Question

A sinusoidal ac current flows through a resistor of resistance R. If the peak current is Ip , then the power dissipated is
(a) $I^2_p R cos \theta$
(b) $\frac{1}{2} I^2_p R$
(c) $\frac{4}{\pi} I^2_p R$
(d) $\frac{1}{\pi} I^2_p R$

Answer 1

Answer: (b) I^2pR/2

Explanation:

A sinusoidal ac current flows through a resistor of resistance R.

the peak current is Ip

To find:

The power dissipated :

We know that

Power dissipated= I^2R

Since the wave is sinuoisidal

I = Ip/√2

This

Power dissipated= ((Ip/√2)^2) *R

= Ip^2*R/2 (b)

A sine wave or sinusoid is a mathematical curve that defines a smooth periodic oscillation.

The graph is sine wave.

The resistors in a circuit that has a voltage drop across it dissipates electrical power. This electrical power is converted into heat energy. Hence all resistors have a power rating. This is the maximum power that can be dissipated from the resistor without it burning out. The rate of conversion is the power of dissipation.