Electromagnetic Induction and Alternating Current 3 Question 7

####8. Two different coils have self-inductances L1=8mH and L2=2mH. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current, the induced voltage and the energy stored in the first coil are i1,V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2,V2 and W2 respectively. Then

(a) i1i2=14

(b) i1i2=4

(c) W1W2=14

(d) V1V2=4

(1994, 2M)

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Answer:

Correct Answer: 8. (a, c, d)

Solution:

  1. From Faraday’s law, the induced voltage

VL, if rate of change of current is constant V=Ldidt

V2V1=L2L1=28=14 or V1V2=4

Power given to the two coils is same, i.e.

V1i1=V2i2 or i1i2=V2V1=14

Energy stored, W=12Li2

W2W1=L2L1i2i12=14(4)2 or W1W2=14



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