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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 L2
When the coils are connected in parallel, let the currents in the two coils be i1 and i2 respectively. Total induced current
i=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=Ldidt or didt=eL ....(3)
From eqs. (1), (2) and (3), we get
eL=eL1eL2 or 1L=1L2+1L2
L=L1L2L1+L2
Here L1=L2=L
L=L×LL+L=L2

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