Question

A capacitor of capacitance 150.0 µF is connected to an alternating source of emf given by E = 36 sin(120πt) V. The maximum value of current in the circuit is approximately equal to

• A. \(\sqrt{2A}\)
• B. \(2\sqrt{2A}\)
• C. \(\frac{1}{\sqrt{2}}A\)
• D. 2A

Answer 1

The Correct Option is D

Given alternating AC source \(E = 36 \sin (120 \pi t)\) V and capacitor \(C = 150 \mu F\), we can write: \[ Q = CV \quad \text{and} \quad i = \frac{dQ}{dt} = C E \omega \cos \omega t \] Maximum value of current: \[ i_0 = C E_0 \omega = 150 \times 10^{-6} \times 36 \times 120 \pi = 2.03 A \]