In the circuit shown in the figure,the maximu | vedclass

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(B) During the positive half cycle of the input voltage,diode $D_1$ is forward-biased and diode $D_2$ is reverse-biased. The circuit simplifies to a v

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$5$

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(B) During the positive half cycle of the input voltage,diode $D_1$ is forward-biased and diode $D_2$ is reverse-biased. The circuit simplifies to a voltage divider network consisting of three $2 \, k\Omega$ resistors. Let the input voltage be $V_i$. The output voltage $V_0$ is taken across one of the $2 \, k\Omega$ resistors. Using the voltage divider rule,the output voltage is $V_0 = V_i 	imes \frac{2 \, k\Omega}{2 \, k\Omega + 2 \, k\Omega} = V_i 	imes \frac{2}{4} = \frac{V_i}{2}$. Given the maximum input voltage $(V_i)_{\max} = 10 \, V$,the maximum output voltage is $(V_0)_{\max} = \frac{10 \, V}{2} = 5 \, V$.

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

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