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
View Answer
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
View Answer
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) ∈1 ∈2
View Answer
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
View Answer
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
View Answer
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 ________
a) IB = IA + 6
b) IB = IA + 2
c) IB = 1.5IA
d) IB = IA
View Answer
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
View Answer
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
View Answer
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
View Answer
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°
View Answer
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
View Answer
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
View Answer
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°
View Answer
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
View Answer
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
View Answer
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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