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NEET 2020 (Phase 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from NEET 2020 (Phase 1)

Q59NEET 2020 · PHASE 1Easy

Pure R, L and C Circuits

A 40 $\mu$ F capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly:

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Answer: (b)

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 40 \times 10^{-6}} \approx 79.6\ \Omega$
$I_{rms} = \frac{V_{rms}}{X_C} = \frac{200}{79.6} \approx 2.5$ A
So the answer is (b) 2.5 A.

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