CBSE Class 12 Physics 2017 Delhi Set 1 Paper

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Question : 24
Total: 26
SECTION -E

(a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.
(b) A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with angular speed of 50rad s1 in a uniform magnetic field of magnitude 3.0× 102T. Calculate the maximum value of the current in the coil.
OR
(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils.
(b) A power transmission line feeds input power at 2200V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220V.
Solution:  
(a) Labelled diagram of AC generator Expression for instantaneous value of induced emf.



(b) Calculation of maximum value of current

When the coil is rotated with constant angular speed the angle θ between the magnetic field and area vector of the coil, at instant t, is given by
θ=ωt,
Therefore, magnets flux, (ϕB), at this instant, is
ϕB=BAcosωt
Induced emf e=N
dϕB
dt

e=NBAωsinωt
e=e0sinωt
where e0=NBAω
(b) Maximum value of emf
e0=NBAω
=20×200×104×3×104×50V
=600mV
Maximum induced current
i0=
e0
R
=
600
R
mA

OR
(a) Labelled diagram of a step up transformer
Derivation of ratio of secondary and primary voltage
(b) Calculation of number of turns in the secondary
(a)

When ac voltage is applied to primary coil the resulting current produces an alternating magnetic flux, which also links the secondary coil.
The induced emf, in the secondary coil, having Ns turns, is
es=Ns
dϕ
dt

This flux, also induces an emf, called back emf, in the primary coil.
e s=Ns
dϕ
dt

But ep=Vp
and e s=V s
Vs
Vp
=
Ns
Np

For an ideal transformer
lpVp=isVs
Vs
Vp
=
ip
is

(b)
Ns
Np
=
Vs
Vp

Ns
3000
=
220
2200

Ns=300
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