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  • Two pure inductors, each of self inductance L are connected in parallel but are well separated from each other, then the total inductance is  Option:

Two pure inductors, each of self inductance L are connected in parallel but are well separated from each other, then the total inductance is
 

Option: 1

L


Option: 2

2 \mathrm{~L}


Option: 3

\mathrm{L} / 2


Option: 4

\mathrm{L} / 4


Answers (1)

best_answer

When the coils are connected in parallel, let the currents in the two coils be \mathrm{ i_1 \, and \, \, i_2} respectively. Total induced current
\mathrm{ \mathrm{I}=\mathrm{i}_1+\mathrm{i}_2 \text { or } \frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{di}_1}{\mathrm{dt}}+\frac{\mathrm{di}}{\mathrm{dt}}}............................(1)
\mathrm{Now\, \, \mathrm{e}=-\mathrm{L}_1\left(\frac{\mathrm{di}_1}{\mathrm{dt}}\right)=-\mathrm{L}_2\left(\frac{\mathrm{di}_2}{\mathrm{dt}}\right)}
because In parallel, induced e.m.f. across each coil will be same
Hence
\mathrm{\frac{\mathrm{di}_1}{\mathrm{dt}}=-\frac{\mathrm{e}}{\mathrm{L}_1} and \frac{\mathrm{di}_2}{\mathrm{dt}}=-\frac{\mathrm{e}}{\mathrm{L}_2} Let \mathrm{L}^{\prime} }.......................(2)
 be the equivalent inductance
Then
\mathrm{\mathrm{e}=-\mathrm{L}^{\prime} \frac{\mathrm{di}}{\mathrm{dt}} or \frac{\mathrm{di}}{\mathrm{dt}}=-\frac{\mathrm{e}}{\mathrm{L}^{\prime}} }....................(3)
From eqs. (1), (2) and (3), we get
\mathrm{-\frac{\mathrm{e}}{\mathrm{L}^{\prime}}=-\frac{\mathrm{e}}{\mathrm{L}_1}-\frac{\mathrm{e}}{\mathrm{L}_2} \text { or } \frac{1}{\mathrm{~L}^{\prime}}=\frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2} }
\mathrm{\begin{aligned} & \therefore \quad \mathrm{L}^{\prime}=\frac{\mathrm{L}_1 \mathrm{~L}_2}{\mathrm{~L}_1+\mathrm{L}_2} \\ & \text { Here } \mathrm{L}_1=\mathrm{L}_2=\mathrm{L} \\ & \therefore \quad \mathrm{L}^{\prime}=\frac{\mathrm{L} \times \mathrm{L}}{\mathrm{L}+\mathrm{L}}=\frac{\mathrm{L}}{2} \end{aligned} }

 

Posted by

Ritika Harsh

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