John Abraham Asks:

“a circuit has a battery of emf E and internal resistance r.two inductors of inductance L1 and L2 are connected in series to the battery but parallel to each other.what is the current through each inductor in steady state?”

**Answer: **A pure inductor does not offer any reactance to pure dc. Therefore the only resistance in circuit is the internal resistance. But since the two inductors are in parallel, current through each of them is half the total current in circuit, = E/2r

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sir,

the answer to this question according to the book from which the question was taken is in terms of L1 ,L2.the currents are not equal unless the inductances are the same.

Please check whether it is dc or ac