In the series R-L circuit shown above, R = 0.250 Ohms, L= 375 mH, and the batter
ID: 2260786 • Letter: I
Question
In the series R-L circuit shown above, R = 0.250 Ohms, L= 375 mH, and the battery's emf is 10.0 V. (Assume that the solenoid and battery both have zero internal resistance.) The switch is closed at time = 0.
a. What will be the value of the current in the circuit after a very long period of time (as t -> infinity)?
b. What is the energy stored in the inductor after a very long period of time?
c. How much time is needed for the current in the above circuit to reach 92% of its eventual maximum value?
d. What is the power being dissipated by the resistor at t = 2.50 s?
Please show work for all the above.
In the series R-L circuit shown above, R = 0.250 Ohms, L= 375 mH, and the battery's emf is 10.0 V. (Assume that the solenoid and battery both have zero internal resistance.) The switch is closed at time = 0. What will be the value of the current in the circuit after a very long period of time (as t -> infinity)? What is the energy stored in the inductor after a very long period of time? How much time is needed for the current in the above circuit to reach 92% of its eventual maximum value? What is the power being dissipated by the resistor at t = 2.50 s?Explanation / Answer
a)
for t = infinity, the inductor will get shorted.So, only resistor remains in the circuit.
So, current,I = voltage/resistance = E/R = 10/0.25 = 40 A <--------answer
2)
energy stored in an inductor is given by:
U = 0.5*L*Im^2
here L = inductance = 375 mH = 375*10^-3 H
Im = maximum current at t ---> infinity = 40 A
So, U = 0.5*375*10^-3*40^2 = 300 J <-------answer
3)
the current in the circuit at a particular time t is given by:
I = Io*(1-e^(-t*R/L))
where, Io = current at t--->infinity = 40 A
t = time
R = resistance = 0.25 ohm
L = inductance = 375*10^-3 H
So, for I to be 92 percent of eventual maximum(ie Io = 40 A)
So, I = (92/100)*Io
So, for I = 0.92*Io = Io*(1-e^(-t*0.25/(375*10^-3)))
=> 0.92 = 1-e^(-t*0.6667)
So, t = 3.79 s <-------------answer
4)
power dissipated at a given time t is given by:
P = I^2*R
but I = Io*(1-e^(-t*R/L))
So, P = (Io*(1-e^(-t*R/L)))^2*R
So, for t = 2.5 s,
P = (40*(1-e^(-2.5*0.25/(375*10^-3))))^2*0.25
= 263.2 W <-------- answer