1
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Question

# Two pure inductors, each of self inductance L are connected in parallel but are well separated from each other, then the total inductance is

A
L
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B
2L
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C
L2
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D
L4
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Solution

## The correct option is D L2When the coils are connected in parallel, let the currents in the two coils be i1 and i2 respectively. Total induced currenti=i1+i2 or didt=di1dt+dt2dt ....(1)Now e=−L1(di1dt)=−L2(di2dt)(Q In parallel, induced e.m.f. across each coil will be same)Hence di1dt=−eL1 and di2dt=−eL2 .....(2)Let L' be the equivalent inductance.Then e=−L′didt or didt=−eL′ ....(3)From eqs. (1), (2) and (3), we get−eL′=−eL1−eL2 or 1L′=1L2+1L2∴L′=L1L2L1+L2Here L1=L2=L∴L′=L×LL+L=L2

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