NEET 2025 Physics Full Wave Rectifier Operation MCQ Question

To determine the state of the diodes (D₁ and D₂) at time t = 15 ms, we first need to calculate the input voltage at that time. The input supply voltage is given by the equation:

$V_n = 220 \sin(100\pi t)$

Substituting t = 15 ms (or 0.015 s):

$V_n = 220 \sin(100\pi \times 0.015) = 220 \sin(4.71)$

Calculating $\sin(4.71)$, we find it is approximately 0.999. Therefore:

$V_n \approx 220 \times 0.999 \approx 219.78 \text{ volts}$

In a full wave rectifier circuit, D₁ conducts when the input voltage is positive, and D₂ conducts when the input voltage is negative. Since at t = 15 ms, the voltage is positive (approximately 219.78 volts), D₁ is forward biased and D₂ is reverse biased. Thus, the correct answer is A.

For options B, C, and D:

• B is incorrect because it states D₁ is reverse biased, which contradicts our calculation.
• C is incorrect as it suggests both diodes are forward biased, which is not possible at the same time in a full wave rectifier.
• D is incorrect as it states both are reverse biased, which is also not possible given the positive voltage at this time.