For the circuit shown below with ideal diodes, the output will be

The correct answer is: V_out = V_in for V_in > 0

To determine the output voltage $V_{\mathrm{out}}$ of the circuit, we analyze the behavior of the two ideal diodes, $D_1$ and $D_2$, under different polarities of the input voltage $V_{\mathrm{in}}$.

1. Case 1: When $V_{\mathrm{in}}>0$ (Positive Half-Cycle)
During the positive half-cycle, the upper terminal of the input is positive (+) and the lower terminal is negative (-):
- The anode of diode $D_1$ is connected to the positive terminal, making $D_1$ forward-biased (ON).
- The cathode of diode $D_2$ is connected to the negative terminal, making $D_2$ forward-biased (ON) as it completes the return path to the negative terminal.
Since the diodes are ideal, they act as short circuits (closed switches) when forward-biased. The input voltage is directly connected across the load resistor $R_1$:
$V_{\mathrm{out}}=V_{\mathrm{in}}$

2. Case 2: When $V_{\mathrm{in}}<0$ (Negative Half-Cycle)
During the negative half-cycle, the upper terminal of the input is negative (-) and the lower terminal is positive (+):
- The anode of diode $D_1$ is connected to the negative terminal, which reverse-biases $D_1$ (OFF).
- The cathode of diode $D_2$ is connected to the positive terminal, which reverse-biases $D_2$ (OFF).
Since both diodes are ideal and reverse-biased, they act as open circuits (open switches). No current flows through the resistor $R_1$, which results in:
$V_{\mathrm{out}}=0$

Conclusion:
Combining the two cases, the output voltage is given by:
$V_{\mathrm{out}}=V_{\mathrm{in}}$ for $V_{\mathrm{in}}>0$