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(B) The input voltage $V_i$ is an alternating voltage with a peak value of $10 \, V$.For the positive half-cycle of $V_i$, the top terminal is at a hi

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$5 \, V$

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(B) The input voltage $V_i$ is an alternating voltage with a peak value of $10 \, V$.For the positive half-cycle of $V_i$, the top terminal is at a higher potential than the bottom terminal. In this case, diode $D_2$ is forward-biased and conducts, while diode $D_1$ is reverse-biased and acts as an open circuit.The circuit simplifies to a voltage divider consisting of the $2 \, k\Omega$ resistor (where $V_0$ is measured) in series with another $2 \, k\Omega$ resistor connected to the negative terminal.The total resistance in the path is $2 \, k\Omega + 2 \, k\Omega = 4 \, k\Omega$.Using the voltage divider rule, the output voltage $V_0$ across the $2 \, k\Omega$ resistor is:$V_0 = V_i 	imes \frac{2 \, k\Omega}{2 \, k\Omega + 2 \, k\Omega} = V_i 	imes \frac{2}{4} = \frac{V_i}{2}$.Since the maximum input voltage is $10 \, V$, the maximum output voltage is:$V_{0, 	ext{max}} = \frac{10 \, V}{2} = 5 \, V$.

In the circuit shown for the given input voltage $V_i$, the maximum output voltage $V_0$ is

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