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
- \(\sqrt{2A}\)
- \(2\sqrt{2A}\)
- \(\frac{1}{\sqrt{2}}A\)
- 2A
The Correct Option is D
Solution and Explanation
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