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 rate of change of current in the circuit 4.0 $\mu$ s after the switch is closed is A) 2.0 * 10³ A/s. B) 1.0 * 10³ A/s. C) 7.2 A/s. D) 3.0 A/s. E) 0.60 A/s.

Question

Quizplus · Accepted Answer

The rate of change of current in an RL circuit is given by $ \frac{di}{dt} = \frac{V}{L} e^{-Rt/L} $. Substituting the given values: $ V = 12.0 \, V $, $ L = 6.0 \, mH = 6.0 \times 10^{-3} \, H $, $ R = 20 \, \Omega $, and $ t = 4.0 \, \mu s = 4.0 \times 10^{-6} \, s $, we get $ \frac{di}{dt} = \frac{12.0}{6.0 \times 10^{-3}} e^{-20 \times 4.0 \times 10^{-6}/6.0 \times 10^{-3}} \approx 2.0 \times 10^{3} \, A/s $.

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