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

Self inductance of the device is 10.4 mH.

Step-by-step explanation:

It is given that,

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

Time,
`\Delta t=0.17\ ms=0.17* 10^(-3)\ s`

Induced EMF, E = 160 V

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

`E=L(\Delta I)/(\Delta t)`

`L=E(\Delta t)/(\Delta I)`

`L=160\ V* (0.17* 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.