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

CBSE Class 12 Physics sample paper 2024-25: question 32(II)

From the official CBSE sample question paper for 2024-25, section E, question 32(II), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(II) · 5 marks · Long answerAlternating Current

a) With the help of a diagram, explain the principle of a device which changes a low ac voltage into a high voltage. Deduce the expression for the ratio of secondary voltage to the primary voltage in terms of the ratio of the number of turns of primary and secondary winding. For an ideal transformer, obtain the ratio of primary and secondary currents in terms of the ratio of the voltages in the secondary and primary coils.
b) Write any two sources of the energy losses which occur in actual transformers.
c) A step-up transformer converts a low input voltage into a high output voltage. Does it violate law of conservation of energy? Explain.

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

Both coils share the same changing flux through the iron core; how does each coil's induced emf depend on its number of turns?

Hint 2

Divide the two Faraday's-law expressions, then use power in = power out for an ideal transformer.

Hint 3

For (c), compare input and output power, not just voltage.

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) Device and principle: The device is a step-up transformer. It works on mutual induction: an alternating current in one coil produces a changing flux that induces an emf in a neighbouring coil.
Diagram (described):
- Primary coil P (few turns) and secondary coil S (many turns) are wound on a closed, laminated soft-iron core.
- P is connected to the a.c. input; S is connected to the load.
1 mark
Working: The a.c. in the primary sets up an alternating flux $ϕ$ in the core. Assuming no flux leakage, the same flux per turn links both coils.
Flux linkages: $ϕ_{P} = N_{P} ϕ$ and $ϕ_{S} = N_{S} ϕ$
½ mark
Voltage ratio:
- $e_{S} = N_{S} \frac{d ϕ}{d t}$
- $e_{P} = N_{P} \frac{d ϕ}{d t}$ (equal to the applied voltage when coil resistance is negligible)
Dividing: $\frac{e _{S}}{e _{P}} = \frac{N _{S}}{N _{P}}$
For step-up, $N_{S} > N_{P}$ , so $e_{S} > e_{P}$ .
½ mark
Current ratio (ideal transformer, no losses):
Input power = output power, so $e_{P} i_{P} = e_{S} i_{S}$ .
$\Rightarrow \frac{i _{P}}{i _{S}} = \frac{e _{S}}{e _{P}} = \frac{N _{S}}{N _{P}}$
So in a step-up transformer the current is stepped down.
½ mark
Use: Step-up transformers raise the voltage at generating stations for long-distance transmission. Step-down transformers lower it at the receiving substations.
½ mark
(b) Energy losses (any two):
1. Eddy current loss in the core, reduced by using a laminated core.
2. Flux leakage, since not all of the primary flux links the secondary.
Other losses: hysteresis loss in the core, and copper ( $I^{2} R$ ) loss in the windings.
1 mark
(c) No, it does not violate the law of conservation of energy. When the voltage is stepped up, the current is stepped down in the same ratio. So the output power $e_{S} i_{S}$ is at most equal to the input power $e_{P} i_{P}$ ; no extra energy is created.
1 mark

Answer: (a) $\frac{e _{S}}{e _{P}} = \frac{N _{S}}{N _{P}}$ ; $\frac{i _{P}}{i _{S}} = \frac{e _{S}}{e _{P}}$ . (b) Eddy current loss and flux leakage (also hysteresis, copper loss). (c) No; the current falls as the voltage rises, so power is not increased.

Where the marks are

Diagram and principle of mutual induction, laminated soft-iron core (1 mark)
Flux linkages $N_{P} ϕ$ and $N_{S} ϕ$ (½ mark)
Derives $e_{S} / e_{P} = N_{S} / N_{P}$ (½ mark)
Uses $e_{P} i_{P} = e_{S} i_{S}$ to get $i_{P} / i_{S} = e_{S} / e_{P}$ (½ mark)
Step-up/step-down use statement (½ mark)
Any two energy losses (1 mark)
Explains no violation: current decreases, power same (1 mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Alternating Current, 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 32(II)

Hints

Step-by-step solution

Revise the chapter

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