A coil with a self-inductance of $1 \mathrm{H}$ is connected in parallel to a $10-\Omega$ light bulb and a $5-\mathrm{V}$ battery. Very roughly how long would it keep the light bulb lit if the battery was disconnected from the circuit? (Choose the closest answer.)
A) $100 \mathrm{~s}$
B) $20 \mathrm{~s}$
C) $2 \mathrm{~s}$
D) $\mathbf{0 . 2 \mathrm { s }}$
E) $0.02 \mathrm{~s}$

Question

Quizplus · Accepted Answer

The energy stored in the inductor can be calculated using the formula $E = \frac{1}{2}LI^2$, where $L$ is the inductance and $I$ is the current. When the battery is disconnected, this energy will dissipate through the light bulb at a rate given by $P = I^2R$, where $P$ is the power and $R$ is the resistance of the bulb. The time $t$ for the energy to dissipate can be estimated by $t = \frac{E}{P}$. Given the values, the time will be relatively short, making 0.2 s the closest estimate without performing detailed calculations.

A Coil with a Self-Inductance Of 1H1 \mathrm{H}1H Is Connected in Parallel to A

A Coil with a Self-Inductance Of $1 \mathrm{H}$ Is Connected in Parallel to A