Question:medium

In an $L - R$ circuit, the inductive reactance is equal to $\sqrt{3}$ times the resistance '$R$' of the circuit. An e.m.f. $\text{E} = \text{E}_0 \sin(\omega t)$ is applied to the circuit. The power consumed in the circuit is

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Power is only dissipated in the resistance $R$, never in the pure inductor $L$.
Updated On: May 12, 2026
  • $\frac{\text{E}_0^2}{4\text{R}}$
  • $\frac{E_0^2}{6R}$
  • $\frac{\text{E}_0^2}{8\text{R}}$
  • $\frac{E_0^2}{12R}$
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The Correct Option is C

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