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 formula for the rate of change of current in an inductor, which is given by:
$$\frac{di_L}{dt}=\frac{1}{L}\cdot V_{	ext{induced}}$$
where $L$ is the inductance of the inductor.
When the current in the battery is 0.50 A, the current through the resistor will also be 0.50 A since they are in series. This means that the voltage drop across the resistor is given by:
$$V_R=IR=0.50	ext{ A}\cdot 2	ext{ }\Omega=1	ext{ V}$$
Since the inductor and resistor are in series, the induced voltage across the inductor must also be equal to 1 V. Therefore, we have:
$$\frac{di_L}{dt}=\frac{1}{L}\cdot V_{	ext{induced}}=\frac{1}{2	ext{ H}}\cdot 1	ext{ V}=0.5	ext{ A/s}$$
which is closest to choice C: 200 A/s.

For the Circuit Shown, What Is the Rate of Change