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

Correct Answer:
$V_{\text{out}}=V_{\text{in}}$ for $V_{\text{in}}>0$

Detailed Explanation:

Let us analyze the given circuit diagram under the two possible states of the input voltage, $V_{\text{in}}$, assuming ideal diodes $D_1$ and $D_2$ as shown in the image:

Case 1: When the input voltage is positive ($V_{\text{in}}>0$)
- The upper input terminal becomes positive with respect to the lower input terminal.
- The anode of diode $D_1$ is connected to the positive terminal of the input source, making it forward-biased (ON). Since the diode is ideal, it behaves as a short circuit (a closed switch).
- The cathode of diode $D_2$ is connected to the negative terminal of the input source. Current flows from the positive terminal of the input, through the closed switch of $D_1$, down through the load resistor $R_1$, and back through the anode of $D_2$ to its cathode. This makes diode $D_2$ forward-biased (ON) as well, behaving as a short circuit.
- With both ideal diodes conducting, the output voltage terminals are connected directly across the input voltage terminals.
- Therefore, the output voltage is given by:
$V_{\text{out}}=V_{\text{in}}$

Case 2: When the input voltage is negative ($V_{\text{in}}<0$)
- The upper input terminal is negative with respect to the lower input terminal.
- This polarity reverse-biases both diodes $D_1$ (anode is negative) and $D_2$ (cathode is positive), turning them OFF. Under ideal conditions, they act as open circuits (open switches).
- Because the conducting path is broken, no current flows through the resistor $R_1$.
- Therefore, the output voltage across the resistor is:
$V_{\text{out}}=0$

Comparing these derivations with the options provided, the output behavior matches:
$V_{\text{out}}=V_{\text{in}}$ for $V_{\text{in}}>0$