CBSE Class 12 Physics 2020 Delhi Set 1 Paper

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Question : 36
Total: 37
(a) Show that an ideal conductor does not dissipate power in an a.c. circuit.
(b) The variation of inductive reactance (XL) of an inductor with the frequency (f) of the a.c. source of 100V and variable frequency is shown in the figure.

(i) Calculate the self-inductance of the inductor.
(ii) When this inductor is used in series with a capacitor of unknown value and a resistor of 10Q at 300 s1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
OR
(a) A conductor of length ' l ' is rotated about one of its ends at a constant angular speed ' ω ' in a plane perpendicular to a uniform magnetic field B. Plot graphs to show variations of the emf induced across the IP variations of the emf induced across the ends of the conductor with (i) angular speed ω and (ii) length of the conductor l.
(b) Two concentric circular loops of radii 1cm and 20cm are placed coaxially.
(i) Find mutual inductance of the arrangement.
(ii) If the current passed through the outer loop is changed at a rate of 5 Ams, find the emf induced in the inner loop. Assume the magnetic field on the inner loop to be uniform.
Solution:  
(a) Power dissipation =P=VRMSIRMScosϕ
cosϕ=
R
Z
.

For ideal inductor R=0
cosϕ=0
P=VRMS IRMS cosϕ=0
Thus, ideal inductor does not dissipate power in an ac circuit.
(b) (i) Inductive reactance =XL=2πfL
L=
XL
2πf

From graph f=100Hz
XL=20
L=
XL
2πf
=
20
2π×100
=0.032H
=32mH

(ii) Power dissipation is maximum when
2πfL=
1
2πfC

f=300 s1
L=0.032H
2πfL=
1
2πfC

Or 2π×300×0.032 =
1
2π×300×C

C=8.8×106F=8.8µF
OR
(a) (i) e=12BωL2
So, eω
So, the graph is :

(ii) e=12BωL2
So, eL2
So, the graph is :

(b) (i) M=
µ0πr2
2R

µ0=4π×107
r=1cm=0.01m
R=20cm=0.2m
Putting the values
M=
4π×107×π(0.01)2
2×0.2

=0.00986×107H=9.98×104H
(ii) |e|=M
i
t

Or, e|=9.98×104H×5000A s
Induced emf =|e|=9.98×0.5 =4.99V
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