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

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

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Network Theory MCQs – Network Theory and Complex Circuit Diagram 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) 121+2
b) ∈1 – ∈2
c) 12
d) ∈1 ∈2
Answer: a
Explanation: The combination is equal to two capacitors in series.
So, C = [01(A0.5d)][02(A0.5d)]01A0.5d+01A0.5d
Hence, C is proportional to 121+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 IA and IB is ________
network-theory-questions-answers-advanced-problems-network-theory-1-q6
a) IB = IA + 6
b) IB = IA + 2
c) IB = 1.5IA
d) IB = IA
Answer: c
Explanation: In the circuit of figure (IB), transforming 3A source into 18 V source, all sources are 1.5 times of that in circuit (IA). Hence, IB = 1.5IA.
network-theory-questions-answers-advanced-problems-network-theory-1-q6a


7. For the circuit given below, the current I in the circuit is ________
network-theory-questions-answers-advanced-problems-network-theory-1-q7
a) –j1 A
b) J1 A
c) Zero
d) 20 A
Answer: a
Explanation: XEQ = sL + R×1/sCR+1/sC=sL+R1+sRC
IO = VXEQ
∴ I = XCXC+R IO
1/sC1sC+R×VsL(1+sRC)+R(1+sRC)
11+sRC×VsL(1+sRC)+R(1+sRC)
VsL(1+sRC)+R
Vj×103×20×103(1+j×103×50×106+1)
V20j(1+j50×103)+1
V20j1+1=2020j = -j1 A.


8. An AC source of RMS voltage 20 V with internal impedance ZS = (1+2j) Ω feeds a load of impedance ZL = (7+4j) Ω in the circuit given below. The reactive power is _________
network-theory-questions-answers-advanced-problems-network-theory-1-q8
a) 8 VAR
b) 16 VAR
c) 28 VAR
d) 32 VAR
Answer: b
Explanation: Current I = VZL+ZS=200°8+6j
2082+62=0°arctan(34)
2010 ∠-arc tan⁡(34)
= 2∠-arc tan⁡(34)
Power consumed by load = |I|2ZL
= 4(7+4j)
= 28 + j16
∴ The reactive power = 16 VAR.


9. In the circuit given below, RI = 1 MΩ, RO = 10 Ω, A = 106 and VI = 1μV. Then the output voltage, input impedance and output impedance respectively are _________
network-theory-questions-answers-advanced-problems-network-theory-1-q9
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: VO (output voltage) = AVI = 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 = RO = 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 __________
network-theory-questions-answers-advanced-problems-network-theory-1-q12
a) – 3V
b) Zero
c) 3 V
d) 5 V
Answer: a
Explanation: By applying KVL, I = 1 A
VAB – 2 × 1 + 5 = 0
Or, VAB = -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 RA, RB and RC are 20 Ω, 10 Ω and 10 Ω respectively. The resistances R1, R2 and R3 in Ω are ________
network-theory-questions-answers-advanced-problems-network-theory-1-q14
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 _________
network-theory-questions-answers-advanced-problems-network-theory-2-q1
a) 10 Ω
b) 18 Ω
c) 24 Ω
d) 12 Ω
Answer: d
Explanation: By KCL,
∴ VP401+VP10014+VP2 = 0
Or, 22 VP = 660
∴ VP = 30 V
Potential difference between node x and y = 60 V
∴ -I – 5 + 4030I = 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 RA and RB are _________
network-theory-questions-answers-advanced-problems-network-theory-2-q2
a) RA = 9 Ω and RB = 7 Ω
b) RA = 2.5 Ω and RB = 4.5 Ω
c) RA = 3 Ω and RB = 3 Ω
d) RA = 5 Ω and RB = 1 Ω
Answer: b
Explanation: Given RA + RB = 7 with C open
RB + RC = 12 with A as open
RA + RC = 10 with B open
Then, RA + RB + RC = 292 = 14.5
Hence, RA = 2.5 Ω, RB = 4.5 Ω and RC = 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 ________
network-theory-questions-answers-advanced-problems-network-theory-2-q4
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 REQ = 6 || 4
6×46+4
2410 = 2.4 Ω
VAB = 10 – 6 × (1510) = 1 V.


5. In the circuit given below, the voltage across the 2Ω resistor is ________
network-theory-questions-answers-advanced-problems-network-theory-2-q5
a) 3.41 V
b) -3.41 V
c) 3.8 V
d) -3.8 V
Answer: b
Explanation: Applying KCL to node A, VA1010+VA20+VA7 = 0
Or, VA (0.1 + 0.05 + 0.143) = 1
Or, VA = 3.41 V
The voltage across the 2 Ω resistor due to 10 V source is V2 = VA7×2 = 0.97 V
V2Ω due to 20 V source, VA10+VA20+VA207 = 0
Or, 0.1 VA + 0.05VA + 0.143VA = 2.86
∴ VA = 2.860.293 = 9.76 V
V2Ω = VA207 × 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
V2Ω = 0.97 – 2.92 -1.46 = -3.41 V.


6. In the circuit given below, the value of IX using nodal analysis is _______
network-theory-questions-answers-advanced-problems-network-theory-2-q6
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
∴ 10V11+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
∴ IX = V26=156 = 2.5 A.


7. In the circuit given below, the values of V1 and V2 respectively are _________
network-theory-questions-answers-advanced-problems-network-theory-2-q7
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 = V1V22
Applying KCL at node 1, 5 = V11+V1V22+V1
10 = 2V1 + V1 – V2 + 2V1
Or, 10 = 5V1 – V2
KCL at node 2, V1V22 + 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 _________
network-theory-questions-answers-advanced-problems-network-theory-2-q8
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 _________
network-theory-questions-answers-advanced-problems-network-theory-2-q9
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, VC = VD
Or, V10 = VC = VD = VA × 1015
VR = VA × R20+R
And V10 = VR
∴ VA × 1015 = VA × 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(124et)
= 6 – 3e-t.


12. A resistance R is connected to a voltage source V having internal resistance RI. 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) VR1R2R1R2R1RRR2
d) VSRR2R1R2+R1R+RR2
Answer: d
Explanation: Effective resistance of R and Rm is
Req = 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 ___________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q2
a) 8.18 A
b) 0 A
c) 9 A
d) 10 A
Answer: c
Explanation: Equivalent Resistance, REQ = R = 1 + (2 || R)
Or, REQ = 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 __________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q3
a) 0°
b) +45°
c) -45°
d) -90°
Answer: d
Explanation: Net voltage applied to the circuit is 200∠0° V
I1 = 2000°10.0
= 20∠0° = 20
I2 = 2000°1090°
= 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 1st Ammeter is in series with a resistance, the 2nd 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 = 13252−−−−−−√
Or, A2 = 16925−−−−−−√
Or, A2 = 144−−−√
Or, A2 = 12 A.


5. In the circuit given below, the value of V1 is __________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q5
a) 32.2 V
b) -25.23 V
c) 29.25 V
d) -29.25 V
Answer: b
Explanation: VA30+VA4012+VAVB8 = 0
Or, 29 VA – 15 VB = 400
Also, VBVA8+VB1208 + 6 = 0
Or, VA = 65.23 V, VB = 99.44 V
V1 = 40-65.23 = -25.23 V.


6. For the three coupled coils shown in figure, KVL equation is ____________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q6
a) V = (L1 + L2 + L3didt
b) V = (L1 – L2 – L3 – M13didt
c) V = (L1 + L2 + L3 + 2M12 – 2M23 – 2M13didt
d) V = (L1 + L2 + L3 – 2M12 + 2M23 + 2M13didt
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 – 2M13didt.


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, ZDY = LRLC
But given that RL = 0
So, ZDY = L0XC = ∞
Therefore current through circuit, I = V = 0
∴ VD = 0.


8. In the circuit given below, the value of resistance R is _________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q8
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, VP – VQ = 14I – (15-I).1
= 70 – 10 = 60 V
∴ R = 6010I=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 ____________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q9
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 = jX4X0.75I13.922.56j
= (0.23 – 0.35j) cos 4t


10. In the circuit given below, the voltage VAB is _________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q10
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
VOC = 50 – (10×4) =10 V
And REQ = 10×510+5=103 Ω
Current I1 = 1010/3+5=65
Hence, VAB = 65×5 = 6 V.


11. In the circuit given below, the magnitudes of VL and VC are twice that of VK. Calculate the inductance of the coil, given that f = 50.50 Hz.
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q11
a) 6.41 mH
b) 5.30 mH
c) 3.18 mH
d) 2.31 mH
Answer: c
Explanation: VL = VC = 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 _________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12
a) 1 A
b) 5 A
c) 6 A
d) 8 A
Answer: b
Explanation: At steady state, the circuit becomes,
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12a
∴ 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 ________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q13
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, qqodq100Cq=1RCt0dt
100C – q = (100C – qo)e-t/RC
I = dqdt=(100Cqo)RCe1/1
∴ 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 ___________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q14
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’.
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q14a
Vxx’ = VN = 48+82418+124 = 5V
∴ RN = 8 || (16 + 8)
8×248+24 = 6 Ω
∴ IN = 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 _________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12
a) 1 A
b) 3 A
c) 2 A
d) 4 A
Answer: d
Explanation: At steady state, the circuit becomes,
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12a
∴ 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 ___________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q1
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,
VC = VR
Now, VR = 0.67 X 10 = 6.7 V
So, VC = 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(π2π3)π
2π+ππ = 3 Volt.


5. For a series RLC circuit excited by a unit step voltage, Vc is __________
a) 1 – e-t/RC
b) e-t/RC
c) et/RC
d) 1
Answer: a
Explanation: At t = 0, Vc = 0 and at t = ∞, Vc = 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 ____________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q6
a) VTH = 5 V, RTH = 0.75 Ω
b) VTH = 0.5 V, RTH = 0.75 Ω
c) VTH = 2.5 V, RTH = 1 Ω
d) VTH = 5 V, RTH = 1 Ω
Answer: b
Explanation: VTH = 1X2040 X 1 = 0.5 V
Also, RTH = 1X3040 = 0.75 Ω.


7. In the circuit given below, the value of R is ____________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q7
a) 12 Ω
b) 6 Ω
c) 3 Ω
d) 1.5 Ω
Answer: b
Explanation: The resistance of parallel combination is given by,
Req = 403 – 10 = 3.33 Ω
Or, 13.33=112+115+115
Or, R = 6 Ω.


8. A circuit consists of an excitation voltage VS, 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, Voc = 5V and the circuit is open
When R = 0, I = 2.5A
Then, Isc = 2.5 and the circuit is short circuited.
So, Req = 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 _________________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q11
a) 1.6 F
b) 3.1 F
c) 0.5 F
d) 4.6 F
Answer: b
Explanation: CCB = (C2C3C2+C3) + C5 = 7.5 F
Now, CAB = (C1CCBC1+CCB) + C6 = 8 F
CXY = CAB×C4CAB+C4 = 3.1 F.


12. In the circuit given below, the equivalent capacitance is ______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q12
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.
∴ CEQ = 0.43 + 5 = 5.43 μF.


13. In the circuit given below, the voltage across AB is _______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q13
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
VAB = (50) (2.5) + 100 + (5) (20)
= 125 + 100 + 100
= 325 V.


14. The number of non-planar graph of independent loop equations is ______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q14
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 _______________
network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q15
a) 4.1 Hz
b) 41 Hz
c) 0.41 Hz
d) 0.041 Hz
Answer: d
Explanation: IEQ = L1 + L2 + 2M
LEQ = 10 + 20 + 2 × 120 = 30.1 H
∴ FO = 12πLC
12π30.1×0.5
= 0.041 Hz.

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

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