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

Using Kirchhoff's voltage law, we can write:

V_battery - L(di/dt) - IR = 0

where V_battery is the emf of the battery, L is the inductance, i is the current in the inductor, R is the resistance, and di/dt is the rate of change of current in the inductor.

When the current in the battery is 0.50 A, we have:

V_battery = IR + L(di/dt)

since there is no voltage drop across the resistor.

Taking the derivative of both sides with respect to time, we get:

0 = R(di/dt)^2 + L(d^2i/dt^2)

Rearranging and solving for d^2i/dt^2, we get:

d^2i/dt^2 = (-R/L)(di/dt)

Substituting the values given in the problem, we have:

d^2i/dt^2 = (-4 Ω)/(0.04 H)(0.50 A) = -200 A/s^2

Taking the absolute value, we get:

|d^2i/dt^2| = 200 A/s^2

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. The correct answer is (C).

For the Circuit Shown,what Is the Rate of Change of the Current