PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2024-25: question 31(I)

From the official CBSE sample question paper for 2024-25, section E, question 31(I), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(I) · 5 marks · Long answerCurrent Electricity

a) Write two limitations of ohm's law. Plot their I-V characteristics.
b) A heating element connected across a battery of 100 V having an internal resistance of 1 Ω draws an initial current of 10 A at room temperature 20.0 °C which settles after a few seconds to a steady value. What is the power consumed by battery itself after the steady temperature of 320.0 °C is attained? Temperature coefficient of resistance averaged over the temperature range involved is $3.70 \times 1 0^{- 4}$ °C $^{- 1}$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

Think of devices like the diode, whose current-voltage graph is not a straight line, or not symmetric about the origin.

Hint 2

Use the initial current to find the element's resistance at 20 °C, then $R = R_{0} (1 + α Δ T)$ .

Hint 3

The battery's own power loss is $I^{2} r$ with the steady current.

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.

(a) Limitation 1: V is not always proportional to I; the V-I graph is non-linear. Example: a p-n junction diode or a filament lamp. Graph: I against V is a curve, not a straight line through the origin.
1 mark
Limitation 2: the relation between V and I depends on the sign of V. Reversing V with the same magnitude does not give the same magnitude of current, e.g. a diode. Graph: a large forward current for small positive V, but almost no current for negative V. (Another limitation: the relation may not be unique, with more than one V for the same I, as in GaAs.)
1 mark
(b) At room temperature (20 °C): $I = \frac{ε}{R _{0} + r} \Rightarrow R_{0} = \frac{ε}{I} - r = \frac{100}{10} - 1$
½ mark
$R_{0} = 9 Ω$
½ mark
At 320 °C: $R = R_{0} (1 + α Δ T) = 9 (1 + 3.70 \times 1 0^{- 4} \times 300)$
½ mark
$R = 9 \times 1.111 \approx 10 Ω$
½ mark
Steady current: $I^{'} = \frac{100}{10 + 1} = \frac{100}{11} = 9.09$ A. Power consumed by the battery itself: $P = I^{'2} r = (\frac{100}{11})^{2} \times 1$
½ mark
$P \approx 82.6 W$
½ mark

Answer: (a) Non-linear V-I relation; dependence on the sign of V (or non-unique relation). (b) R₀ = 9 Ω, R = 10 Ω, power in battery ≈ 82.6 W

Where the marks are

First limitation with its V-I graph (1 mark)
Second limitation with its V-I graph (1 mark)
Finds $R_{0} = 9 Ω$ from $I = ε / (R_{0} + r)$ (1 mark)
Finds R ≈ 10 Ω at 320 °C (1 mark)
Power in internal resistance $I^{2} r \approx 82.6$ W (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

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CBSE Class 12 Physics sample paper 2024-25: question 31(I)

Hints

Step-by-step solution

Revise the chapter

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