In figure shown below, the Peak Inverse Voltage (PIV) required for diode is [Figure description: A series loop with an AC source 300 sin(ωt) (+ at top), a diode D in the top wire, a 10 Ω resistor, and a 100 V DC battery in series on the right side (battery + terminal toward the resistor? marked 100 V), closing back to the source.]
- A. 300 V
- B. 200 V
- C. 100 V
- D. 400 V
Answer: D. 400 V
Explanation: During the negative half-cycle of the AC supply, the peak supply voltage (300 V) and DC battery voltage (100 V) act in series against the diode, requiring PIV = 300 V + 100 V = 400 V.