Question

Three identical inductors are connected as shown in the figure. At the instant when current in inductor P is increasing at the rate of $0.0320$ $A{s}^{-1}$ then magnitude of induced emf in it is $0.080$ V. The equivalent inductance between the points A and B will be?

• A. $\frac{15}{2}$H
• B. $\frac{5}{6}$H
• C. $\frac{15}{4}$H
• D. $\frac{5}{3}$H

Answer 1

The correct option is C $\frac{15}{4}$H

We know that, $emf=L\frac{di}{dt}$

emf= 0.080V, and rate of chage of current,$\frac{di}{dt}=0.0320A{s}^{-1}$

$L=\frac{e}{\left(\frac{di}{dt}\right)}$

L= $\frac{0.080}{0.0320}=2.56H$

Since all the three inductance are identical.

So P=Q=S= 2.56 H

and S are in parallel. So equiv. inductance of Q and S

$\frac{2.5\times 2.5}{2.5+2.5}=1.25H$

Now, it is in series with P,

So Equiv. inductance is= $2.5+1.25=3.75H=\frac{15}{4}H$