Question

In an a.c. circuit, voltage and current are given by:

V = 100 sin (100 t) V and I = `100 sin (100 t + pi/3)` mA respectively.

The average power dissipated in one cycle is:

पर्याय

• 10 W

• 2.5 W

• 25 W

• 5 W

Answer 1

2.5 W

Explanation:

From the equations V = 100 sin (100 t) V and I = `100 sin (100 t + pi/3)` mA:

Peak voltage (V₀) = 100 V

Peak current (I₀) = 100 mA = 0.1 A

Phase Difference (Φ) = `pi/3` = 60°

The average power (P_avg) dissipated in an AC circuit is:

P_avg = V_rms . I_rms . cos Φ

Since, V_rms = `V_0/sqrt 2` and I_rms = `I_0/sqrt 2`, the formula becomes:

P_avg = `(V_0 I_0)/2 cos Phi`

= `(100 xx 0.1)/2 xx cos (60^circ)`

= `10/2 xx 1/2`

= 5 × 0.5

= 2.5 W