Question

A source of ac voltage v = v₀ sin ωt, is connected across a pure inductor of inductance L. Derive the expressions for the instantaneous current in the circuit. Show that average power dissipated in the circuit is zero.

Answer 1

Using Kirchhoff’s loop rule in the above circuit,

\[V - L\frac{dI}{dt} = 0\]

\[\text { where, I is the instantaneous current flowing in the circuit }\]

\[ \Rightarrow L\frac{dI}{dt} = V_o \sin\omega t\]

\[ \Rightarrow \int dI = \frac{V_o}{L}\int\sin\omega tdt\]

\[ \Rightarrow I = \frac{V_o}{L}[ -\frac{\cos\omega t}{\omega}]\]

\[ \Rightarrow I = - \frac{V_o}{\omega L}\cos\omega t\]

\[ \Rightarrow I = \frac{V_o}{X_L}\sin(\omega t - \frac{\pi}{2})\]

\[\text { where, }X_L = \omega L = \text { inductance reactance }\]

\[ \therefore I = I_o \sin(\omega t - \frac{\pi}{2})\]

\[\text { where,} I_o = \frac{V_o}{X_L} =\text {  peak value of ac}\]

\[\text { Now instantaneous power supplied by the source is } \]

\[P = VI = V_o I_o \sin\omega t \sin(\omega t - \frac{\pi}{2})\]

\[\text { Now the average power } ( P_{avg} ) \text { supplied over a complete cycle} ( 0 to 2\pi) \text { is }\]

\[ P_{avg} = \int_o^{2\pi} P = \int_o^{2\pi} V_o I_o \sin\theta \sin(\theta - \frac{\pi}{2})d\theta [\text { where } \theta = \omega t]\]

\[\text { Now over a complete cycle } \int_o^{2\pi} \sin\theta\sin(\theta - \frac{\pi}{2})d\theta = 0\]

\[\text { Therefore, }P_{avg} = \int_o^{2\pi} V_o I_o \text { sin }\theta\sin(\theta - \frac{\pi}{2})d\theta = 0\]

\[ P_{avg} = 0\]

Hence, the average power dissipated in the circuit is zero.