Question

A 100 mH inductor whose windings have a resistance of 4.0 Ω is connected across a 12 V battery having an internal resistance of 2.0 Ω.

How much energy is stored in the inductor?

Answer 1

Concepts and reason

The concepts required to solve this problem are ohm’s law and energy stored in an inductor.

Initially, calculate the current flowing through the inductor by using ohm’s law. Finally, calculate the energy stored in the inductor by using the relation between energy, inductance, and current.

Fundamentals

The expression for the energy stored in an inductor is as follows:

$E = \frac{1}{2}L{I^2}$

Here, L is the in inductance and I is the current.

The Ohm’s law states that the potential difference across an ideal conductor is proportional to the current flowing through it. The constant of proportionality is called as resistance.

The expression of the Ohm’s law is as follows:

$V = IR$

Here, V is the potential difference.

Substitute 12 V for V, $4.0{\rm{ }}\Omega$ for R, and $2.0{\rm{ }}\Omega$ for r in the equation $I = \frac{V}{{R + r}}$.

$\begin{array}{c}\\I = \frac{{12{\rm{ V}}}}{{4.0{\rm{ }}\Omega + 2.0{\rm{ }}\Omega }}\\\\ = \frac{{12{\rm{ V}}}}{{6.0{\rm{ }}\Omega }}\\\\ = 2.0{\rm{ A}}\\\end{array}$

Substitute 12 V for V, $4.0{\rm{ }}\Omega$ for R, and $2.0{\rm{ }}\Omega$ for r, and 100 mH for L in the equation $E = \frac{1}{2}L{\left( {\frac{V}{{R + r}}} \right)^2}$.

$\begin{array}{c}\\E = \frac{1}{2}\left( {100{\rm{ mH}}} \right){\left( {\frac{{12.0{\rm{ V}}}}{{4.0{\rm{ }}\Omega + 2.0{\rm{ }}\Omega }}} \right)^2}\\\\ = \frac{1}{2}\left( {100{\rm{ mH}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ H}}}}{{1.00{\rm{ mH}}}}} \right){\left( {2.0{\rm{ A}}} \right)^2}\\\\ = 200 \times {10^{ - 3}}{\rm{ J}}\\\\ = {\rm{0}}{\rm{.20 J}}\\\end{array}$

Ans:

The magnitude of the energy stored in the inductor is 0.20 J.