For the L-R series circuit connected to an ac mains \( e = e_0 \sin\omega t \), the average power consumed is:
(Symbols have their usual meanings)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{e_0^{2}\,R}{2\,(R^{2} + \omega^{2}L^{2})} \)
The average power in an ac circuit is \( P = E_{\mathrm{rms}}\,I_{\mathrm{rms}}\cos\phi \). In an L-R series circuit the impedance is \( Z = \sqrt{R^{2} + \omega^{2}L^{2}} \) and the power factor is \( \cos\phi = R/Z \).
\( \displaystyle E_{\mathrm{rms}} = \frac{e_0}{\sqrt{2}} \qquad I_{\mathrm{rms}} = \frac{E_{\mathrm{rms}}}{Z} \)
\( \displaystyle P = \frac{E_{\mathrm{rms}}^{2}}{Z}\cos\phi = \frac{E_{\mathrm{rms}}^{2}\,R}{Z^{2}} \)
\( \displaystyle P = \frac{(e_0^{2}/2)\,R}{R^{2} + \omega^{2}L^{2}} = \frac{e_0^{2}\,R}{2\,(R^{2} + \omega^{2}L^{2})} \)
Shortcut: Check the dimensions: power has the unit \( \mathrm{V^{2}/\Omega} \). Option (1) gives \( \mathrm{V^{2}\,\Omega/\Omega^{2}} = \mathrm{V^{2}/\Omega} \). Option (2) has the unit \( \mathrm{V^{2}\,s/\Omega} \), since \( 1\,\mathrm{H} = 1\,\Omega\,\mathrm{s} \). Option (3) has the unit \( \mathrm{V^{2}/\Omega^{2}} \). Option (4) adds \( R^{2} \) to \( \omega^{2}R^{2} \), which have different units, so it cannot be right.