An alternating emf E = 110√(2) sin 100t volt is applied to a capacitor of 2 μ F, the rms value of current in the circuit is ________ mA. Numerical answer type. Work it out, then open the answer.

Answer: 22. Irms = ErmsXC = Erms ω C Erms = 110 V, ω = 100 rad/s, C = 2 × 10⁻⁶ F. Irms = 110 × 100 × 2 × 10⁻⁶ = 0.022 A So the rms current is 22 mA.

An alternating emf E = 110√(2) sin 100t volt is applied to a… | JEE Main 2024 (8 Apr, Shift 2)

Q29JEE Main 2024 · 8 APR · SHIFT 2Easy

Pure R, L and C Circuits

An alternating emf $E = 110\sqrt{2} \sin 100t$ volt is applied to a capacitor of $2\ \mu F$ , the rms value of current in the circuit is ________ $\text{mA}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 22

$I_{rms} = \frac{E_{rms}}{X_C} = E_{rms}\,\omega C$
$E_{rms} = 110$ V, $\omega = 100$ rad/s, $C = 2 \times 10^{-6}$ F.
$I_{rms} = 110 \times 100 \times 2 \times 10^{-6} = 0.022$ A
So the rms current is 22 mA.

Report a mistake

Pure R, L and C Circuits question from JEE Main 2024 (8 Apr, Shift 2)

More questions on Pure R, L and C Circuits