A full wave rectifier circuit with diodes D₁ and D₂ is shown in the figure. If input supply voltage

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Question

A full wave rectifier circuit with diodes D₁ and D₂ is shown in the figure. If input supply voltage Vin = 220 sin(100πt) volt, then at t = 15 ms:

Accepted Answer

Correct Answer: C — D₁ is reverse biased, D₂ is forward biasedStep-by-step Solution:Vin = 220 sin(100πt) VAngular frequency: ω = 100π rad/sTime period: T = 2π/ω = 2π/(100π) = 1/50 s = 0.02 s = 20 msAt t = 15 ms = 0.015 s:t/T = 0.015/0.02 = 0.75 = 3/4 → t = 3T/4At t = 3T/4, the sinusoidal waveform is in the negative half cycle.Vin = 220 sin(100π × 0.015) = 220 sin(1.5π) = 220 × (−1) = −220 V (negative)In a full wave rectifier during the negative half cycle: D₁ is reverse biased, D₂ is forward biased.Why Options A and D are wrong: In a full wave rectifier, at any instant during either half cycle, one diode conducts and the other does not. Both simultaneously ON or OFF is impossible (they complement each other).Why Option B is wrong: D₁ forward and D₂ reverse corresponds to the positive half cycle (t = T/4), not the negative half cycle at t = 3T/4.Key Concepts from Semiconductor Electronics (NCERT Class 12, Chapter 14):Full wave rectifier: Uses two diodes and a centre-tap transformer. Each diode conducts during opposite half cycles, giving rectified output for both halves.During positive half cycle: D₁ forward biased, D₂ reverse biased.During negative half cycle: D₁ reverse biased, D₂ forward biased.Ripple frequency of full wave rectifier output = 2 × input frequency = 2f.A half wave rectifier uses one diode and has ripple frequency = f.Diode in forward bias: p-side at higher potential; in reverse bias: n-side at higher potential.

A full wave rectifier circuit with diodes D₁ and D₂ is shown in the figure. If input supply voltage Vin = 220 sin(100πt) volt, then at t = 15 ms:

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A full wave rectifier circuit with diodes D₁ and D₂ is shown in the figure. If input supply voltage Vin = 220 sin(100πt) volt, then at t = 15 ms:

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