1) (a) In a series RL circuit with an R = 1.8 ohms and an inductance of 30 mH co
ID: 1290489 • Letter: 1
Question
1) (a) In a series RL circuit with an R = 1.8 ohms and an inductance of 30 mH connected to a voltage of 4.2 V at t = 0, find the time constant for this RL circuit as the inductor is energized.
(b) What is the value of the maximum current that will flow in this RL circuit?
2)A 4.2 mH inductor is initially connected to a 5.8 V battery at t = 0 through a 2.3 ohms resistor. Once the inductor is fully energized, the battery is removed and the 2.3 ohms resistor is connected between the two terminals of the inductor. Find the current flowing through the resistor 11.0e-6 sec after the inductor begins to be de-energized
Explanation / Answer
1.
a)
Time Constant
T=L/R =0.03/1.8 =0.01667 s or 16.67 ms
b)
Maximum current
Im=V/R=4.2/1.8=2.33 A
2,
Maximum Current
Io=V/R =5.8/2.3 =2.52 A
Time Constant
T=L/R =4.2*10-3/2.3=1.826*10-3s
In a RL Circuit current as a function of time is given by
I=Ioe-t/T=2.52*e-(11*10^-6/1.826*10^-3)
I=2.5 A
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