# Network Theory Questions and Answers – Advanced Problems on Network Theory – 1

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “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

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

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) $$\frac{∈_1 ∈_2}{∈_1+∈_2}$$
b) ∈1 – ∈2
c) $$\frac{∈_1}{∈_2}$$
d) ∈12

Explanation: The combination is equal to two capacitors in series.
So, C = $$\frac{\Big[∈_0 ∈_1 \left(\frac{A}{0.5d}\right)\Big]\Big[∈_0 ∈_2 \left(\frac{A}{0.5d}\right)\Big]}{∈_0 ∈_1 \frac{A}{0.5d} + ∈_0 ∈_1 \frac{A}{0.5d}}$$
Hence, C is proportional to $$\frac{∈_1 ∈_2}{∈_1+∈_2}$$.

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

Explanation: We know that,
Dynamic resistance = $$\frac{L_1}{R_1 C_2}$$
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

Explanation: We know that,
Energy, E = 0.5 CV2
= 0.5 X 1 X 10-6 X 144
= 72 x 10-6 J.
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6. For the two circuits shown below, the relation between IA and IB is ________

a) IB = IA + 6
b) IB = IA + 2
c) IB = 1.5IA
d) IB = IA

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.

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

a) –j1 A
b) J1 A
c) Zero
d) 20 A

Explanation: XEQ = sL + $$\frac{R×1/sC}{R+1/sC} = sL + \frac{R}{1+sRC}$$
IO = $$\frac{V}{X_{EQ}}$$
∴ I = $$\frac{X_C}{X_C+R}$$ IO
= $$\frac{1/sC}{\frac{1}{sC}+R} × \frac{V}{\frac{sL(1+sRC)+R}{(1+sRC)}}$$
= $$\frac{1}{1+sRC} × \frac{V}{\frac{sL(1+sRC)+R}{(1+sRC)}}$$
= $$\frac{V}{sL(1+sRC)+R}$$
= $$\frac{V}{j×10^3×20×10^{-3} (1+j×10^3×50×10^{-6}+1)}$$
= $$\frac{V}{20j(1+j50×10^{-3})+1}$$
= $$\frac{V}{20j-1+1} = \frac{20}{20j}$$ = -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 _________

a) 8 VAR
b) 16 VAR
c) 28 VAR
d) 32 VAR

Explanation: Current I = $$\frac{V}{Z_L+Z_S} = \frac{20∠0°}{8+6j}$$
= $$\frac{20}{\sqrt{8^2+6^2}} = \frac{∠0°}{∠arc tan⁡(\frac{3}{4})}$$
= $$\frac{20}{10}$$ ∠-arc tan⁡($$\frac{3}{4}$$)
= 2∠-arc tan⁡($$\frac{3}{4}$$)
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 _________

a) 1 V, ∞ and 10 Ω
b) 1 V, 0 and 10 Ω
c) 1V, 0 and ∞
d) 10 V, ∞ and 10 Ω

Explanation: VO (output voltage) = AVI = 106 × 10-6 = 1 V
V1 = Z11 I1 + Z12 I2
V2 = Z21 I1 + Z22 I2
Here, I1 = 0
Z11 = $$\frac{V_1}{I_1} = \frac{V_O}{0}$$ = ∞
Z22 = $$\frac{V_2}{I_2} = \frac{AV_I}{I_2}$$
Or, Z22 = $$\frac{1}{I_2}$$ = RO = 10 Ω.

10. If operator ‘a’ = 1 ∠120°. Then (1 – a) is equal to ____________
a) $$\sqrt{3}$$
b) $$\sqrt{3}$$∠-30°
c) $$\sqrt{3}$$∠30°
d) $$\sqrt{3}$$∠60°

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

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

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°

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 ________

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

Explanation: R1 = $$\frac{R_B R_C}{R_A+R_B+R_C} = \frac{100}{40}$$ = 2.5 Ω
R2 = $$\frac{R_A R_C}{R_A+R_B+R_C} = \frac{200}{40}$$ = 5 Ω
R3 = $$\frac{R_B R_A}{R_A+R_B+R_C} = \frac{200}{40}$$ = 5 Ω.

15. The resistance of a thermistor decreases with increases in __________
a) temperature
b) circuit
c) light control
d) sensors

Explanation: The resistance of a thermistor decreases with increases in temperature. Hence, it is used to monitor hot spot temperature of electric machines.

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