CBSE Class 12 Physics sample paper 2025-26: question 29
From the official CBSE sample question paper for 2025-26, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.
Q29 · 4 marks · Case-basedCompetency-basedSemiconductor Electronics: Materials, Devices and Simple Circuits
When an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. When an external voltage is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. The applied voltage mostly drops across the depletion region.
- (I)1 mark
Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of changes by (assume that the Ge diode has large breakdown voltage)
- (A)0.2 V
- (B)0.4 V
- (C)0.6 V
- (D)0.8 V
- (II)1 mark
The value of and for the network are :
- (A)13 V, 2.32mA
- (B)11.7 V, 2.08mA
- (C)11.3V, 2.01mA
- (D)11V, 1.96mA
- (III)1 mark
The circuit shown below contains two ideal diodes, each with a forward resistance of . If the battery voltage is 6 V, the current through the resistance (in amperes) is
- (A)0.036
- (B)0.020
- (C)0.030
- (D)0.027
- (IV)1 mark
In the figure, potential difference between A and B is
- (A)Zero
- (B)5 V
- (C)10 V
- (D)15 V
Figure in the paper: (I) 12 V battery in series with parallel Ge and Si diodes (opposite diodes shown parallel, same direction) then 5 kΩ resistor to ground; V0 at the output. (II) 12 V source, Si diode then Ge diode in series, V0 across 5.6 kΩ resistor to ground. (III) Branches: D1 with 150 Ω, D2 (reversed) with 75 Ω in parallel, in series with 100 Ω and a 6 V battery. (IV) 30 V battery, 10 kΩ in series, then 10 kΩ across A-B, and a diode from A to a 10 kΩ resistor connected to B. See it in the official paper.
Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.
Hints
Open one hint at a time and try again before the next.
Hint 1
Decide first which diodes are forward biased; a reverse-biased ideal diode is an open circuit.
Hint 2
In parallel diodes, the one with the lower cut-in voltage fixes the drop.
Hint 3
Subtract cut-in voltages from the supply and apply Ohm's law to the rest.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(I) With both diodes forward, the Ge diode turns on first and clamps the drop at 0.3 V: V. With Ge reversed, only Si conducts: V. Change V → option (B).
- 1 mark
(II) Si and Ge in series both conduct: V; mA → option (D).
- 1 mark
(III) is reverse biased, so its 75 Ω branch carries no current. Current flows through (50 Ω), 150 Ω and 100 Ω: A → option (B).
- 1 mark
(IV) The diode is forward biased (acts as a wire), so the two 10 kΩ resistors across AB are in parallel: 5 kΩ. Total resistance 15 kΩ, mA, V → option (C).
Answer: (I) B – 0.4 V; (II) D – 11 V, 1.96 mA; (III) B – 0.020 A; (IV) C – 10 V
Where the marks are
- (I) 0.4 V, option B (1 mark)
- (II) 11 V, 1.96 mA, option D (1 mark)
- (III) 0.020 A, option B (1 mark)
- (IV) 10 V, option C (1 mark)
Revise the chapter
- Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits (notes, formulas, where marks are lost and every past question)
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The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.