current I = VZL+ZS=20∠0°8+6j
= 2082+62√=∠0°∠arctan(34)
= 2010 ∠-arc tan⁡(34)
= 2∠-arc tan⁡(34)
Power consumed by load = |I|²Z_L
= 4(7+4j)
= 28 + j16
∴ The reactive power = 16 VAR.

9. In the circuit given below, R_I = 1 MΩ, R_O = 10 Ω, A = 10⁶ and V_I = 1μV. Then the output voltage, input impedance and output impedance respectively are _________

a) 1 V, ∞ and 10 Ω
b) 1 V, 0 and 10 Ω
c) 1V, 0 and ∞
d) 10 V, ∞ and 10 Ω
Answer: a
Explanation: V_O (output voltage) = AV_I = 10⁶ × 10^-6 = 1 V
V₁ = Z₁₁ I₁ + Z₁₂ I₂
V₂ = Z₂₁ I₁ + Z₂₂ I₂
Here, I₁ = 0
Z₁₁ = V1I1=VO0 = ∞
Z₂₂ = V2I2=AVII2
Or, Z₂₂ = 1I2 = R_O = 10 Ω.

10. If operator ‘a’ = 1 ∠120°. Then (1 – a) is equal to ____________
a) 3–√
b) 3–√∠-30°
c) 3–√∠30°
d) 3–√∠60°
Answer: b
Explanation: Given that, ‘a’ = 1 ∠120°
So, 1 – a = 1 – 1∠120°
= 1 + 0.5 – j 0.866
= 1.5 – j 0.866
= 3∠-30°.

11. For making a capacitor, the dielectric should have __________
a) High relative permittivity
b) Low relative permittivity
c) Relative permittivity = 1
d) Relative permittivity neither too high nor too low
Answer: a
Explanation: Relative permittivity is for ideal dielectric which is air. Achieving such a precise dielectric is very difficult.
Low relative permittivity will lead to low value of capacitance.
High relative permittivity will lead to a higher value of capacitance.

12. In the circuit shown below, the voltage V will be __________

a) – 3V
b) Zero
c) 3 V
d) 5 V
Answer: a
Explanation: By applying KVL, I = 1 A
V_AB – 2 × 1 + 5 = 0
Or, V_AB = -3 V.

13. If A = 3 + j1, then A⁴ is equal to __________
a) 3.16 ∠18.4°
b) 100 ∠73.72°
c) 100 ∠18.4°
d) 3.16 ∠73.22°
Answer: b
Explanation: Given A = 3 + j1
So, 3 + j1 = 10∠18.43°
Or, 3 + j1 = (10)⁴ ∠4 X 18.43°
= 100∠73.72°.

14. In the figures given below, Value of R_A, R_B and R_C are 20 Ω, 10 Ω and 10 Ω respectively. The resistances R₁, R₂ and R₃ in Ω are ________

a) 2.5, 5 and 5
b) 5, 2.5 and 5
c) 5, 5 and 2.5
d) 2.5, 5 and 2.5
Answer: a
Explanation: R₁ = RBRCRA+RB+RC=10040 = 2.5 Ω
R₂ = RARCRA+RB+RC=20040 = 5 Ω
R₃ = RBRARA+RB+RC=20040 = 5 Ω.

15. The resistance of a thermistor decreases with increases in __________
a) temperature
b) circuit
c) light control
d) sensors
Answer: b
Explanation: The resistance of a thermistor decreases with increases in temperature. Hence, it is used to monitor hot spot temperature of electric machines.

Advanced Problems on Network Theory – 2

1. In the circuit given below, the value of R is _________

a) 10 Ω
b) 18 Ω
c) 24 Ω
d) 12 Ω
Answer: d
Explanation: By KCL,
∴ VP−401+VP−10014+VP2 = 0
Or, 22 V_P = 660
∴ V_P = 30 V
Potential difference between node x and y = 60 V
∴ -I – 5 + 40−30I = 0
Or, I = 5 A
∴ R = 605= 12 Ω.

2. In the circuit given below, the resistance between terminals A and B is 7Ω, between terminals B and C is 12Ω and between terminals C and A is 10Ω. The remaining one terminal in each case is assumed to be open. Then the value of R_A and R_B are _________

a) R_A = 9 Ω and R_B = 7 Ω
b) R_A = 2.5 Ω and R_B = 4.5 Ω
c) R_A = 3 Ω and R_B = 3 Ω
d) R_A = 5 Ω and R_B = 1 Ω
Answer: b
Explanation: Given R_A + R_B = 7 with C open
R_B + R_C = 12 with A as open
R_A + R_C = 10 with B open
Then, R_A + R_B + R_C = 292 = 14.5
Hence, R_A = 2.5 Ω, R_B = 4.5 Ω and R_C = 7.5 Ω.

3. Currents I₁, I₂ and I₃ meet at a junction in a circuit. All currents are marked as entering the node. If I₁ = -6sin(ωt) mA and I₂ = 8 cos(ωt) mA, the I₃ is ________
a) 10 cos(ωt + 36.87) mA
b) 14 cos(ωt + 36.87) mA
c) -14 sin(ωt + 36.87) mA
d) -10 cos(ωt + 36.87) mA
Answer: d
Explanation: Applying KCL, we get, I₁ + I₂ + I₃ = 0
∴ -6 sin(ωt) + 8 cos(ωt) + I₃ = 0
∴ I₃ = 6 sin (ωt) – 8 cos (ωt)
= 10[sin (ωt).sin (36.86) – cos (ωt) cos (36.86)]
=-10[cos (ωt) cos (36.86) – sin (ωt) sin (36.86)]
= -10 cos (ωt + 36.86)
[As, cos (A+B) = cosA.cosB – sinA.sinB].

4. Viewed from the terminals A and B the circuit given below can be reduced to an equivalent circuit with a single voltage source in series with a resistor with ________

a) 5 V source in series with 10 Ω resistor
b) 1 V source in series with 2.4 Ω resistor
c) 15 V source in series with 2.4 Ω resistor
d) 1 V source in series with 10 Ω resistor
Answer: b
Explanation: Applying Thevenin’s Theorem R_EQ = 6 || 4
= 6×46+4
= 2410 = 2.4 Ω
V_AB = 10 – 6 × (1510) = 1 V.

5. In the circuit given below, the voltage across the 2Ω resistor is ________

a) 3.41 V
b) -3.41 V
c) 3.8 V
d) -3.8 V
Answer: b
Explanation: Applying KCL to node A, VA−1010+VA20+VA7 = 0
Or, V_A (0.1 + 0.05 + 0.143) = 1
Or, V_A = 3.41 V
The voltage across the 2 Ω resistor due to 10 V source is V₂ = VA7×2 = 0.97 V
V_2Ω due to 20 V source, VA10+VA20+VA−207 = 0
Or, 0.1 V_A + 0.05V_A + 0.143V_A = 2.86
∴ V_A = 2.860.293 = 9.76 V
V_2Ω = VA−207 × 2 = -2.92 V
The current in 2 Ω resistor = 2 × 55+8.67
= 1013.67 = 0.73 A
The voltage across the 2 Ω resistor = 0.73 × 2 = 1.46 V
V_2Ω = 0.97 – 2.92 -1.46 = -3.41 V.

6. In the circuit given below, the value of I_X using nodal analysis is _______

a) -2.5 A
b) 2.5 A
c) 5 A
d) -5 A
Answer: b
Explanation: Applying KCL, we get, I₁ + 5 = I₂ + I₃
∴ 10–V11+5=V12+V26
∴ 30 – 3V₁ + 15 = 3V₁ + V₂
∴ 6 V₁ + V₂ = 45
From voltage source, V₂ – V₁ = 10
Now, 7 V₁ = 35, V₁ = 5 V
And V₂ = 15 V
∴ I_X = V26=156 = 2.5 A.

7. In the circuit given below, the values of V₁ and V₂ respectively are _________

a) 0 and 5V
b) 5 and 0 V
c) 5 and 5 V
d) 2.5 and 2.5 V
Answer: d
Explanation: I₁ = V1−V22
Applying KCL at node 1, 5 = V11+V1−V22+V1
10 = 2V₁ + V₁ – V₂ + 2V₁
Or, 10 = 5V₁ – V₂
KCL at node 2, V1−V22 + V₁ + 2I₁ = V₂
∴ 1.5 V₁ – V₂ = 0
∴ V₁ = V₂ = 2.5 V.

8. In the circuit given below, the voltage across the 18 Ω resistor is 90 V. The voltage across the combined circuit is _________

a) 125 V
b) 16 V
c) 24 V
d) 40 V
Answer: a
Explanation: Current through the 18 Ω resistance = 9018 = 5 A
Equivalent resistance of 3 Ω and 7 Ω banks = 3×63+6 = 2 Ω
Since, this 2 Ω resistance is in series with 18 Ω resistance, therefore total resistance = 18 + 2 = 20 Ω
This 20 Ω resistance is in parallel with 5 Ω resistance = 5×205+20 = 4 Ω
Hence, total resistance of the circuit = 1 + 4 = 5 Ω
Current through this branch = 5 A
∴ Voltage across dc= 5 × 20 = 100 V
Hence current through 5 Ω resistance = 1005 = 20 A
∴ Total current = 20 + 5 = 25 A
Since, total resistance of the circuit is 5 Ω therefore, voltage E = 25 × 5 = 125 V.

9. In the circuit given below, the value of R in the circuit, when the current is zero in the branch CD is _________

a) 10 Ω
b) 20 Ω
c) 30 Ω
d) 40 Ω
Answer: d
Explanation: The current in the branch CD is zero if the potential difference in the branch CD is zero.
That is, V_C = V_D
Or, V₁₀ = V_C = V_D = V_A × 1015
V_R = V_A × R20+R
And V₁₀ = V_R
∴ V_A × 1015 = V_A × RR+20 ∴ R = 40 Ω.

Elements of Realizability and Synthesis of One-Port Networks

10. Two capacitors of 0.5 μF and 1.5 μF capacitance are connected in parallel across a 110 V dc battery. The charges across the two capacitors after getting charged is ___________
a) 55 μC each
b) 275 μC each
c) 55 μC and 275 μC respectively
d) 275 μC and 55 μC respectively
Answer: c
Explanation: Q₁ = 0.5 x 10^-6 x 110
= 55 μC.
Also, Q₂ = 2.5 x 10^-6 x 110
= 275 μC.

11. Consider a circuit having resistances 2 Ω and 2 Ω in series with an inductor of inductance 2 H. The circuit is excited by a voltage of 12 V. A switch S is placed across the first resistance. Battery has remained switched on for a long time. The current i(t) after switch is closed at t=0 is _____________
a) 6
b) 6 – 3e^-t
c) 6 + 3e^-t
d) 3 – 6e^-t
Answer: b
Explanation: From the figure, we can infer that,
I(t) = 1212(1–24e−t)
= 6 – 3e^-t.

12. A resistance R is connected to a voltage source V having internal resistance R_I. A voltmeter of resistance R₂ is used to measure the voltage across R. The reading of the voltmeter is _____________
a) VSR1R2R2R1+R1R+RR2
b) VsRR+Rs
c) VR1R2R1R2–R1R–RR2
d) VSRR2R1R2+R1R+RR2
Answer: d
Explanation: Effective resistance of R and R_m is
R_eq = RRmR+Rm
Therefore the reading is VR1+RR+R2[RR2R+R2]
= VSRR2R1R2+R1R+RR2.

13. When a lead acid battery is being charged, the specific gravity of the electrolyte will ___________
a) Decrease
b) Increase
c) Either Increase or Decrease
d) Neither Increase nor Decrease
Answer: b
Explanation: We know that the specific gravity of electrolyte is highest when battery is fully charged and is lowest when discharged. So, the specific gravity of the electrolyte will increase when a lead acid battery is being charged.

14. A series RLC circuit has a resonant frequency of 550 Hz. The maximum voltage across C is likely to occur at a frequency of ___________
a) 1000 Hz
b) 2000 Hz
c) 1025 Hz
d) 500 Hz
Answer: d
Explanation: We know that, maximum voltage across capacitance occurs at a frequency slightly less than resonant frequency.
Here, given that resonant frequency = 550 Hz.
So, out of the given options 500 is the lowest nearest integer to 550.
Hence, 975 Hz.

Advanced Problems Involving Complex Circuit Diagram – 1

1. The current wave shape is in the form of a square terminating at t = 4sec. The voltage across the element increases linearly till t = 4 sec and then becomes constant. The element is ____________
a) Resistance
b) Inductance
c) Capacitance
d) Semi-conductor
Answer: c
Explanation: We know that, when a current pulse is applied to a capacitor, the voltage will have a waveform which rises linearly and then becomes constant towards the end of pulse. Hence, the element is a capacitor.

2. An infinite ladder is constructed with 1 Ω and 2 Ω resistor shown below. The current I flowing through the circuit is ___________

a) 8.18 A
b) 0 A
c) 9 A
d) 10 A
Answer: c
Explanation: Equivalent Resistance, R_EQ = R = 1 + (2 || R)
Or, R_EQ = 2
So, I = 182 A = 9 A.

3. In the circuit given below, the phase angle of the current I with respect to the voltage V₁ is __________

a) 0°
b) +45°
c) -45°
d) -90°
Answer: d
Explanation: Net voltage applied to the circuit is 200∠0° V
I₁ = 200∠0°10.0
= 20∠0° = 20
I₂ = 200∠0°10∠90°
= 20∠-90° = -j20
I = I₁ + I₂ = 20(1-j) = 202–√∠45°
Voltage V₁ = 100(1+j)
= 1002–√∠45°
∴ Required phase angle = -45° – 45° = -90°.

4. Consider a circuit having 3 identical Ammeters A₁, A₂, A₃ parallel to one another. The 1^st Ammeter is in series with a resistance, the 2^nd Ammeter is in series with a capacitor and the circuit is excited by a voltage V. If A₁ and A₃ read 5 and 13 A respectively, reading of A₂ will be?
a) 8 A
b) 13 A
c) 18 A
d) 12 A
Answer: d
Explanation: We can infer from the circuit,
A₂ = 132–52−−−−−−√
Or, A₂ = 169–25−−−−−−√
Or, A₂ = 144−−−√
Or, A₂ = 12 A.

5. In the circuit given below, the value of V₁ is __________

a) 32.2 V
b) -25.23 V
c) 29.25 V
d) -29.25 V
Answer: b
Explanation: VA30+VA−4012+VA−VB8 = 0
Or, 29 V_A – 15 V_B = 400
Also, VB−VA8+VB−1208 + 6 = 0
Or, V_A = 65.23 V, V_B = 99.44 V
V₁ = 40-65.23 = -25.23 V.

6. For the three coupled coils shown in figure, KVL equation is ____________

a) V = (L₁ + L₂ + L₃) didt
b) V = (L₁ – L₂ – L₃ – M₁₃) didt
c) V = (L₁ + L₂ + L₃ + 2M₁₂ – 2M₂₃ – 2M₁₃) didt
d) V = (L₁ + L₂ + L₃ – 2M₁₂ + 2M₂₃ + 2M₁₃) didt
Answer: c
Explanation: M₁₂ is positive while M₂₃ and M₁₃ are negative because of dots shown in figure.
So, the KVL equation is given by,
V = (L₁ + L₂ + L₃ + 2M₁₂ – 2M₂₃ – 2M₁₃) didt.

7. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________
a) 0
b) V2
c) V3
d) V
Answer: a
Explanation: Dynamic resistance of the tank circuit, Z_DY = LRLC
But given that R_L = 0
So, Z_DY = L0XC = ∞
Therefore current through circuit, I = V∞ = 0
∴ V_D = 0.

8. In the circuit given below, the value of resistance R is _________

a) 10 Ω
b) 18 Ω
c) 24 Ω
d) 12 Ω
Answer: d
Explanation: Using KVL in loop 1, we get, 100 – 14I – 30 = 0
Or, I = 7014 = 5 A
Then, V_P – V_Q = 14I – (15-I).1
= 70 – 10 = 60 V
∴ R = 6010−I=605 = 12 Ω.

9. The current flowing through the resistance R in the circuit in the figure has the form 2 cos 4t, where 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: Inductor is not given, hence ignoring the inductance. Let I₁ and I₂ are currents in the loop then,
I₁ = 2cos4t3
= 0.66 cos 4t
Again, I₂ = −jX4X0.75I13.92−2.56j
= (0.23 – 0.35j) cos 4t

10. In the circuit given below, the voltage V_AB is _________

a) 6 V
b) 25 V
c) 10 V
d) 40 V
Answer: a
Explanation: For finding the Thevenin Equivalent circuit across A-B we remove the 5 Ω resistor.
Then, I = 10+5015 = 4 A
V_OC = 50 – (10×4) =10 V
And R_EQ = 10×510+5=103 Ω
Current I₁ = 1010/3+5=65
Hence, V_AB = 65×5 = 6 V.

11. In the circuit given below, the magnitudes of V_L and V_C are twice that of V_K. Calculate the inductance of the coil, given that f = 50.50 Hz.

a) 6.41 mH
b) 5.30 mH
c) 3.18 mH
d) 2.31 mH
Answer: c
Explanation: V_L = V_C = 2 VR
∴ Q = VLVR = 2
But we know, Q = ωLR=1ωCR
∴ 2 = 2πf×L5
Or, L = 3.18 mH.

12. 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.

13. 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.

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

a) 6 Ω and 1.333 A
b) 6 Ω and 0.833 A
c) 32 Ω and 0.156 A
d) 32 Ω and 0.25 A
Answer: b
Explanation: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

V_xx’ = V_N = 48+82418+124 = 5V
∴ R_N = 8 || (16 + 8)
= 8×248+24 = 6 Ω
∴ I_N = VNRN=56 = 0.833 A.

15. 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.

Advanced Problems Involving Complex Circuit Diagram – 2

1. In the circuit given below, the KVL for first loop is ___________

a) V (t) = R₁i₁ + L₁ di1dt + M di2dt
b) V (t) = R₁i₁ – L₁ di1dt – M di2dt
c) V (t) = R₁i₁ + L₁ di1dt – M di2dt
d) V (t) = R₁i₁ – L₁ di1dt + M di2dt
Answer: c
Explanation: We know that, in general, the KVL is of the form V (t) = R₁i₁ + L₁ di1dt + M di2dt
But here, M term is negative because i₁, is entering the dotted terminal and i₂, is leaving the dotted terminal.
So, V (t) = R₁i₁ + L₁ di1dt – M di2dt.

2. In a parallel RL circuit, 12 A current enters into the resistor R and 16 A current enters into the Inductor L. The total current I the sinusoidal source is ___________
a) 25 A
b) 4 A
c) 20 A
d) Cannot be determined
Answer: c
Explanation: Currents in resistance and inductance are out of phase by 90°.
Hence, I = I21+I22
Or, I = [12² + 16²]^0.5
Or, I = 144+256−−−−−−−−√=400−−−√
= 20 A.

3. Consider a series RLC circuit having resistance = 1Ω, capacitance = 1 F, considering that the capacitor gets charged to 10 V. At t = 0 the switch is closed so that i = e^-2t. When i = 0.37 A, the voltage across capacitor is _____________
a) 1 V
b) 6.7 V
c) 0.37 V
d) 0.185 V
Answer: b
Explanation: We know that, during discharge of capacitor,
V_C = V_R
Now, V_R = 0.67 X 10 = 6.7 V
So, V_C = 6.7 V.

4. A waveform is of the form of a trapezium, which increases linearly with the linear slope till θ = π3, constant till θ = π2 and again linearly decreases to 0 till θ = π. The average value of this waveform is ______________
a) 2 V
b) 0 V
c) 4 V
d) 3 V
Answer: d
Explanation: The average value of the waveform = 2XAreaof1sttriangle+Areaof2ndtriangleπ
= 2Xπ3X12X6+6(π2–π3)π
= 2π+ππ = 3 Volt.

5. For a series RLC circuit excited by a unit step voltage, V_c is __________
a) 1 – e^-t/RC
b) e^-t/RC
c) e^t/RC
d) 1
Answer: a
Explanation: At t = 0, V_c = 0 and at t = ∞, V_c = 1.
This condition can be satisfied only by (1 – e^-t/RC).

6. In the circuit given below, a dc circuit fed by a current source. With respect to terminals AB, Thevenin’s voltage and Thevenin’s resistance are ____________

a) V_TH = 5 V, R_TH = 0.75 Ω
b) V_TH = 0.5 V, R_TH = 0.75 Ω
c) V_TH = 2.5 V, R_TH = 1 Ω
d) V_TH = 5 V, R_TH = 1 Ω
Answer: b
Explanation: V_TH = 1X2040 X 1 = 0.5 V
Also, R_TH = 1X3040 = 0.75 Ω.

7. In the circuit given below, the value of R is ____________

a) 12 Ω
b) 6 Ω
c) 3 Ω
d) 1.5 Ω
Answer: b
Explanation: The resistance of parallel combination is given by,
R_eq = 403 – 10 = 3.33 Ω
Or, 13.33=112+115+115
Or, R = 6 Ω.

8. A circuit consists of an excitation voltage V_S, a resistor network and a resistor R. For different values of R, the values of V and I are as given, R = ∞, V = 5 volt; R = 0, I = 2.5 A; when R = 3 Ω, the value of V is __________
a) 1 V
b) 2 V
c) 3 V
d) 5 V
Answer: c
Explanation: When R = ∞, V = 5v,
Then, V_oc = 5V and the circuit is open
When R = 0, I = 2.5A
Then, I_sc = 2.5 and the circuit is short circuited.
So, R_eq = VOCISC
= 52.5 = 2 Ω
Hence the voltage across 3 Ω is 3 volt.

9. Three inductors each 30 mH are connected in delta. The value of inductance or each arm of equivalent star is _____________
a) 10 mH
b) 15 mH
c) 30 mH
d) 90 mH
Answer: a
Explanation: We know that if an inductor L is connected in delta, then the equivalent star of each arm = LXLL+L+L
Given that, L = 30 mH
= 30X3030+30+30
= 90090 = 10 mH.

10. In a series RLC circuit having resistance R = 2 Ω, and excited by voltage V = 1 V, the average power is 250 mW. The phase angle between voltage and current is ___________
a) 75°
b) 60°
c) 15°
d) 45°
Answer: d
Explanation: VI cos θ = 0.25 or I cos θ = 0.25
Or, Z cosθ = 2
Or, VI cos⁡θ = 2
Or, cos θ = 12√
So, from the above equations, cos θ = 0.707 and θ = 45°.

11. In the circuit given below, the equivalent capacitance is _________________

a) 1.6 F
b) 3.1 F
c) 0.5 F
d) 4.6 F
Answer: b
Explanation: C_CB = (C2C3C2+C3) + C₅ = 7.5 F
Now, C_AB = (C1CCBC1+CCB) + C₆ = 8 F
C_XY = CAB×C4CAB+C4 = 3.1 F.

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

a) 5.43 μF
b) 4.23 μF
c) 3.65 μF
d) 5.50 μF
Answer: a
Explanation: The 2 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 0.5 μF.
Hence, C₁ = 0.5(2+1)0.5+2+1
= 1.53.5 = 0.43
Now, C₁ is in parallel with the 5 μF capacitor.
∴ C_EQ = 0.43 + 5 = 5.43 μF.

13. In the circuit given below, the voltage across AB is _______________

a) 250 V
b) 150 V
c) 325 V
d) 100 V
Answer: c
Explanation: Loop current I₁ = 5020 = 2.5 A
I₂ = 10020 = 5 A
V_AB = (50) (2.5) + 100 + (5) (20)
= 125 + 100 + 100
= 325 V.

14. The number of non-planar graph of independent loop equations is ______________

a) 8
b) 12
c) 3
d) 5
Answer: c
Explanation: The total number of independent loop equations are given by L = B – N + 1 where,
L = number of loop equations
B = number of branches = 10
N = number of nodes = 8
∴ L = 10 – 8 + 1 = 3.

15. In the circuit given below, M = 20. The resonant frequency is _______________

a) 4.1 Hz
b) 41 Hz
c) 0.41 Hz
d) 0.041 Hz
Answer: d
Explanation: I_EQ = L₁ + L₂ + 2M
L_EQ = 10 + 20 + 2 × 120 = 30.1 H
∴ F_O = 12πLC√
= 12π30.1×0.5√
= 0.041 Hz.