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

View Answer

Explanation: By KCL,

∴ \(\frac{V_P-40}{1} + \frac{V_P-100}{14} + \frac{V_P}{2}\) = 0

Or, 22 V

_{P}= 660

∴ V

_{P}= 30 V

Potential difference between node x and y = 60 V

∴ -I – 5 + \(\frac{40-30}{I}\) = 0

Or, I = 5 A

∴ R = \(\frac{60}{5}\)= 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

View Answer

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}= \(\frac{29}{2}\) = 14.5

Hence, R

_{A}= 2.5 Ω, R

_{B}= 4.5 Ω and R

_{C}= 7.5 Ω.

3. Currents I_{1}, I_{2} and I_{3} meet at a junction in a circuit. All currents are marked as entering the node. If I_{1} = -6sin(ωt) mA and I_{2} = 8 cos(ωt) mA, the I_{3} 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

View Answer

Explanation: Applying KCL, we get, I

_{1}+ I

_{2}+ I

_{3}= 0

∴ -6 sin(ωt) + 8 cos(ωt) + I

_{3}= 0

∴ I

_{3}= 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

View Answer

Explanation: Applying Thevenin’s Theorem R

_{EQ}= 6 || 4

= \(\frac{6 × 4}{6 + 4}\)

= \(\frac{24}{10}\) = 2.4 Ω

V

_{AB}= 10 – 6 × (\(\frac{15}{10}\)) = 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

View Answer

Explanation: Applying KCL to node A, \(\frac{V_A-10}{10} + \frac{V_A}{20} + \frac{V_A}{7}\) = 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 V

_{2}= \(\frac{V_A}{7} × 2\) = 0.97 V

V

_{2Ω}due to 20 V source, \(\frac{V_A}{10} + \frac{V_A}{20} + \frac{V_A-20}{7}\) = 0

Or, 0.1 V

_{A}+ 0.05V

_{A}+ 0.143V

_{A}= 2.86

∴ V

_{A}= \(\frac{2.86}{0.293}\) = 9.76 V

V

_{2Ω}= \(\frac{V_A-20}{7}\) × 2 = -2.92 V

The current in 2 Ω resistor = 2 × \(\frac{5}{5+8.67}\)

= \(\frac{10}{13.67}\) = 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

View Answer

Explanation: Applying KCL, we get, I

_{1}+ 5 = I

_{2}+ I

_{3}

∴ \(\frac{10 – V_1}{1} + 5 = \frac{V_1}{2} + \frac{V_2}{6}\)

∴ 30 – 3V

_{1}+ 15 = 3V

_{1}+ V

_{2}

∴ 6 V

_{1}+ V

_{2}= 45

From voltage source, V

_{2}– V

_{1}= 10

Now, 7 V

_{1}= 35, V

_{1}= 5 V

And V

_{2}= 15 V

∴ I

_{X}= \(\frac{V_2}{6} = \frac{15}{6}\) = 2.5 A.

7. In the circuit given below, the values of V_{1} and V_{2} respectively are _________

a) 0 and 5V

b) 5 and 0 V

c) 5 and 5 V

d) 2.5 and 2.5 V

View Answer

Explanation: I

_{1}= \(\frac{V_1-V_2}{2}\)

Applying KCL at node 1, 5 = \(\frac{V_1}{1} + \frac{V_1-V_2}{2} + V_1 \)

10 = 2V

_{1}+ V

_{1}– V

_{2}+ 2V

_{1}

Or, 10 = 5V

_{1}– V

_{2}

KCL at node 2, \(\frac{V_1-V_2}{2}\) + V

_{1}+ 2I

_{1}= V

_{2}

∴ 1.5 V

_{1}– V

_{2}= 0

∴ V

_{1}= V

_{2}= 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

View Answer

Explanation: Current through the 18 Ω resistance = \(\frac{90}{18}\) = 5 A

Equivalent resistance of 3 Ω and 7 Ω banks = \(\frac{3 × 6}{3 + 6}\) = 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 = \(\frac{5 × 20}{5 + 20}\) = 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 = \(\frac{100}{5}\) = 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

View Answer

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, V

_{10}= V

_{C}= V

_{D}= V

_{A}× \(\frac{10}{15}\)

V

_{R}= V

_{A}× \(\frac{R}{20+R} \)

And V

_{10}= V

_{R}

∴ V

_{A}× \(\frac{10}{15}\) = V

_{A}× \(\frac{R}{R+20} \) ∴ R = 40 Ω.

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

View Answer

Explanation: Q

_{1}= 0.5 x 10

^{-6}x 110

= 55 μC.

Also, Q

_{2}= 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}

View Answer

Explanation: From the figure, we can infer that,

I(t) = \(\frac{12}{12} \left(1 – \frac{2}{4} e^{-t}\right)\)

= 6 – 3e

^{-t}.

12. A resistance R is connected to a voltage source V having internal resistance R_{I}. A voltmeter of resistance R_{2} is used to measure the voltage across R. The reading of the voltmeter is _____________

a) \(\frac{V_S R_1 R_2}{R_2 R_1 + R_1 R + RR_2}\)

b) \(\frac{V_s R}{R + R_s}\)

c) \(\frac{VR_1 R_2}{R_1 R_2 – R_1 R – RR_2}\)

d) \(\frac{V_S RR_2}{R_1 R_2 + R_1 R + RR_2}\)

View Answer

Explanation: Effective resistance of R and R

_{m}is

R

_{eq}= \(\frac{RR_m}{R + R_m}\)

Therefore the reading is \(\frac{V}{R_1 + \frac{R}{R + R_2}}\Big[\frac{RR_2}{R+R_2}\Big]\)

= \(\frac{V_S RR_2}{R_1 R_2 + R_1 R + RR_2}\).

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

View Answer

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

View Answer

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.

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