A $1.50 - \mathrm { H }$ inductor is connected in series with a $200 - \Omega$ resistor through an ideal $15.0$ - $\mathrm { V }$ dc power supply and an open switch. How much energy is contained in the inductor $20.0 \mathrm {~ms}$ after closing the switch?
A) $3.65 \mathrm {~mJ}$
B) $0.910 \mathrm {~mJ}$
C) $1.83 \mathrm {~mJ}$
D) $5.48 \mathrm {~mJ}$
E) $7.31 \mathrm {~mJ}$

Question

Quizplus · Accepted Answer

The energy stored in an inductor is given by $E = \frac{1}{2} L I^2$, where $L$ is the inductance and $I$ is the current through the inductor. After closing the switch, the current through the inductor increases according to $I = I_0(1 - e^{-\frac{Rt}{L}})$, where $I_0 = \frac{V}{R}$ is the final steady-state current, $V$ is the voltage of the power supply, $R$ is the resistance, and $t$ is the time after the switch is closed. Substituting the given values, $I_0 = \frac{15.0 \, \text{V}}{200 \, \Omega} = 0.075 \, \text{A}$, and for $t = 20.0 \, \text{ms} = 0.020 \, \text{s}$, $I = 0.075(1 - e^{-\frac{200 \times 0.020}{1.50}})$. Calculating $I$ and then substituting into the energy formula gives approximately $3.65 \, \text{mJ}$.

A 1.50−H1.50 - \mathrm { H }1.50−H Inductor Is Connected in Series with A

A $1.50 - \mathrm { H }$ Inductor Is Connected in Series with A