A sinusoidal AC current flows through a resistor of resistance R . If the peak current is Ip, then the power dissipated is

Question

A sinusoidal AC current flows through a resistor of resistance R . If the peak current is Ip, then the power dissipated is (A) Ip2 R cos θ (B) (1/2) Ip2 R (C) (4/π) Ip2 R (D) (1/π) Ip2 R

Tardigrade Admin · Accepted Answer

Power =I2 R=((Ip/√2))2 R=(Ip2 R/2)

Q. A sinusoidal AC current flows through a resistor of resistance $R .$ If the peak current is $I_{p}$ , then the power dissipated is

$I_{p}^{2} R cos θ$

$\frac{1}{2} I_{p}^{2} R$

$\frac{4}{π} I_{p}^{2} R$

$\frac{1}{π} I_{p}^{2} R$

Power $= I^{2} R = (\frac{I _{p}}{2})^{2} R = \frac{I _{p}^{2} R}{2}$

Q. A sinusoidal AC current flows through a resistor of resistance R. If the peak current is Ip​, then the power dissipated is

Ip2​Rcosθ

21​Ip2​R

π4​Ip2​R

π1​Ip2​R