A battery is connected in series with a 029 ohm resistor and

A battery is connected in series with a 0.29. ohm resistor and an inductor, as shown in the figure below. The switch is closed at r = 0. The time constant of the circuit is 0.24 s, and the maximum current in the circuit is 7.7 A. Find the emf of the battery. Find the current in the circuit after one time constant has elapsed. Find the inductance of the circuit.

Solution

a)

Emf of the battery

E=I_maxR =7.7*0.29

E=2.233 Volts

b)

In a RL circuit current as a function of time for charging is given by

I=I_max[1-e^-t/T]

at t=T

I=7.7[1-e^-1]

I=4.867 A

c)

Time Constant

T=L/R

So Inductance

L=TR =0.24*0.29

L=0.0696 H