Question

The sinusoidal A.C. current flows through a resistor of resistance $R$. If the peak current is $I_p$, then power dissipated is

• A. ${I_p}^{2}R\cos \theta$
• B. $\frac{1}{2}{I_p}^{2}R$
• C. $\frac{4}{\pi }{I_p}^{2}R$
• D. $\frac{1}{{\pi }^2}{I_p}^{2}R$

Answer 1

The correct option is B $\frac{1}{2}{I_p}^{2}R$

The value of r.m.s current is, $I_{max}=\frac{I_P}{\sqrt{2}}$
So power dissipated is, $P=I_{rms}^{2}R=\frac{1}{2}{I}_P^{2}R$