Question

A device is turned on and 2.60 A flows through it 0.170 ms later. What is the self-inductance of the device (in mH) if an induced 160 V emf opposes this?

Answer 1

Answer:

Self inductance of the device is 10.4 mH.

Explanation:

It is given that,

Change in current, $\Delta I=2.6\ A$

Time, $\Delta t=0.17\ ms=0.17\times 10^{-3}\ s$

Induced EMF, E = 160 V

Let L is the self inductance of the device. It is given by :

$E=L\dfrac{\Delta I}{\Delta t}$

$L=E\dfrac{\Delta t}{\Delta I}$

$L=160\ V\times \dfrac{0.17\times 10^{-3}}{2.6}}$

L = 0.0104 H

or

L = 10.4 mH

So, the self inductance of the device is 10.4 mH. Hence, this is the required solution.