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
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.
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 =
CORRECTION: T = 1.15 Milliseconds.
To solve the given problem, we need to understand the concept of inductive time constant and the behavior of RL circuits. Let's break down the problem step by step:
(a) In order to find the inductive time constant (τ) of the circuit, we need to use the formula τ = L/R, where L is the inductance and R is the resistance. In this case, L = 5.00 mH and R = 7.00 Ω.
Substituting these values into the formula, we get:
τ = 5.00 mH / 7.00 Ω = 0.714 ms
So the inductive time constant of the circuit is 0.714 ms.
(b) To calculate the current in the circuit 250 µs after the switch is closed, we can use the equation for the current in an RL circuit:
I = (e/R) * (1 - e^(-t/τ))
Since we're given e = 6 V and τ = 0.714 ms, and t = 250 µs = 0.25 ms, we can substitute these values into the equation to find the current.
I = (6 V / 7.00 Ω) * (1 - e^(-0.25 ms / 0.714 ms))
Using a calculator to evaluate the exponential term, we find:
I ≈ 4.989 A
Therefore, the current in the circuit 250 µs after the switch is closed is approximately 4.989 A.
(c) To find the value of the final steady-state current in an RL circuit, we need to consider the behavior of the inductor and the resistor. In the steady state, the inductor acts like a short circuit, allowing current to flow freely through it. Therefore, the current will eventually settle to a constant value determined by the voltage source and the resistance.
The final steady-state current (I_ss) can be found using Ohm's Law: I_ss = e / R
Substituting the given values, we have:
I_ss = 6 V / 7.00 Ω ≈ 0.857 A
So, the value of the final steady-state current is approximately 0.857 A.
(d) To find the time it takes for the current to reach 80% of its maximum value, we need to use the time constant (τ) and the exponential decay equation for an RL circuit:
I(t) = I_ss * (1 - e^(-t/τ))
We want to solve for t when I(t) equals 80% of the maximum current, i.e., 0.80 * I_ss.
0.80 * I_ss = I_ss * (1 - e^(-t/τ))
Dividing both sides by I_ss and rearranging the equation, we get:
0.80 = 1 - e^(-t/τ)
Rearranging further, we find:
e^(-t/τ) = 0.20
Taking the natural logarithm (ln) of both sides, we have:
-t/τ = ln(0.20)
Solving for t, we get:
t = -τ * ln(0.20)
Substituting the given value of τ = 0.714 ms into the equation and evaluating, we find:
t ≈ -0.714 ms * ln(0.20) ≈ 1.28 ms
Therefore, it takes approximately 1.28 ms for the current to reach 80% of its maximum value.