A circuit contains two inductors of self-inductance L1 and L2 in series (figure). If M is the mutual inductance, then the effective inductance of the circuit shown will be
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a
L1 + L2
b
L1+L2−2M
c
L1+L2+M
d
L1+L2+2M
answer is D.
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Detailed Solution
When inductances are connected like this in series, then L=L1+L2+2M