Solve the problem.
-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 $35 \%$ of its original value?
A) -4.60 L/R seconds
B) -4.40 L/R seconds
C) 1.25 L/R seconds
D) 1.05 L/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 is of the form $i(t) = i_0 e^{-\frac{R}{L}t}$, where $i_0$ is the initial current. To find the time $t$ when the current $i(t)$ falls to 35% of its original value, we set $i(t) = 0.35i_0$ and solve for $t$:$$0.35i_0 = i_0 e^{-\frac{R}{L}t}$$Dividing both sides by $i_0$ and taking the natural logarithm of both sides gives:$$\ln(0.35) = -\frac{R}{L}t$$Solving for $t$ gives:$$t = \frac{L}{R} \ln(0.35)$$Using the value of $\ln(0.35) \approx -1.05$, we get:$$t = -1.05 \frac{L}{R}$$Therefore, the correct answer is $1.05 L/R$ seconds, which corresponds to choice D.

Solve the Problem Ldidt+Ri=0\mathrm { L } \frac { \mathrm { di } } { \mathrm { dt } } + \mathrm { Ri } = 0Ldtdi​+Ri=0

Solve the Problem $\mathrm { L } \frac { \mathrm { di } } { \mathrm { dt } } + \mathrm { Ri } = 0$