In the circuit shown in the figure, the maxim | 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

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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 $2 	ext{ k}\Omega$ resistors. The input voltage $V_i$ is applied across a series combination of two $2 	ext{ k}\Omega$ resistors, while the output voltage $V_0$ is taken across one of these $2 	ext{ k}\Omega$ resistors. Using the voltage divider rule: $V_0 = V_i 	imes \frac{2 	ext{ k}\Omega}{2 	ext{ k}\Omega + 2 	ext{ k}\Omega} = V_i 	imes \frac{2}{4} = \frac{V_i}{2}$. Given the maximum input voltage $(V_i)_{\max} = 10 	ext{ V}$, the maximum output voltage is: $(V_0)_{\max} = \frac{(V_i)_{\max}}{2} = \frac{10 	ext{ V}}{2} = 5 	ext{ V}$.

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

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