Q.In a series LR circuit $X_L = R$ and the power factor of the circuit is $P_1$. When a capacitor with capacitance C such that $X_L = X_C$ is put in series, the power factor becomes $P_2$. Calculate $P_1/P_2$.

Because both instantaneous voltage e and instantaneous current i constantly change in an AC circuit, the instantaneous power $P=ei$ fluctuates continuously and is not, by itself, a physically meaningful description of energy consumption; what matters instead is the AVERAGE power over one complete cycle. Working this out separately for each basic element: a pure resistor (e and i in phase) gives $P_{av}=\frac12 e_0i_0=e_{rms}i_{rms}$, the familiar 'apparent power' relation; a pure inductor or pure capacitor (each with a $\pi/2$ phase difference) gives EXACTLY ZERO average power, since the power expression reduces to a zero-mean sinusoid at twice the original frequency -- energy is merely exchanged cyclically with the source, never permanently consumed.

For the general case of a phase difference $\phi$ between voltage and current (as in a series LCR circuit), the average power is $P_{av}=e_{rms}\,i_{rms}\cos\phi$, where $\cos\phi$ -- the POWER FACTOR -- can also be read geometrically off the impedance triangle as $R/Z$. A power factor of 1 (purely resistive, or exactly at resonance) means all the apparent power is genuinely dissipated; a power factor of 0 (purely reactive, L or C alone) means the current, however large, dissipates no real power at all -- such a current is called WATTLESS or IDLE current. In practice, a low power factor forces a disproportionately larger current to deliver the same real power, causing much greater $I^2R$ transmission losses, which is why keeping the power factor close to 1 matters for real electrical systems.