Question

The equivalent inductance of two inductors is 2.4 mH when connected in parallel and 10 mH when connected in series. The difference between two inductance is $($neglecting mutual induction between coils$)$.
(A) 3mH
(B) 2mH
(C) 4mH
(D) 16mH

Answer 1

Hint: When inductor connected in series combination then equivalent inductance will be given as
$L_{eq}=L_1+L_2+L_3+....$
When inductors are connected in parallel combination then equivalent inductance will be given as
${\displaystyle \frac{1}{L_{eq}}}={\displaystyle \frac{1}{L_1}}+{\displaystyle \frac{1}{L_2}}+{\displaystyle \frac{1}{L_3}}+.....$

Step by step answer: Given that 2 inductors $L_1$ and $L_2(Let)$ are connected in parallel then their equivalent inductance is 2.4 mH.
i.e., ${\displaystyle \frac{1}{L_{eq}}}={\displaystyle \frac{1}{L_1}}+{\displaystyle \frac{1}{L_2}}$

So, $${\displaystyle \frac{1}{L_1}}+{\displaystyle \frac{1}{L_2}}={\displaystyle \frac{1}{2.4}}$$
${\displaystyle \frac{L_2+L_1}{L_1{L}_2}}={\displaystyle \frac{10}{24}}$
$L_1{L}_2={\displaystyle \frac{24}{10}}(L_1+L_2)$ …..(1)
When $L_1$ and $L_2$ are connected in series then their equivalent inductance is 10 mH.
i.e., $L_{eq}=L_1+L_2$

$10=L_1+L_2$ …..(2)
From equation 1 & 2 we get
$L_1{L}_2={\displaystyle \frac{24}{10}}\times 10$
$L_1{L}_2=24$
So, $L_1={\displaystyle \frac{24}{L_2}}$ …..(3)
On putting the value of $L_1$ in equation 2
${\displaystyle \frac{24}{L_2}}+L_2=10$
${\displaystyle \frac{24+L_2^2}{L_2}}=10$
$24+L_2^2=10L_2$
$L_2^2-10L_2+24=0$
$L_2^2-6L_2-4L_2+24=0$
$L_2(L_2-6)-4(L_2-6)=0$
$(L_2-6)(L_2-4)=0$
$L_2=4,6$ ….(4)
Now put the value of $L_2$ in equation 3
We will get $L_1$.
Here we have 2 values of $L_2$. So, put one by one each value and will get $L_1$.
When $L_2=4mH$
Then from equation 3
$L_1={\displaystyle \frac{24}{4}}$
$L_1=6mH$ …..(6)
When $L_2=6mH$
Then from equation 3
$L_1={\displaystyle \frac{24}{6}}$
$L_1={\displaystyle \frac{24}{6}}$
$L_1=4mH$ …..(6)
So, we will get 2 combinations of $L_1$ & $L_2$ which are
If $L_1=4mH$ If $L_1=6mH$
Then $L_2=6mH$ then $L_2=4mH$
Now, we have to calculate the difference between $L_1$ and $L_2$. Which is
$L_1-L_2=6-4=2mH$
So, from both combinations of $L_1$ and $L_2$ we will get a difference between them of 2 mH.
Hence, option B is the correct answer. 2mH


Note: In many problems of inductors, they can ask for current and voltage.
In series combination, the value of current in each inductor is the same. But voltage is different.
In parallel combination, the potential difference i.e., voltage across each inductor is same but current is different.