If the switch is thrown open after the current in an RL circuit has built up to its steady-state value, the decaying current obeys the equation $\mathrm { L } \frac { \mathrm { di } } { \mathrm { dt } } + \mathrm { Ri } = 0$. How long after the switch is thrown open will it take the current to fall to $40 \%$ of its original value?
A) $- 4.60 \mathrm {~L} / \mathrm { R }$ seconds
B) $0.92 \mathrm {~L} / \mathrm { R }$ seconds
C) $1.12 \mathrm {~L} / \mathrm { R }$ seconds
D) $- 4.40 \mathrm {~L} / \mathrm { R }$ seconds

Question

Quizplus · Accepted Answer

The equation given is a first-order linear differential equation representing the decay of current in an RL circuit. The solution to this equation gives the current $i(t) = i_0 e^{-\frac{R}{L}t}$, where $i_0$ is the initial current. To find the time it takes for the current to fall to 40% of its original value, set $i(t) = 0.4i_0$, leading to $0.4 = e^{-\frac{R}{L}t}$. Solving for $t$ gives $t = \frac{\ln(0.4)}{-\frac{R}{L}} = 0.91629 \frac{L}{R}$, which is approximately $0.92 \frac{L}{R}$ seconds.

If the Switch Is Thrown Open After the Current in an RL