Answer the following. (i) Describe the construction and working of a transformer and hence obtain the relation for ((_s)/(_p) ) in terms of number of turns of primary and secondary. (ii) Discuss four main causes of energy loss in a real transformer.

(v_s)/(v_p)=(N_s)/(N_p); energy losses: flux leakage, copper (I²R) loss, eddy currents, hysteresis (also magnetostriction/humming).

PhysicsCBSE Class 12 · Board exam 2024, set 55/5/1 · 5 marks · Case-based

CBSE Class 12 Physics 2024: question 32(b)

Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Case-basedCompetency-basedAlternating Current

Answer the following.

(i)
Describe the construction and working of a transformer and hence obtain the relation for $(\frac{υ _{s}}{υ _{p}})$ in terms of number of turns of primary and secondary.
0 marks
(ii)
Discuss four main causes of energy loss in a real transformer.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/5/1.

Hints

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

Hint 1

Think about what links the two coils: they are not connected electrically, so what carries the effect from one to the other?

Hint 2

In an ideal transformer the same rate of change of flux per turn, $d ϕ / d t$ , appears in both coils, so apply Faraday's law to each coil.

Hint 3

For the losses, think about where energy can turn into heat (the wires, the core) and where flux can fail to reach the secondary.

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.

(i) Construction: A transformer has two coils, the primary ( $N_{p}$ turns) and the secondary ( $N_{s}$ turns), insulated from each other. Both are wound on a common laminated soft-iron core, either one over the other or on separate limbs of the core. *(Diagram: a rectangular laminated soft-iron core with the primary on one limb connected to the AC input $v_{p}$ and the secondary on the other limb connected to the output/load $v_{s}$ .)*
1 mark
Working: An alternating voltage applied to the primary drives an alternating current in it. This current sets up an alternating magnetic flux in the core. The core carries this flux to the secondary, so the flux linked with the secondary changes continuously and an alternating emf is induced in the secondary by mutual induction.
1 mark
Relation: For an ideal transformer there is no flux leakage and the coil resistances are negligible. The flux $ϕ$ through each turn is then the same in both coils. Induced emf in the secondary: $ε_{s} = - N_{s} \frac{d ϕ}{d t}$ , and $v_{s} = ε_{s}$ . Back emf in the primary: $ε_{p} = - N_{p} \frac{d ϕ}{d t}$ , and $v_{p} = ε_{p}$ (since primary resistance $\approx 0$ ).
½ mark
Dividing the two equations: $\frac{v _{s}}{v _{p}} = \frac{N _{s}}{N _{p}}$ . If $N_{s} > N_{p}$ it is a step-up transformer; if $N_{s} < N_{p}$ it is a step-down transformer.
½ mark
(ii) Cause 1 – Flux leakage: Not all the flux produced by the primary passes through the secondary, because of poor core design or air gaps. It is reduced by winding the primary and secondary one over the other.
½ mark
Cause 2 – Resistance of windings (copper loss): The windings have resistance, so energy is lost as heat ( $I^{2} R$ ). It is reduced by using thick copper wires.
½ mark
Cause 3 – Eddy currents (iron loss): The alternating flux induces eddy currents in the iron core, which heat it. It is reduced by using a laminated core.
½ mark
Cause 4 – Hysteresis: The core is magnetised and demagnetised repeatedly by the alternating field, and energy is lost as heat in each cycle. It is reduced by using a material of low hysteresis loss, such as soft iron.
½ mark

Answer: $\frac{v _{s}}{v _{p}} = \frac{N _{s}}{N _{p}}$ ; energy losses: flux leakage, copper ( $I^{2} R$ ) loss, eddy currents, hysteresis (also magnetostriction/humming).

Where the marks are

Describes construction: two insulated coils on a laminated soft-iron core (1 mark)
Explains working through alternating flux and mutual induction (1 mark)
Writes $ε_{s} = - N_{s} d ϕ / d t$ and $ε_{p} = - N_{p} d ϕ / d t$ and obtains $v_{s} / v_{p} = N_{s} / N_{p}$ (1 mark)
States any four energy losses, such as flux leakage, copper loss, eddy currents, hysteresis or magnetostriction (½ mark each, 2 marks)

Revise the chapter

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

Practise this chapterBoard-style questions from Alternating Current, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2024: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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