A circuit contains two inductors of self-inductance L1 and L2 in series as shown in the figure. If M is the mutual inductance, then the effective inductance of the circuit shown will be
A
L1+L2
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B
L1+L2−M
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C
L1+L2+3M
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D
L1+L2+2M
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Solution
The correct option is CL1+L2+2M if a current i passes through the series combination, induced emf in L1=L1didt+Mdidt induced emf in L2=L2didt+Mdidt total induced emf =(L1+L2+2M)didt