Electric circuits miscellaneous Easy Questions and Answers | Page - 6

Electric circuits miscellaneous

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

In the series Rc circuit shown in the figure, the voltage across C starts increasing when the d.c. source is switched on. The rate of increase of voltage across C at the instant just after the switch is closed (i.e. at t = 0⁺) will be

1. zero
1. infinity
1. RC
1. 1
  RC

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Voltage across the capacitor at any time t, V_c = V (l – e^{– t / RC})
= l – e^{– t / RC} ... (since V = 1 volt)
∴ dV_c = 1 e^{– t / RC}
dt RC

At t = 0⁺ , dV_c = 1
dt RC

Correct Option: D

Voltage across the capacitor at any time t, V_c = V (l – e^{– t / RC})
= l – e^{– t / RC} ... (since V = 1 volt)
∴ dV_c = 1 e^{– t / RC}
dt RC

At t = 0⁺ , dV_c = 1
dt RC

A periodic rectangular signal X(t) has the wave form shown in the figure. Frequency of the fifth harmonic of its spectrum is

1. 40 Hz
1. 200 Hz
1. 250 Hz
1. 1250 Hz

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Periodic time = 4 ms = 4 × 10^{– 3} sec.
Fundamental frequency = 10³ = 250 Hz
4

∴ Frequency of the 5th harmonic = 250 × 5 = 1250 Hz

Correct Option: D

Periodic time = 4 ms = 4 × 10^{– 3} sec.
Fundamental frequency = 10³ = 250 Hz
4

∴ Frequency of the 5th harmonic = 250 × 5 = 1250 Hz

In the given figure, A₁, A₂ and A₃ are ideal ammeters. If A₁ reads 5A, A₂ reads 12 A, then A₃ should read

1. 7 A
1. 12 A
1. 13 A
1. 17 A

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Since the source is an a.c voltage, therefore
I _3rms = √I_1rms ² + I_2rms ²
= √5² +12² = 13 A

Correct Option: C

Since the source is an a.c voltage, therefore
I _3rms = √I_1rms ² + I_2rms ²
= √5² +12² = 13 A