Two inductors having self inductances $L_{1}$ and $L_{2}$ and mutual inductances M are arranged in parallel as shown in the figure. The sense of the helix is same for both the inductors. Then

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Effective inductance of the parallel combination is $\frac{L_{1} L_{2} - M^{2}}{L_{1} + L_{2} + 2 M}$

If the separation between the coils is large then effective inductance of the combination will be close to $\frac{L_{1} L_{2}}{L_{1} + L_{2}}$ because then M will become negligible.

The effective inductance of the combination will be $\frac{L_{1} L_{2}}{L_{1} + L_{2}}$ irrespective of the value of M

Effective inductance of the combination is $L_{1} + L_{2} .$

answer is B.

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Detailed Solution

Applying Kirchoff law equations, we get $L_{1} \frac{d i_{1}}{d t} - M \frac{d i_{2}}{d t} = ε$
$L_{2} \frac{d i_{2}}{d t} - M \frac{d i_{1}}{d t} = ε$
Solving we get
$\frac{d i_{2}}{d t} = \frac{(M + L_{1}) ε}{L_{1} L_{2} - M^{2}} \frac{d i_{1}}{d t} = \frac{(M + L_{2}) ε}{L_{1} L_{2} - M^{2}} \Rightarrow \frac{d i}{d t} = \frac{(M + L_{1}) ε}{L_{1} L_{2} - M^{2}} = \frac{(L_{1} + L_{2} + 2 M) ε}{L_{1} L_{2} - M^{2}} \Rightarrow ε = L_{e f} \frac{d i}{d t} = (\frac{L_{1} L_{2} - M^{2}}{L_{1} + L_{2} + 2 M}) \frac{d i}{d t} \Rightarrow L_{e f} = (\frac{L_{1} L_{2} - M^{2}}{L_{1} + L_{2} + 2 M})$