A $1.5 - \mathrm { H }$ inductor is connected in series with a $200 - \Omega$ resistor through an ideal $15 - \mathrm { V }$ dc power supply and an open switch. After closing the, what is the maximum energy that will be contained in the inductor?
A) $3.2 \mathrm {~mJ}$
B) $2.2 \mathrm {~mJ}$
C) $1.2 \mathrm {~mJ}$
D) $4.2 \mathrm {~mJ}$
E) $5.2 \mathrm {~mJ}$

Question

Quizplus · Accepted Answer

The maximum energy stored in an inductor is given by the formula $E = \frac{1}{2} L I^2$, where $L$ is the inductance and $I$ is the current. In a DC circuit, the inductor will eventually act like a short circuit after a long time, so the current through the inductor can be found using Ohm's Law, $I = \frac{V}{R}$, where $V$ is the voltage and $R$ is the resistance. Here, $I = \frac{15V}{200\Omega} = 0.075A$. Substituting $I$ and $L = 1.5H$ into the energy formula gives $E = \frac{1}{2} \times 1.5 \times (0.075)^2 = 0.00421875 \mathrm{J}$ or $4.2 \mathrm{mJ}$.

A 1.5−H1.5 - \mathrm { H }1.5−H Inductor Is Connected in Series with A

A $1.5 - \mathrm { H }$ Inductor Is Connected in Series with A