Electric circuits miscellaneous Easy Questions and Answers | Page - 16

Electric circuits miscellaneous

When a voltage V₀ sin ω₀ t is applied to the pure inductor, the ammeter shown in the figure reads I₀ . If the voltage applied is
– V₀ sin ω₀ t + 2 V₀ sin 2ω₀ t + 3 V₀ sin 3ω₀ t + 4 V₀ sin 4ω₀ t
the ammeter reading would be

1. 0
1. 10 I₀
1. √4² + 3² + 2² + 1 I₀
1. 2 I₀

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The importance of inductor for n^th harmonic is n times the impedance for the fundamental. Each harmonic gives the same current I₀. Therefore, A reads
√I₀² + I₀² + I₀² + I₀² = 2I₀

Correct Option: D

The importance of inductor for n^th harmonic is n times the impedance for the fundamental. Each harmonic gives the same current I₀. Therefore, A reads
√I₀² + I₀² + I₀² + I₀² = 2I₀

A series LCR circuit consisting of R = 10 ohms [X_L] = 20 ohms is connected across an a.c. supply of 200 V rms. The rms voltage across the capacitor is

1. 200 ∠– 90° V
1. 200 ∠+ 90° V
1. 400 ∠+ 90° V
1. 400 ∠– 90° V

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Z = √R² + (X_L ² − X_C ²)
= R (since | X_L | = | X_C | = 20 ohms)
∴ Z = 10 ohm
I = 200 = 20 A
10

∴ V_C = – jIX_L = 400 ∠– 90° V

Correct Option: D

Z = √R² + (X_L ² − X_C ²)
= R (since | X_L | = | X_C | = 20 ohms)
∴ Z = 10 ohm
I = 200 = 20 A
10

∴ V_C = – jIX_L = 400 ∠– 90° V

After closing the switch ‘S’ at t = 0, the current i(t) at any instant ‘t’ in the network shown in the given figure will be

1. 10 + 10 e^{100 t}
1. 10 – 10 e^{100 t}
1. 10 + 10 e^{–100 t}
1. 10 – 10 e^{–100 t}

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It can be verified by checking the value of the current at t = 0 and at t = ∞ which
should be zero and 10 = 10 A respectively .
1

Correct Option: D

It can be verified by checking the value of the current at t = 0 and at t = ∞ which
should be zero and 10 = 10 A respectively .
1

In the circuit shown in the figure, the switch S has been opened for long time. It is closed at t = 0. For t > 0, the current flowing through the inductor will be given by

1. i_L (t) = 1.2 + 0.8 e^{– 2t}
1. i_L (t) = 0.8 + 1.2 e^{– 2t}
1. i_L (t) = 1.2 – 0.8 e^{– 2t}
1. i_L (t) = 0.8 – 1.2 e^{– 2t}

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At the instant of closing the switch, current through the inductor
= 36 = 2 A
12 + 6

Under steady state conditions after closing of the switch,
current = 36 + 6 = 6 A
15 12 5

Here the equation which satisfies is
i_L (t) = 1.2 + 0.8 e^{– 2t}

Correct Option: A

At the instant of closing the switch, current through the inductor
= 36 = 2 A
12 + 6

Under steady state conditions after closing of the switch,
current = 36 + 6 = 6 A
15 12 5

Here the equation which satisfies is
i_L (t) = 1.2 + 0.8 e^{– 2t}

In the network shown in the figure, the circuit was initially in the steady-state condition with the switch K closed. At the instant when the switch is opened, the rate of decay of current through the inductance will be

1. Zero
1. 0.5 A/s
1. 1 A/s
1. 2 A/s

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Current in the inductor of the instant of opening the switch in 1 A. Thereafter
i(t) = 1e^{(– L / R)C} = e^{– 2t}
⇒ di(t) = 2 e^{– 2t}
dt

Correct Option: D

Current in the inductor of the instant of opening the switch in 1 A. Thereafter
i(t) = 1e^{(– L / R)C} = e^{– 2t}
⇒ di(t) = 2 e^{– 2t}
dt