To solve this problem, let's analyze the given circuit and determine the maximum supply voltage \(V_{s(max)}\) and the Peak Inverse Voltage (PIV) of the diode.

First, identify the conditions under which the maximum output voltage \(V_o(max)\) is achieved. According to the problem, \(V_{o(max)} = 12 \text{ V}\).

Taking into account the diode's cut-in voltage of \(0.7 \text{ V}\), the maximum supply voltage can be calculated as:

1. \(V_{s(max)} = V_{o(max)} + V_d = 12 \text{ V} + 0.7 \text{ V} = 12.7 \text{ V}\)

The PIV of the diode is the maximum reverse voltage the diode can withstand without breaking down. This is essentially twice the maximum output voltage in this type of rectifier circuit. Therefore, the PIV is:

1. \(PIV = 2 \times V_{o(max)} = 2 \times 12 \text{ V} = 24.7 \text{ V}\)

Therefore, the correct answer is:

• \(V_{s(max)} = 12.7 \text{ V}\)
• PIV of the diode = 24.7 V

So, the correct answer is 12.7V, 24.7V, which corresponds to the given answer choice.

Note: In rectifier circuits, always consider the diode's cut-in voltage and understand that the PIV is a key factor in determining diode specifications to avoid breakdown during operation.