A battery is connected in series with a 0.20 - Ohm resistor and an inductor, as

Question

A battery is connected in series with a 0.20 - Ohm resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is 0.18 s, and the maximum current in the circuit is 7.6 A. (a) Find the emf of the battery. 2.184 v (b) Find the inductance of the circuit 50.4 mH (c) Find the current in the circuit after one time constant has elapsed. (d) Find the voltage across the resistor after one time constant has elapsed. v (e) Find the voltage across the inductor after one time constant has elapsed. v

Explanation / Answer

a.) emf E = maximum current x resistance = 7.8 x 0.28 = 2.184 A

b.) time constant of an LR circuit = L/R

0.18 = L / 0.28

L = 0.0504 H = 50.4 mH

c.) current in a charging L/R circiut is given by I = V/R ( 1 - e-Rt/L )

putting t = 0.18 (one time constant) , we get

I = 7.8 ( 1 - e-1 ) = 4.930540359 A

d.) voltage across the resistor = current through the resistor x resistance

after one time constant current = 4.930540359

So, voltage across resistor at that moment = 4.930540359 x 0.28 = 1.3805513 volts

e.) Out of the net emf E = 2.184, if 1.3805513 volts is the potential difference across the resistor at the instant after one time constant, the remaining 2.184 - 1.3805513 = 0.8034486995 volts will be the potential difference across the inductor.

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