A 14.0 \( \mu\mathrm{F} \) capacitor is connected to a 220 V, 50 Hz source. The value of peak current in the circuit is
(Given: \( \pi = 22/7 \) and \( \sqrt{2} = 1.414 \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.37\,\mathrm{A} \)
A capacitor in an ac circuit offers the capacitive reactance \( X_{C} = 1/(2\pi f C) \). The rms current is \( V_{\mathrm{rms}}/X_{C} \), and the peak current is \( \sqrt{2} \) times the rms current.
\( C = 14.0\,\mu\mathrm{F} = 14\times10^{-6}\,\mathrm{F} \)
\( f = 50\,\mathrm{Hz} \)
\( V_{\mathrm{rms}} = 220\,\mathrm{V} \)
\( \displaystyle X_{C} = \frac{1}{2\pi f C} \;\Rightarrow\; \frac{1}{X_{C}} = 2\pi f C = 2 \times \frac{22}{7} \times 50 \times 14\times10^{-6} = 4.4\times10^{-3}\,\mathrm{S} \)
\( \displaystyle I_{\mathrm{rms}} = V_{\mathrm{rms}} \times \frac{1}{X_{C}} = 220 \times 4.4\times10^{-3} = 0.968\,\mathrm{A} \)
\( \displaystyle I_{0} = \sqrt{2}\,I_{\mathrm{rms}} = 1.414 \times 0.968 = 1.37\,\mathrm{A} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – …
