A series RL circuit with inductance 6.0 mH and resistance 20 $\Omega$ is connected to a 12.0-V battery of negligible internal resistance. The current in the circuit 4.0 $\mu$ s after the switch is closed is
A) 1.0 mA.
B) 4.0 mA.
C) 7.9 mA.
D) 0.40 A.
E) 0.59 A.

Question

Quizplus · Accepted Answer

The current in an RL circuit after a time $t$ can be found using the formula $I = \frac{V}{R}(1 - e^{-\frac{Rt}{L}})$, where $V$ is the voltage, $R$ is the resistance, $L$ is the inductance, and $t$ is the time. Plugging in the given values: $I = \frac{12.0}{20}(1 - e^{-\frac{20 \times 4.0 \times 10^{-6}}{6.0 \times 10^{-3}}})$, we get approximately 7.9 mA.

A Series RL Circuit with Inductance 6 Ω\OmegaΩ Is Connected to a 12

A Series RL Circuit with Inductance 6 $\Omega$ Is Connected to a 12