Topic: Reactance, impedance, LCR series circuit, resonance· Electromagnetic Induction and Alternating Currents

A 10 μ F capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π=3.14):

A 10 \(\mu F\) capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (\(\pi=3.14\)):

capacitor C=10 microF connected to 210 V, 50 Hz AC source

1. 0.35 \(A\)

2. 0.58 \(A\)

3. 0.93 \(A\)

4. 1.20 \(A\)

Doubtly Asked 30 Oct 2026 NEET UG 2024 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 3. 0.93 \(A\)

Key idea: For a capacitor in an AC circuit the capacitive reactance is \(X_C=\frac{1}{\omega C}\) and the peak current is \(I_0=\frac{V_0}{X_C}=V_0\omega C\), where \(V_0=\sqrt2\,V_{rms}\).

\(\omega=2\pi f=2\times3.14\times50=314\) rad/s, \(V_0=210\sqrt2\approx297\) V.

\(I_0=297\times314\times10\times10^{-6}\approx0.93\) A.

Option 3. (Using 210 V directly would give the rms current 0.66 A, which is not an option.)

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

A 10 μ F capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π=3.14):

1 answer

Your answer