A series LR circuit includes a 9.0 V battery, a resistance of 0.50 Ω, and an inductance of 0.80 H. What is the induced emf 2.0 s after the switch has been closed?
A) 0.0
B) 9.0 V
C) 5.2 V
D) 2.6 V
E) 19 mV

Question

Quizplus · Accepted Answer

After closing the switch, the current in the circuit gradually increases, so there is an induced emf in the inductor that opposes this change. The formula for the induced emf in an inductor is given by:
emf = -L (dI/dt)
where L is the inductance and dI/dt is the rate of change of current.
At t=0 (right after the switch is closed), the current is zero, so the induced emf is also zero. 
At t=2.0 s, the current is:
I = V/R (1 - e^(-Rt/L))
where V is the battery voltage, R is the resistance, and e is the base of the natural logarithm. 
Plugging in the values given in the problem, we get:
I = 9.0/0.50 (1 - e^(-0.50*2.0/0.80)) = 9.0/0.50 (1 - e^(-1.25)) = 18 (1 - 0.2865) = 12.17 A
Using the formula for induced emf with dI/dt = I/2.0, we get:
emf = -0.80 (12.17/2.0) = -4.86 V
However, the question asks for the magnitude of the induced emf, so the answer is 4.86 V rounded to 2 significant figures, which is approximately 2.6 V (choice D).

A Series LR Circuit Includes a 9