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Q.

Two different coils have self-inductances L1 = 8 mH, L2 =2 mH. The current in one coil is increased at aconstant 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 V1and W1 respectively. Corresponding value for the second coil at the same instant are i2, V2 and W2 respectively. Then

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a

i1i2=14

b

V1V2=4

c

W2W1=4

d

ALL THE ABOVE

answer is D.

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Detailed Solution

We know thatP=dWdt=ddt12Li2=LididtAs P and didt are same, hence L1i1=L2i2 or i1i2=L2L1=2mH8mH=14 ∴ (a) is correct.Voltage induced V=−Ldidt∴V2V1=L2L1=2mH8mH=14 ∵didt is that sameEnergy stored = Work done∴W=12Li2  or W1=12L1i12 and W2=12L2i22 ∴W2W1=L2L1i2i12=2mH8mH×(4)2=14×(4)2=4
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Two different coils have self-inductances L1 = 8 mH, L2 =2 mH. The current in one coil is increased at aconstant 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 V1and W1 respectively. Corresponding value for the second coil at the same instant are i2, V2 and W2 respectively. Then