In the ac circuit an alternating voltage e = 200√(2) sin100t volts is connected to a capacitor of capacity 1 μ F. The r.m.s. value of the current in the circuit is (a) 10 mA (b) 100 mA (c) 200 mA (d) 20 mA

Answer: (d). Irms = ermsXC = erms ω C erms = 200 V, ω = 100 rad/s: Irms = 200 × 100 × 10⁻⁶ = 0.02 A So the answer is (d) 20 mA.

In the ac circuit an alternating voltage e = 200√(2) sin100t volts is… | NEET 2011 (AIPMT, Prelims)

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Pure R, L and C Circuits

In the ac circuit an alternating voltage e = $200\sqrt{2}$ sin100t volts is connected to a capacitor of capacity 1 $\mu$ F. The r.m.s. value of the current in the circuit is

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Answer: (d)

$I_{rms} = \frac{e_{rms}}{X_C} = e_{rms}\,\omega C$
$e_{rms} = 200$ V, $\omega = 100$ rad/s: $I_{rms} = 200 \times 100 \times 10^{-6} = 0.02$ A
So the answer is (d) 20 mA.

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Pure R, L and C Circuits question from NEET 2011 (AIPMT, Prelims)

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