Question

In the circuit shown, the average power dissipated in the resistor is (assume diode to be ideal)

• A. $\frac{E_0^2}{2R}$
• B. $\frac{E_0^2}{4R}$
• C. $\frac{E_0^2}{R}$
• D. Zero

Answer 1

Answer: (B)

It's a half wave rectification $E_{\text{rms }}=\frac{E_0}{2}$
$P_{av}=\frac{{\left(E_{\text{rms }}\right)}^2}{R}$
$=\frac{E_0^2}{4R}$