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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 isOption: 1 <p

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

2L


Option: 3

\mathrm{\text { L/2 }}


Option: 4

\mathrm{\text { L/4 }}


Answers (1)

When the coils are connected in parallel, let the currents in the two coils be\mathrm{\mathrm{i}_1 \text { and } \mathrm{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}}{\mathrm{dt}}+\frac{\mathrm{di}_2}{\mathrm{dt}}}                 ...[1]

\mathrm{\text { Now e }=-\mathrm{L}_1\left(\frac{\mathrm{di}_1}{\mathrm{dt}}\right)=-\mathrm{L}_2\left(\frac{\mathrm{di}_2}{\mathrm{dt}}\right)}

( In parallel, induced e.m.f. across each coil will be same)

\mathrm{\text { Hence } \frac{\mathrm{di}_1}{\mathrm{dt}}=-\frac{\mathrm{e}}{\mathrm{L}_1} \text { and } \frac{\mathrm{di}_2}{\mathrm{dt}}=-\frac{\mathrm{e}}{\mathrm{L}_2}}                 ...[2]

Let \mathrm{L^{\prime}} be the equivalent inductance. Then

\mathrm{\mathrm{e}=-\mathrm{L}^{\prime} \frac{\mathrm{di}}{\mathrm{dt}} \text { or } \frac{\mathrm{di}}{\mathrm{dt}}=-\frac{\mathrm{e}}{\mathrm{L}^{\prime}}}                                    ...[3]

From eqs. (1), (2) and (3), we get

\mathrm{\begin{aligned} & -\frac{e}{L^{\prime}}=-\frac{e}{L_1}-\frac{e}{L_2} \text { or } \frac{1}{L^{\prime}}=\frac{1}{L_1}+\frac{1}{L_2} \\ & \therefore L^{\prime}=\frac{L_1 L_2}{L_1+L_2} \end{aligned}}

\mathrm{\begin{aligned} & \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}}

 

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Kshitij

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