Exam 4 Problem 1 Solution
1250
F 16
EX:
In the circuit shown above, the switch opens at t = 0.
a)
Find the numerical value of the energy stored in the inductor as t approaches infinity,
wL (t → ∞) .
b)
For the circuit shown above, find the initial current through the inductor, iL (t = 0− ) .
c)
Write a numerical expression for iL(t) for t > 0.
SOL'N: a) As time approaches infinity, there is no power source, and the current in
the L will drop to zero.
wL =
1 2 1
LiL = ⋅ 4mH ⋅(0A)2 = 0J
2
2
b) To find the initial value of iL, we use t = 0–. We treat the L as a wire, and
the switch is closed.
The 42 V source is across two circuits: R2, and R1 plus L. So we can solve
for iL using only the 42 V source and R1 plus L, but the L acts like a wire.
v
42 V
iL (t = 0− ) = s =
= 75mA
R1 560Ω
c) We use the standard solution for iL.
iL (t > 0) = iL (t → ∞) + [iL (t = 0+ ) − iL (t → ∞)]e−t/τ
where τ =
L
where RTh is the resistance seen looking into the circuit
RTh
from where the L is connected (after time zero) with the independent vs
source turned off.
We have R1 and R2 in series.
RTh = R1 + R2 = 560Ω + 240Ω = 800Ω
The time constant is
τ=
L
4mH
=
= 5µs
RTh 800Ω
The current in the L cannot change instantly, so we have
iL (0+ ) = iL (0− ) = 75mA
From part (a), we have iL (t → ∞) = 0A
Substituting values into the general solution yields our answer.
iL (t > 0) = 0A + [75mA − 0A]e−t/5µs