Question

In the circuit shown in figure the maximum output voltage $V_0$ is

• A. 0 V
• B. 5 V
• C. 10 V
• D. $\frac{5}{\sqrt{2}} V$

Answer 1

Answer: (B)

For the positive half cycle of input the resulting network is shown below

$\Rightarrow (V_0)_{\text{max}} =\frac{1}{2}(V_i)_{\text{max}}$
$= \frac{1}{2} \times 10 = 5\,V$