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Q.
In the circuit shown in figure the maximum output voltage $V_0$ is
Solution:
For the positive half cycle of input the resulting network is shown below
$\Rightarrow (V_0)_{\text{max}} =\frac{1}{2}(V_i)_{\text{max}}$
$= \frac{1}{2} \times 10 = 5\,V$
