A series LR circuit includes a 9.0 V battery,a resistance of 0.50 $\Omega$,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

The induced emf can be found using the formula:

emf = - L * (dI/dt)

where L is the inductance of the circuit, and dI/dt is the rate of change of current in the circuit.

At t = 2.0 s, the current in the circuit can be found using Ohm's Law:

V = IR

9.0 V = I * 0.5 Ω

I = 18.0 A

The rate of change of current can be found using the formula:

dI/dt = V/L

dI/dt = 9.0 V / 0.80 H

dI/dt = 11.25 A/s

Substituting these values into the formula for emf:

emf = - L * (dI/dt)

emf = - 0.80 H * (11.25 A/s)

emf = - 9.0 V

The negative sign indicates that the induced emf opposes the applied voltage from the battery. Therefore, the best choice is D: 2.6 V.

A Series LR Circuit Includes a 9 Ω\OmegaΩ ,And an Inductance of 0

A Series LR Circuit Includes a 9 $\Omega$ ,And an Inductance of 0