QUIZ 4 PROBLEM 1 SOLUTION
1250
F 14
EX:
In the circuit shown below, the switch closes at t = 0.
a)
Find iL (t → ∞) .
b)
Write a numerical expression for iL (t > 0)
c)
Find the time, t1, when the energy stored in the inductor has dropped by 1/e2
compared to its initial value at t = 0. That is, find t1 such that
w (t = 0)
wL (t1 ) = L 2
.
e
SOL'N: a) As time approaches infinity, the switch is closed, and the L acts like a wire
(because current is no longer changing implies v = Ldi/dt = 0 V).
The switch bypasses the current source and resistor. If we consider a
voltage loop around the outside, we have only R2. Thus, the voltage drop
across R2 must be zero. It follows from Ohm's law that the current in R2
must be zero. Since the current in R2 is the same as the current in L, we
must have zero current in L as time approaches infinity.
iL (t → ∞) = 0 A
b) We use the general solution for iL(t):
i(t) = [i(t = 0+ ) − i(t → ∞)]e−t / τ + i(t → ∞)
where τ = L/RTh where RTh is the Thevenin equivalent resistance as seen
from the terminals where L is connected.
Since we have only an independent source, we may turn off the source and
look into the circuit to find the Thevenin equivalent resistance:
We see only R2, since the switch, being closed, bypasses the current
source and R1.
RTh = R2 and τ = L / RTh =
200 µH 20
=
ns ! 6.7 ns
30 kΩ
3
To find the initial current in L, we consider the circuit at time t = 0–.
The switch is open just before the switch is closed, and the L acts like a
wire. Inspection of the circuit reveals that it is a current-divider.
iL (t = 0 − ) = is
R1
20 kΩ
= 10 mA
= 4 mA
R1 + R2
30 kΩ + 20 kΩ
We substitute into the general solution to obtain our answer:
iL (t) = [iL (t = 0 + ) − iL (t → ∞)]e−t /τ + iL (t → ∞)
or
iL (t) = [4 mA − 0]e−t /6.67 ns + 0 = 4e−t /6.67 ns mA
c) The energy stored in L is proportional to the square of iL(t):
wL (t) =
1 2
LiL (t)
2
Given the answer to part (b), we can write the energy as follows:
wL (t) =
2 1
2
1 ⎡
L iL (t = 0)e−t /τ ⎤ = LiL2 (t = 0) ⎡ e−t /τ ⎤
⎦
⎣
⎦
2 ⎣
2
wL (t) =
1 2
LiL (t = 0)e−2t /τ .
2
or
If we want to scale the energy by a factor of 1/e2 = e–2, we need only
choose t = 2τ. Thus, our answer is
t = τ ! 6.67 ns