Remarks The values used in this problem were taken from actual components salvaged from the starter system of a car. Because the current in such an RL circuit is initially zero, inductors are sometimes referred to as "chokes" because they temporarily choke off the current. In solving part $($d$),$ we traversed the circuit in the direction of positive current, so the voltage difference across the battery was positive and the differences across the resistor and inductor were negative.

Question After two time constants of time have elapsed, the current in the circuit is how many times the final current?

EXERCISEHINTS:$$GETTING STARTED $|$ I$\text{'}$M STUCK!

A $12\mathrm{.}6-$V battery is in series with a resistance of $0\mathrm{.}333$ and an inductor.

$($a$)$ After a long time, what is the current in the circuit?

I$($t $=$ $)=$ A

$($b$)$ What is the current after one time constant?

I$($t $=)=$ A

$($c$)$ What's the voltage drop across the inductor at this time?

$$VL $=$ V

$($d$)$ Find the inductance if the time constant is $0\mathrm{.}105$ s$.$

L $=$ H