An 8.0-mH inductor and a 2.0-$\Omega$resistor are wired in series to an ideal battery.A switch in the circuit is closed at time t = 0, at which time the current is zero.The current reaches half its final value at a time of:
A)2.8 ms
B)4.0 ms
C)3.0 s
D)170 s
E)250 s

Question

Quizplus · Accepted Answer

The time constant for an LR circuit is given by $\tau = \frac{L}{R}$, where $L$ is the inductance and $R$ is the resistance. Here, $L = 8.0 \, \text{mH} = 8.0 \times 10^{-3} \, \text{H}$ and $R = 2.0 \, \Omega$, so $\tau = \frac{8.0 \times 10^{-3}}{2.0} = 4.0 \times 10^{-3} \, \text{s} = 4.0 \, \text{ms}$. The current in an LR circuit reaches (1 - 1/e) of its final value after one time constant, which is approximately 63%. To reach half (50%) of its final value, it would take less time than one time constant. Therefore, the only option less than 4.0 ms is 2.8 ms.

An 80-MH Inductor and a 2 Ω\OmegaΩ Resistor Are Wired in Series to an Ideal Battery

An 80-MH Inductor and a 2 $\Omega$ Resistor Are Wired in Series to an Ideal Battery