Question

In an RL series circuit, an inductor of 3.54 H and a resistor of 7.76 Ω are connected to a 26.6 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor?

Answer 1

Answer:

Energy stored in inductor will be 20.797 J

Explanation:

We have given inductance L = 3.54 H

And resistance R = 7.76 ohm

Battery voltage V = 26.6 VOLT

After very long time means at steady state inductor behaves as short circuit

So  current $i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp$

Now energy stored in inductor $E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J$

So energy stored in inductor will be 20.797 J