A 1.50-H inductor is connected in series with a 200-Ω resistor through a 15.0-V dc power supply and a switch. If the switch is closed at t = 0 s, what is the energy stored by the inductor 20.0 ms after the switch is closed?
A) 0.910 mJ
B) 1.83 mJ
C) 3.65 mJ
D) 7.31 mJ
E) 5.48 mJ

Question

Quizplus · Accepted Answer

The energy stored in an inductor $W_L$ is given by:
$$W_L = \frac{1}{2}LI^2$$
where $L$ is the inductance and $I$ is the current through the inductor.

In this circuit, the switch is closed at $t=0$ s, and therefore a current starts to flow through the circuit. The time constant of the circuit $	au$ is given by:
$$	au = \frac{L}{R} = \frac{1.50\:\mathrm{H}}{200\:\Omega} = 0.0075\:\mathrm{s}$$
This means that the current will reach its steady-state value after a time of approximately $5	au = 37.5$ ms. Therefore, 20.0 ms after the switch is closed, the current is still changing and is given by:
$$I = \frac{\mathcal{E}}{R}(1-e^{-t/	au}) = \frac{15.0\:\mathrm{V}}{200\:\Omega}(1-e^{-20\:\mathrm{ms}/0.0075\:\mathrm{s}}) = 31.2\:\mathrm{mA}$$
where $\mathcal{E}$ is the emf of the power supply.

Finally, the energy stored by the inductor is:
$$W_L = \frac{1}{2}LI^2 = \frac{1}{2}(1.50\:\mathrm{H})(31.2	imes10^{-3}\:\mathrm{A})^2 = 3.65	imes10^{-3}\:\mathrm{J} = \boxed{\mathrm{(C)}}$$

A 150-H Inductor Is Connected in Series with a 200-Ω Resistor