Question

For a given ac i=im sin wt show that average power dissipated in resistor r over complete cycle is 1/2 im2r

Answer 1

The average power dissipated in resistor R over a  complete cycle is

P = Im²R/2

Explanation :

Given the current flowing in the circuit,

i = Im sinwt

Resistance of the circuit = R

For AC the average power dissipated over a cycle is given by,

P = (1/T) $\int\limits^T_0 {i^2 R} \, dt$

putting i = Im sinwt we get,

P = (Im²R/T) $\int\limits^T_0 {sin^2 wt} \, dt$

=> P = (Im²R/2T) ∫(1 - cos2wt)dt

=> P = (Im²R/2T) ∫(dt - cos2wtdt)

=> P = (Im²R/2T) ( T - 0)

=> P = Im²R/2

which is the required expression for average power consumption in an AC circuit.

Answer 2

try to best so give like