Question

Consider a series circuit including a 4Ω, a 16mH-inductor, and a

power supply of 9 volts.

(a) Find the time constant of the RL series circuit?

(b) Find the current in the circuit 100μs after the switch is closed?

(c) How many times constant does it take for the current to reach 99.9% of its

final value?

Answer 1

(a) The time constant is

$$\tau=\frac LR=0.004\text{ s}.$$

(b) Current:

$$I=\frac{V}{R}\Big(1-e^{(-t/\tau)}\Big)=0.056\text{ A}.$$

(c) The number of time constants:

$$0.999=\Big(1-e^{(-N)}\Big),\\\space\\ N=7.$$