Q.(a) An emf $e = e_0\sin\omega t$ applied to a series L-C-R circuit drives a current $i = i_0\sin\omega t$ in the circuit. Deduce the expression for the average power dissipated in the circuit.

[!TLDR] (a) $P_{av}=\frac12 e_0i_0\cos\phi=e_{rms}i_{rms}\cos\phi$, derived by averaging $P=ei$ for $i=i_0\sin\omega t$ and $e=e_0\sin(\omega t\pm\phi)$. …

(a) Let the current in the series LCR circuit be $i=i_0\sin\omega t$, driven by an emf $e=e_0\sin(\omega t\pm\phi)$ that leads or lags it by phase angle $\phi$ (as derived in section 13.5.4). The instantaneous power is $P=ei=e_0i_0\sin(\omega t\pm\phi)\sin\omega t$. Expanding $\sin(\omega t\pm\phi)=\sin\omega t\cos\phi\pm\cos\omega t\sin\phi$ and multiplying out gives $P=e_0i_0\left[\sin^2\omega t\cos\phi\pm\sin\omega t\cos\omega t\sin\phi\right]$. Averaging term by term over one full cycle: $\langle\sin^2\omega t\rangle=\dfrac12$ (survives), while $\langle\sin\omega t\cos\omega t\rangle=0$ (vanishes, as in Long Answer 4-5). So $P_{av}=e_0i_0\cos\phi\times\dfrac12=\dfrac12 e_0i_0\cos\phi=e_{rms}\,i_{rms}\cos\phi$.\n\n(b) For a transmission line delivering a FIXED amount of real power P to a load at a fixed rms voltage $e_{rms}$, rearranging $P=e_{rms}i_{rms}\cos\phi$ gives $i_{rms}=\dfrac{P}{e_{rms}\cos\phi}$. If the power factor $\cos\phi$ is LOW, the current $i_{rms}$ required to deliver the same power P must be correspondingly LARGE (since $\cos\phi$ is in the denominator). But the power actually LOST as heat in the resistance $R_{line}$ of the transmission wires themselves is $P_{loss}=i_{rms}^2\,R_{line}=\dfrac{P^2R_{line}}{e_{rms}^2\cos^2\phi}$ -- which grows as the INVERSE SQUARE of the power factor. So a low power factor, even while delivering …

(a) Expand $P=ei$ using the phase-shifted current and average term by term, keeping only $\langle\sin^2\omega t\rangle=1/2$. …

In part (a), forgetting to average and instead leaving the instantaneous (time-dependent) expression as the final 'average power'. In part (b), reasoning only qualitatively ('low power factor is b …