CBSE Class 12 Physics 2026: question 25

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Step-by-step solution

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1. (a) In the positive half-cycle A is at a higher potential than B.

   • $D_2$ (anode at A, cathode towards R) is forward biased and conducts.
   • $D_1$ (anode at P, cathode at A) is reverse biased, since its cathode is at the higher potential, and does not conduct.
   1 mark

2. (b) Equivalent circuit: $D_2$ is replaced by a short wire and $D_1$ by an open switch. Current flows from A through $D_2$ to R. At R it divides:

   • one part through 3 kΩ directly to B;
   • the other through 1 kΩ (R to P) and then 2 kΩ (P to B).

   So 3 kΩ is in parallel with the series pair (1 kΩ + 2 kΩ) = 3 kΩ.

   1 mark

3. (c) At the peak, $V_i=12$ V and the ideal diode drops nothing, so $V_{RB}=12$ V.

   • Across 3 kΩ: $V=12$ V (current 4 mA).
   • Current in the 1 kΩ–2 kΩ branch: $I=\frac{12}{3\text{k}\Omega }=4$ mA.
   • $V_{1k}=4\text{ mA}\times 1\text{ k}\Omega =4$ V; $V_{2k}=4\text{ mA}\times 2\text{ k}\Omega =8$ V.
   1 mark

Answer: D₂ conducts (forward biased) and D₁ does not. At peak: 12 V across 3 kΩ, 4 V across 1 kΩ, 8 V across 2 kΩ.

Revise the chapter

• Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits (notes, formulas, where marks are lost and every past question)
• Topic: Diode as a Rectifier