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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Infinity Learn · Accepted Answer

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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