Tough Network Theory Questions and Answers on “Calculation of I(0+), di/dt(0+), d2I/dt2(0+) in Circuits Involving both Capacitor and Inductor”.

1. A coil of inductance 10 H, resistance 40 Ω is connected as shown in the figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. If, at t=0^{+}, the voltage across the coil is 120 V, the value of the resistance R is __________

A. Zero

B. 20 Ω

C. 40 Ω

D. 60 Ω

Answer: A

Clarification: I_{L} at 0^{–} = (frac{120}{60}) = 2A

120 = 2(R + 40 + 20)

∴ R = 0.

2. A coil of inductance 10 H, resistance 40 Ω is connected as shown in figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. For the value of R obtained, the time taken for 95 % of the stored energy dissipated is ________

A. 0.10 s

B. 0.15 s

C. 0.50 s

D. 1.0 s

Answer: C

Clarification: For source free circuit,

I (t) = I_{o}(e^{-frac{R}{T} t})

∴ I (t) = 0.05 = 2 × (e^{-frac{60}{10} t})

Or, t = 0.61 ≈ 0.5 s.

3. In the circuit shown below steady state is obtained before the switch closes at t = 0. The switch is closed for 1.5 s and is then opened. At t = 1 s, V (t) will be?

A. – 3.24 V

B. 1.97 V

C. 5.03 V

D. 13.24 V

Answer: C

Clarification: V (0^{–}) = 10 V = V (0^{+})

For 0<t≤1.5 s, τ = 4 × 0.05 = 0.2 s

V_{OC} = 5 V

V (t) = 5 + (10 – 5)(e^{frac{-t}{0.2}}) = 5 + 5e^{-5t}

V (1s) = 5.03 V.

4. The circuit shown below is at steady state before the switch closes at t=0. The switch is closed for 1.5 s and is then opened. At t = 2 s, V (t) will be?

A. 5.12 V

B. 6.43 V

C. 8.57 V

D. 9.88 V

Answer: C

Clarification: V (1.5s) = 5.002 V

For t > 1.5 s, τ = 8 × 0.05 = 0.4

V (t) = 10+ (5-10)(e^{frac{-(t-1.5)}{0.4}}) = 10 – 5e^{-2.5(t-1.5)}

For t≥1.5 s, V (2 s) = 8.57 V.

5. In the circuit given below, the voltage across capacitor when switch is closed at t = ∞ is ____________

A. 50 V

B. 20 V

C. 30 V

D. 7.5 V

Answer: A

Clarification: From the figure, we can infer that,

Voltage across capacitor = voltage across 10 Ω resistance.

Now, voltage across the 10 Ω resistance = (frac{100}{10+10}) X 10

= (frac{100}{20}).10

= 50 V.

6. In the circuit given below, the current source is 1 A, voltage source is 5 V, R_{1} = R_{2} = R_{3} = 1 Ω, L_{1} = L_{2} = L_{3} = 1 H, C_{1} = C_{2} = 1 F. The current through R_{3} is _________

A. 1 A

B. 5 A

C. 6 A

D. 8 A

Clarification: At steady state, the circuit becomes,

∴ The current through R_{3} = (frac{5}{1}) = 5 A.

7. 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

Clarification: Using KVL, 100 = R(frac{dq}{dt} + frac{q}{C})

100 C = RC(frac{dq}{dt}) + q

Or, (int_{q_o}^q frac{dq}{100C-q} = frac{1}{RC} int_0^t ,dt)

100C – q = (100C – q_{o})e^{-t/RC}

I = (frac{dq}{dt} = frac{(100C – q_o)}{RC} e^{-1/1})

∴ e^{-t/RC} = 40e^{-1} = 14.7 A.

8. In the circuit given below, the current source is 1 A, voltage source is 5 V, R_{1} = R_{2} = R_{3} = 1 Ω, L_{1} = L_{2} = L_{3} = 1 H, C_{1} = C_{2} = 1 F. The current through the voltage source V is _________

A. 1 A

B. 3 A

C. 2 A

D. 4 A

Clarification: At steady state, the circuit becomes,

∴ The current through the voltage source V = 5 – 1 = 4 A.

9. In the circuit given below, the voltage across R by mesh analysis is _________

A. 1.59 V

B. 1 V

C. 2.54 V

D. 1.54 V

Answer: D

Clarification: In loop 1, by KVL, (10 + 2) I_{1} + (-2) I_{2} + 0I_{3} = 5

Or, 12 I_{1} – 2I_{2} = 5

In loop 2, -2I_{1} + (10+2+20+2) I_{2} + (-2) I_{3} = 0

Or, -2I_{1} + 34 I_{2} – 2I_{3} = 0

In loop 3, 0I_{1} + – 2I_{2} + (2+10) I_{3} = 10

Or, 0I_{1} – 2I_{2} + 12I_{3} = 10

Now, the voltage across R is V_{R} = (I_{2} – I_{3}) R

Now, I_{2} = (frac{360}{4800} = frac{3}{40})

Now, I_{3} = (frac{4060}{4800} = frac{203}{240}) A

Therefore, V_{R} = (I_{2} – I_{3}) R = (Big[frac{3}{40} – frac{203}{240}Big] 2) = 1.54 V.

10. In a dual slop integrating type digital voltmeter, the first integrating is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V, the total conversion time for an input of 1 V is _________

A. 0.01 s

B. 0.05 s

C. 0.1 s

D. 1 s

Answer: C

Clarification: In a dual slope integrating digital voltmeter,

((frac{t_1}{t_2})) V_{in} = V_{ref}

Where, t_{1} = first integration time = 10 × (frac{1}{50}) = 0.25

But V_{in} = 1 V and V_{ref} = 2 V

∴ t_{2} = (frac{V_{in} t_1}{V_{ref}}) = 0.1 s.

11. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. The internal resistance and a full-scale deflection produced by a DC current are 100 Ω and 1 mA respectively. A voltage of 100 V (rms) is applied to the input terminals. The value of R required is _________

A. 63.56 Ω

B. 89.83 Ω

C. 89.83 kΩ

D. 141.3 kΩ

Answer: C

Clarification: V_{OAverage} = 0.636 × (sqrt{2})V_{rms} = 0.8993 V_{rms}

The deflection with AC is 0.8993 times that with DC for the same value of voltage V

S_{AC} = 0.8993 S_{DC}

S_{DC} of a rectifier type instrument is (frac{1}{I_{fs}}) where I_{fs} is the current required to produce full scale deflection, I_{fs} = 1 mA; R_{m} = 100 Ω; S_{DC} = 10^{3} Ω/V

S_{AC} = 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R_{S} = S_{AC} V – R_{m} – 2R_{d}, where R_{d} is the resistance of diode, for ideal diode R_{d} = 0

∴ R_{S} = 899.3 × 100 – 100 = 89.83 kΩ.

12. A 1 μF capacitor is connected across a 50 V battery. The battery is kept closed for a long time. The circuit current and voltage across capacitor is __________

A. 0.5 A and 0 V

B. 0 A and 50 V

C. 20 A and 5 V

D. 0.05 A and 5 V

Answer: B

Clarification: We know that, when the capacitor is fully charged, it acts as an open circuit.

That is when the capacitor is fully charged.

So, the circuit current and voltage across capacitor are 0 A and 50 V respectively.

13. In the circuit given below, the switch is closed at t = 0. At t = 0^{+} the current through C is ___________

A. 2 A

B. 3 A

C. 4 A

D. 5 A

Answer: D

Clarification: We know that,

When the capacitor is fully charged, it acts as a short circuit.

The equivalent resistance R_{eq} = (10 + 10) Ω

= 20 Ω

Given voltage = 100 V

So, current through C = (frac{100}{20}) A = 5 A.

14. A digital Multimeter reads 10 V when fed with a triangular wave, which is symmetric about the time-axis. If the input is same, the rms reading meter will read?

A. (frac{20}{sqrt{3}})

B. –(frac{10}{sqrt{3}})

C. –(frac{20}{sqrt{3}})

D. (frac{10}{sqrt{3}})

Answer: D

Clarification: For triangular wave Average value = (frac{V_m}{3}), rms value = (frac{V_m}{sqrt{3}})

∴ (frac{V_m}{3}) = 10 V or, V_{m} = 30 V.

15. A resistance R is measured using the connection shown in the below figure.

The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω. The value of R is _________

A. 12.75 Ω

B. 12.0 Ω

C. 12.25 Ω

D. 12.5 Ω

Answer: C

Clarification: Percentage error in ammeter = ± (frac{1}{10×100}) × 100 = ± 0.1%

Percentage error in voltmeter= ±(frac{1}{10 ×150}) × 100 = ± 0.067%

So, δI = ± 0.3 ± 0.1 = ± 0.4%

δV = ± 0.4 ± 0.067 = ± 0.467%

R = (frac{V}{I})

So, error = ± δV ± δI = ± 0.867

Measured value of resistance = R_{m} = (frac{125}{10}) = 12.5

∴ True value = R_{m}(1-(frac{R_a}{R_m})) = 12.25 Ω.