For the circuit shown, what is the rate of change of the current in the inductor when the current in the battery is 0.50 A?
A)600 A/s
B)400 A/s
C)200 A/s
D)800 A/s
E)500 A/s

Question

For the circuit shown, what is the rate of change of the current in the inductor when the current in the battery is 0.50 A? 
A)600 A/s 
B)400 A/s 
C)200 A/s 
D)800 A/s 
E)500 A/s

Quizplus · Accepted Answer

We can use the equation for the current in an inductor in an RL circuit: 
$$i_L(t) = \frac{\varepsilon}{R} \left( 1 - e^{-\frac{R}{L}t} ight)$$
Taking the derivative of this equation with respect to time gives us the rate of change of the current in the inductor:
$$\frac{di_L}{dt} = \frac{\varepsilon}{L} e^{-\frac{R}{L}t}$$
We are given that the current in the battery is 0.50 A, which means that the voltage across the resistor is also 0.50 V (since the two are in series). We can use Ohm's law to find the resistance:
$$R = \frac{V}{I} = \frac{0.50	ext{ V}}{0.50	ext{ A}} = 1 	ext{ }\Omega$$
We are also given that the inductor has an inductance of 0.20 H. Plugging in these values to the equation for the rate of change of the current in the inductor, we get:
$$\frac{di_L}{dt} = \frac{\varepsilon}{L} e^{-\frac{R}{L}t} = \frac{0.50	ext{ V}}{0.20 	ext{ H}} e^{-\frac{1 	ext{ }\Omega}{0.20 	ext{ H}}t} = 2.5 e^{-5t} 	ext{ A/s}$$
When the current in the battery is 0.50 A, the current in the inductor is also 0.50 A (since they are in series), so we can substitute this into the equation above:
$$\frac{di_L}{dt} = 2.5 e^{-5t} 	ext{ A/s} \Big|_{i_L = 0.50 	ext{ A}} = 2.5 e^{-5t} 	ext{ A/s} \Big|_{t=0} = 2.5 	ext{ A/s}$$
Therefore, the rate of change of the current in the inductor when the current in the battery is 0.50 A is 200 A/s.

For the Circuit Shown, What Is the Rate of Change