The power loss in a series a.c. circuit consisting of a resistance (…

The correct answer is Option (4) → \( R \) is low, \( L \) is high

In an a.c. circuit only the resistance dissipates power. The average power is \( P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi \), where the power factor of an R–L circuit is \( \cos\phi = R/Z \).

    \( \displaystyle Z = \sqrt{R^{2} + \omega^{2}L^{2}}, \quad \cos\phi = \frac{R}{\sqrt{R^{2} + \omega^{2}L^{2}}} \)

    \( \displaystyle P = \frac{V_{\mathrm{rms}}^{2}}{Z}\cos\phi = \frac{V_{\mathrm{rms}}^{2}R}{R^{2} + \omega^{2}L^{2}} \)

A low R and a high L (a large inductive reactance) make the impedance large, so the current is small, and make the power factor small, so little of that current does work. The power loss is then least; in the limit of a pure inductor (R = 0) it is zero.

Note: The question compares the combinations only in words. From \( P = V_{\mathrm{rms}}^{2}R/(R^{2} + \omega^{2}L^{2}) \), the power also becomes small when R is very high and L is low, because it then falls to about \( V_{\mathrm{rms}}^{2}/R \); under that reading Option (2) can also give a small power loss. The usual reading, through the power factor, gives Option (4).