$(a)$ For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
$(b)$ Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

(N/A) We know that $P = IV \cos \phi$, where $\cos \phi$ is the power factor. To supply a given power $P$ at a given voltage $V$, if $\cos \phi$ is small, the current $I = P / (V \cos \phi)$ must be large. This leads to a large power loss $I^2 R$ in the transmission lines.
$(b)$ Suppose in a circuit, the current $I$ lags the voltage $V$ by an angle $\phi$. Then the power factor is $\cos \phi = R / Z$. We can improve the power factor (making it tend to $1$) by making the impedance $Z$ tend to the resistance $R$. This is achieved by using a capacitor. Let us resolve the current $I$ into two components: $I_p$ (power component) along the applied voltage $V$, and $I_q$ (wattless component) perpendicular to the applied voltage. $I_q$ is called the wattless component because it contributes no power loss. $I_p$ is in phase with the voltage and corresponds to power loss. To improve the power factor, we must neutralize the lagging wattless current $I_q$ by an equal leading wattless current $I_q'$ provided by a capacitor connected in parallel. Thus, $I_q$ and $I_q'$ cancel each other, and the power becomes $P = I_p V$.