Two different coils have self-inductance L1=8mH, 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 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
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a
i1i2=14
b
i1i2=4
c
W2W1=4
d
V2V1=14
answer is A.
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Detailed Solution
From Faraday’s Law, the induced voltage V∝Lrate of change of current is constant V=−Ldidt∴V2V1=L2L1=28=14⇒V1V2=4Power given to the two coils is same, i.e., V1i1=V2i2⇒i1i2=V2V1=14Energy stored W=12Li2⇒W2W1=L2L1i2i12=14(4)2=4⇒W1W2=14