Question

An inductor with inductance L=25.4 mH and a resistor with resistance R=5.6 are wired in series to an ideal battery with voltage V=12 Volts. A switch in the circuit is closed at time t=0 seconds, at which time the current is zero. What is the energy stored in the inductor?

Answer 1

Answer:

Current in the circuit at the very instant after the switch is closed will be also 0

Energy stored  = 58.324 mJ

Explanation:

Inductor is a device which resists change in current in the circuit. Current in the circuit was 0 before the switch is closed. Therefore current in the circuit at the very instant after the switch is closed will be also 0. Since inductor resists change in current and likes to remain as it is.

After the switch is closed, current will be growing exponentially and finally will reach a steady state. Since we are considering the inductor as ideal, the resistance in the circuit will be only due to the resistor.

By ohm's law

V = IR

I = $\frac{V}{R}$

V = 12 volts

R = 5.6 ohm

Substituting in the equation,

I = $\frac{12}{ 5.6}$

I = 2.143 A

Again, we know that the energy stored in an inductor is given by

E = $\frac{1}{2}LI^{2}$

L = 25.4 mH

I = 2.143 A

Substituting in the equation,

E = $\frac{1}{2}25.4 mH X 2.143 ^{2}$

E = 58.324 mJ