CBSE Class 12 · 2008 · Alternating CurrentShort Answer

Prove that an ideal inductor does not dissipate power in an a.c. circuit.

Answer: Average power $= V_{r m s} I_{r m s} cos \frac{π}{2} = 0$ .

Solution

Let the applied voltage be

V = V_{0} sin ω t

. In a pure inductor the current lags the voltage by

\frac{π}{2}

I = I_{0} sin (ω t - \frac{π}{2}) = - I_{0} cos ω t

The instantaneous power is

P = V I = - V_{0} I_{0} sin ω t cos ω t = - \frac{V _{0} I _{0}}{2} sin 2 ω t

Averaged over a complete cycle,

⟨ sin 2 ω t ⟩ = 0

, so

P_{a v} = \frac{1}{T} \int_{0}^{T} P d t = 0

Equivalently

P_{a v} = V_{r m s} I_{r m s} cos \frac{π}{2} = 0

. Hence an ideal inductor dissipates no power.

P_{a v} = 0

CBSE Class 12 Board · 2008 · Subjective