Question

At the instant when the current in an inductor is increasing at a rate of 6.35×10−2 A/s , the magnitude of the self-induced emf is 1.65×10−2 V . You may want to review (Pages 994 - 997) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Calculating self-inductance.

Answer 1

Answer:

Self inductance, L = 0.259 Henry

Explanation:

It is given that,

Increase in current, $\dfrac{dI}{dt}=6.35\times 10^{-2}\ A/s$

Self induced emf, $\epsilon=1.65\times 10^{-2}\ V$

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

$\epsilon=L\dfrac{dI}{dt}$

$L=\dfrac{\epsilon}{dI/dt}$

$L=\dfrac{1.65\times 10^{-2}}{6.35\times 10^{-2}}$

L = 0.259 Henry

So, the self inductance of the inductor is 0.259 Henry. Hence, this is the required solution.