Answer:
Self inductance of the device is 10.4 mH.
Explanation:
It is given that,
Change in current, [tex]\Delta I=2.6\ A[/tex]
Time, [tex]\Delta t=0.17\ ms=0.17\times 10^{-3}\ s[/tex]
Induced EMF, E = 160 V
Let L is the self inductance of the device. It is given by :
[tex]E=L\dfrac{\Delta I}{\Delta t}[/tex]
[tex]L=E\dfrac{\Delta t}{\Delta I}[/tex]
[tex]L=160\ V\times \dfrac{0.17\times 10^{-3}}{2.6}}[/tex]
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.
