a. T.C.=L/R=5.0/7=0.714 Milliseconds
b. T/TC = 0.25mS/0.714mS = 0.350
Vi = V/e^(T/TC) = 6/e^0.350 = 4.23 Volts
after 250 uS.
I=Vr/R = (V-Vi)/R = (6-4.23)/7= 0.253A.
c. I=Vr/R = (V-Vi)/R = (6-0)/7 = 0.857A
d. Vr=I*R = (0.8*0.857)*7 = 4.80 Volts.
Vi = 6-4.8 = 1.2 Volts. = Voltage across inductor.
V/e^(T/0.35) = 1.2
6/e^2.857T = 1.2
e^2.857T = 6/1.2 = 5
2.857T*Ln e = Ln 5
2.857T = 1.609
T = 0.563 s.
Consider the circuit in Figure P32.17, taking script e = 6 V, L = 5.00 mH, and R = 7.00 .
Figure P32.17
(a) What is the inductive time constant of the circuit?
ms
(b) Calculate the current in the circuit 250 µs after the switch is closed.
A
(c) What is the value of the final steady-state current?
A
(d) How long does it take the current to reach 80% of its maximum value?
ms
3 answers
CORRECTION:
d. Vr = I*R = (0.8*0.857)*7 = 4.8 Volts
Vi=6-4.8=1.2 Volts across the inductor.
V/e^(T/0.714) = 1.2
6/e^(1.40T) = 1.2
e^1.40T = 6/1.2 = 5
1.40T*Ln e = Ln 5
1.40T = 1.609
T = 1.15 s.
7 =
d. Vr = I*R = (0.8*0.857)*7 = 4.8 Volts
Vi=6-4.8=1.2 Volts across the inductor.
V/e^(T/0.714) = 1.2
6/e^(1.40T) = 1.2
e^1.40T = 6/1.2 = 5
1.40T*Ln e = Ln 5
1.40T = 1.609
T = 1.15 s.
7 =
CORRECTION: T = 1.15 Milliseconds.
1 answer