CBSE Class 12 Physics 2016 Outside Delhi Set 1 Paper

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Question : 24
Total: 26
SECTION - E

(i) Draw a labelled diagram of a step-down transformer. State the principle of its working.
(ii) Express the turn ratio in terms of voltages.
(iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.
(iv) How much current is drawn by the primary of a transformer connected to 220V supply when it delivers power to a 110V−550W refrigerator?
OR
(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r1 and the other of radius r2(r1<r2) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
(b) A rectangular coil of area A, having number of turns N is rotated at ' f ' revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2Ï€fNBA.
Solution:  
(i) Principle of working: It works on the principle of mutual induction i.e. if two coils are inductively coupled and when current or magnetic flux is changed through one of the two coils, then induced e.m.f. is produced in the other coil.
(ii) Turns ratio is
K=‌
ns
np
=‌
Es
Ep

A transformer with a primary winding of 1000 turns and secondary winding of 100 turns has a turns ratio of 1000:100 or 10:1. Therefore, 100 volts applied to primary winding will produce a secondary voltage of 10 volts.
(iii) EsIs=‌EpIp
‌‌ (Input Power = Output Power) ‌
⇒‌‌‌
Es
Ep
=‌
Ip
Is

⇒‌‌‌
Ip
Is
=‌
ns
np
=K

(iv) ep=220v;es=110v,esIs=550W
Now,
epIp‌=esIs
Ip‌=‌
esIs
ep
=‌
550
220
=2.5A

OR
(a) Meaning of Mutual Inductance
Expression
(b) Proof
Diagram
(a) Mutual Inductance is the property of a pair of coils due to which an emf induced in one of the coils due to the change in the current in the other coil.
‌‌ Mathematically ‌e2‌=‌
Mdi1
dt

∴‌‌‌M‌=−‌
e2
di1∕dt

Let a current I2 flows through the outer circular coil. Then
B2=‌
µI2
2r2

∴‌‌ϕ1=πr2B2=‌
µπr12
2r2
I2
=M12I2

Thus ‌‌M12=‌
µπr12
2r2
I2
=M21

(b)

Flux at any time ' t '.
ϕB=BA‌cos‌θ=BA‌cos‌ω‌t
From Faraday's Law, induced emf
e=−N‌
dϕB
dt
=NBA‌
d
dt
(cos‌ω‌t)

Thus the instantaneous value of emf is
E=NBAωsin‌ωt
For maximum value of emf sin‌ωt=±1
i.e., ‌‌e0=NBAω=2πfNBA
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