CBSE Class 12 · 2013 · Alternating CurrentShort Answer

(a) For a given a.c.,

i = i_{m} sin ω t

, show that the average power dissipated in a resistor R over a complete cycle is

\frac{1}{2} i_{m}^{2} R

.
(b) A light bulb is rated at 125 W for a 250 V a.c. supply. Calculate the resistance of the bulb.

Answer: (a) $⟨ P ⟩ = \frac{1}{2} i_{m}^{2} R$ .
(b) $R = 500 Ω$ .

Solution

(a) The instantaneous power in the resistor is

P = i^{2} R = i_{m}^{2} R sin^{2} ω t

Averaging over one complete cycle and using

⟨ sin^{2} ω t ⟩ = \frac{1}{2}

⟨ P ⟩ = i_{m}^{2} R ⟨ sin^{2} ω t ⟩ = \frac{1}{2} i_{m}^{2} R

(b) Using rated values with

P = \frac{V _{r m s}^{2}}{R}

R = \frac{V _{r m s}^{2}}{P} = \frac{( 250 ) ^{2}}{125} = 500 Ω

⟨ P ⟩ = \frac{1}{2} i_{m}^{2} R, R = 500 Ω

CBSE Class 12 Board · 2013 · Subjective