Question

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

• A. $L$
• B. $2L$
• C. $\frac{L}{2}$
• D. $\frac{L}{4}$

Answer 1

Answer: (C)

$\frac{L}{2}$

When the coils are connected in parallel, let the currents in the two coils be $i_1$ and $i_2$ respectively. Total induced current

$i=i_1+i_2$ or $\frac{di}{dt}=\frac{d{i}_1}{dt}+\frac{d{t}_2}{dt}$ ....(1)

Now $e=-L_1\left(\frac{d{i}_1}{dt}\right)=-L_2\left(\frac{d{i}_2}{dt}\right)$
(Q In parallel, induced e.m.f. across each coil will be same)

Hence $\frac{d{i}_1}{dt}=-\frac{e}{L_1}$ and $\frac{d{i}_2}{dt}=-\frac{e}{L_2}$ .....(2)

Let L' be the equivalent inductance.

Then $e=-L^{\prime }\frac{di}{dt}$ or $\frac{di}{dt}=-\frac{e}{L^{\prime }}$ ....(3)

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

$-\frac{e}{L^{\prime }}=-\frac{e}{L_1}-\frac{e}{L_2}$ or $\frac{1}{L^{\prime }}=\frac{1}{L_2}+\frac{1}{L_2}$

$\therefore L^{\prime }=\frac{L_1{L}_2}{L_1+L_2}$

Here $L_1=L_2=L$

$\therefore L^{\prime }=\frac{L\times L}{L+L}=\frac{L}{2}$