Question

A circuit contains two inductors of self-inductance $L_1$ and $L_2$ in series. If $M$ is the mutual inductance, then the effective inductance of the circuit shown will be

• A. $L_1+L_2$
• B. $L_1+L_2-2M$
• C. $L_1+L_2+M$
• D. $L_1+L_2+2M$

Answer 1

The correct option is D $L_1+L_2+2M$

Let, the current flowing in the circuit is $i$.

If the current in the circuit increases then due to self inductance both inductors will resist the increase in current.

Now, as the current increases in inductor $L_1$ flux linked with $L_2$ will also increase and vice versa. Therefore, due to their mutual inductance, both the inductors will also oppose the increasing current through the circuit.

Therefore,

Induced emf in $L_1$,

$|E_1|=L_1\frac{di}{dt}+M\frac{di}{dt}$

Induced emf in $L_2$

$|E_2|=L_2\frac{di}{dt}+M\frac{di}{dt}$

Total induced emf,

$=(L_1+L_2+2M)\frac{di}{dt}$

On comparing with, standard equation , $|E|=L_{eff}\frac{di}{dt}$ , the effective inductance of the circuit be

$L_{eff}=L_1+L_2+2M$

Hence, $\left(D\right)$ is the correct answer.