For an ideal inductor of inductance $L$, the voltage and current relation is: $E(t) = L \frac{dI}{dt}$

Solving gives: $I(t) = \frac{E_0}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right)$

which shows that the current lags the voltage by $\frac{\pi}{2}$.

Inductive Reactance: $X_L = \omega L = 2\pi f L$

Instantaneous Power: $P(t) = E(t) I(t) = E_0 I_0 \sin(\omega t) \sin\left(\omega t - \frac{\pi}{2}\right)$

$= E_0 I_0 \sin(\omega t) \cdot (-\cos(\omega t)) = -E_0 I_0 \sin(\omega t) \cos(\omega t)$

Using identity: $\sin(2\omega t) = 2 \sin(\omega t) \cos(\omega t)$

So, $P(t) = -\frac{E_0 I_0}{2} \sin(2\omega t)$

and hence, $P_{\text{avg}} = 0$

Energy is stored in the magnetic field and returned to the circuit cyclically.

For a pure capacitor of capacitance $C$, the current is: $I(t) = C \frac{dE}{dt} = C \frac{d}{dt}[E_0 \sin(\omega t)] = \omega C E_0 \cos(\omega t)$

$= I_0 \sin\left(\omega t + \frac{\pi}{2}\right)$

Hence, current leads voltage by $\frac{\pi}{2}$.

Capacitive Reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$

Instantaneous Power: $P(t) = E_0 \sin(\omega t) \cdot I_0 \sin\left(\omega t + \frac{\pi}{2}\right)$

$= E_0 I_0 \sin(\omega t) \cos(\omega t) = \frac{E_0 I_0}{2} \sin(2\omega t)$

Average Power: $P_{\text{avg}} = 0$

Energy is alternately stored and released by the electric field in the capacitor.

i. Impedance in RLC Circuit

For a series RLC circuit: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

The current is: $I = \frac{E_0}{Z} \sin(\omega t - \phi)$

where $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$ is the phase angle.

ii. RMS and Peak Relationships

RMS (Root Mean Square) values relate to peak values: $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}, \quad E_{\text{rms}} = \frac{E_0}{\sqrt{2}}$

These give equivalent d.c. values in terms of heating effect.

iii. Power Factor

The power factor in an RLC circuit is: $\text{Power Factor} = \cos(\phi)$

where $\phi$ is the phase angle. Power factor affects real power: $P_{\text{avg}} = E_{\text{rms}} I_{\text{rms}} \cos(\phi)$