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Question: Two coils of self inductance \(L_{1}\) and \(L_{2}\) are placed closer to each other so that total f...

Two coils of self inductance L1L_{1} and L2L_{2} are placed closer to each other so that total flux in one coil is completely linked with other. If MM is mutual inductance between them, then

A

M=L1L2M = L_{1}L_{2}

B

M=L1/L2M = L_{1}/L_{2}

C

M=L1L2M = \sqrt{L_{1}L_{2}}

D

M=(L1L2)2M = (L_{1}L_{2})^{2}

Answer

M=L1L2M = \sqrt{L_{1}L_{2}}

Explanation

Solution

M=e2di1/dt=e1di2/dtM = - \frac{e_{2}}{di_{1}/dt} = - \frac{e_{1}}{di_{2}/dt}

Also e1=L1di1dt.e2=L2di2dte_{1} = - L_{1}\frac{di_{1}}{dt}.e_{2} = - L_{2}\frac{di_{2}}{dt}

M2=e1e2(di1dt)6mu(di2dt)=L1L2M=L1L2M^{2} = \frac{e_{1}e_{2}}{\left( \frac{di_{1}}{dt} \right)\mspace{6mu}\left( \frac{di_{2}}{dt} \right)} = L_{1}L_{2} \Rightarrow M = \sqrt{L_{1}L_{2}}