Question

The average output voltage of the circuit shown below is

• A. $\frac{10}{\sqrt{2}}$V
• B. 3.18 V
• C. $\frac{10\sqrt{2}}{\pi }$V
• D. 2.5 V

Answer 1

The correct option is B 3.18 V

During the positive half cycle the circuit behaves as,



$V_0=\frac{V_i\times 2k}{4k}=5sin\omega tV$

During negative half cycle it is same, therefore it works as full wave rectifier with $V_m=5$ V

$V_0=\frac{2V_m}{\pi }=\frac{2\times 5}{\pi }=3.18V$