4.Electromagnetic Induction Alternating Currents Rammohan Mudgal
12 Physics Electromagnetic Induction and Alternating Current Test 02 Answer 4f7g
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Transcript of 12 Physics Electromagnetic Induction and Alternating Current Test 02 Answer 4f7g
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7/28/2019 12 Physics Electromagnetic Induction and Alternating Current Test 02 Answer 4f7g
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Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in
Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks
CBSE TEST PAPER-02
CLASS - XII PHYSICS (Electromagnetic Induction and Alternating current)
[ANSWERS]
Ans 1: X is a purely capacitive circuit
1 1
2C
Xc wC
= =
Ans 2:2200
10220
Ns EsK
Np Ep= = = =
Ans 3: P = E rms I rms cos
But for an ideal inductor2
=
cos cos 02
= =
Ans 4: The capacitive reactance
1 1
2C
XwC Vc
= =
For d.c. = 0
CX =
Since capacitor offers infinite resistance to the flow of d.c. so d.c. cannot pass
through the capacitor.
Ans 5: Consider a long solenoid of area A through which current I is flowing
Let N be the total number of turns in the solenoid
Total flux = NBA
Here = B = onI
Where n is no. of turns per unit length of the solenoid
N = nlnl onIA =
2(1)
=LI-------------(2)
on AlI
Also
=
Equation (1) & (2)2on AlI LI =
2L on Al=
[n = N/ ]
P = 0
2oN A
L
l
=
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Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in
Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks
One henry if current is changing at a rate of 1A/s in a coil induces an emf. of 1
volt in it then the inductance of the coil is one henry.
Ans 6: consider a circular loop connect the centre with point P with a resistor.
The potential difference across the resistor = induced emf.B= Rate of change of area of loop.
If the resistor QP is rotated with angular velocity w and turns by an angle in
time t then
Area swept1
2A R R=
2
2
2 2
1
2
cos
1
2
1 1
2 2
o
A R
BA BA
B R
d d dBR BR
dt dt dt
=
= =
=
= = =
Ans 7: (a)1
2C
Xc
=
For a.c. when
XC = 0Thus at a very high frequency of a.c. capacitor behaves as a conductor
(b)1
2C
Xc
=
1
2
C
L
X
X WL LXL
= =
Ans 8: (1) k = 100, Ep = 200V
EpIp = 1000 W, Np = 100
100
100
100 100
NsK
Np
Ns Np
Ns
= =
=
=
21
2BwR=
NS = 10,000
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Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in
Portal for CBSE Notes, Test Papers, Sample Papers, Tips and Tricks
(2) EpIp = 1000W
1000
10005
200
Ip Ep
Ip A
=
= =
(3)Es Ns
Ep Np=
200 100
NsEs Ep
Np
Es
=
=
(4)Es Ip
Ep Is=
1000 1
20000 20
IpEpIs
Es
Is
=
= =
(5) For step up Trans former k > 1
Ip = 5A
Es = 20000 Volt
Is = 0.05 A
NsK
Np=
