For the L-R series circuit connected to an ac mains $e=e_0\sin ωt$, the average power consumed is: (Symbols have their usual meanings)

$\frac{e_0^2R}{2(R^2+ω^2L^2)}$

The correct answer is Option (1) → $\frac{e_0^2R}{2(R^2+ω^2L^2)}$

For an L-R series circuit connected to AC source $e = e_0 \sin \omega t$:

Average power consumed: $P_\text{avg} = I_\text{rms}^2 R$

RMS current: $I_\text{rms} = \frac{e_0}{\sqrt{R^2 + (\omega L)^2}}$

∴ $P_\text{avg} = \frac{e_0^2 R}{2(R^2 + (\omega L)^2)}$