Download EX: In the circuit shown below, the switch closes at time t = 0. a) Find

PRACTICE EXAM 4 PROBLEM 1 SOLUTION
1250
F 14
EX:
In the circuit shown below, the switch closes at time t = 0.
a)
Find the time constant for the circuit.
b)
Write a numerical expression for vC(t) for t > 0 in the above circuit. Note:
vC(t = 0) = 0 V.
c)
At what time is the energy stored on the capacitor, wC, equal to 10 µJ?
SOL'N: a) The time constant is τ = RThC where RTh is the Thevenin equivalent
resistance seen looking into the terminals where the C is attached. Here,
we may turn off the independent source and look into the circuit to find
RTh. Note that the switch is closed, since we are finding the behavior of
the circuit after time zero.
We have
τ = RThC = R1 || R2 ⋅C .
or
6
τ = R1 || R2 ⋅C = 60 kΩ ⋅ 2||3⋅ 0.5 µF = 60 kΩ ⋅ ⋅ 0.5 µF = 36 ms
5
b) For vC, we use the general solution for RC circuits:
v(t) = [v(t = 0+ ) − v(t → ∞)]e−t / τ + v(t → ∞)
The problem statement gives v(t = 0) = 0V , which is also the value of
vC(0+) since vC cannot change instantly.
The final value of vC is the value of the Thevenin equivalent voltage seen
from the terminals where C is attached.
This is a voltage-divider problem.
v(t → ∞) = vTh = vs
R2
180 kΩ
= 5V
= 3V
R1 + R2
120 kΩ + 180 kΩ
We now plug the values into the general solution.
v(t) = [v(t = 0 + ) − v(t → ∞)]e−t /τ + v(t → ∞)
or
v(t) = [0 − 3V]e−t /τ + 3V = 3V[1− e−t /τ ]
c) The energy stored in a capacitor depends only on the capacitor voltage and
not the history of the voltage.
1
wC = CvC2 = 10 µ J
2
We use the expression for vC from part (b).
1
wC = (0.5 µF)[3V(1− e−t /36 ms )]2 = 10 µ J
2
or, if we cancel the µJ from both sides,
9
(1− e−t /36 ms )2 = 10
4
or
⎛ 4⎞
(1− e−t /36 ms )2 = 10 ⎜ ⎟
⎝ 9⎠
Now we take the square root of both sides.
1− e−t /36 ms =
40 2
=
10
9
3
Next, we isolate the exponential term.
−e−t /36 ms =
2
10 − 1
3
or
e−t /36 ms = 1−
2
10
3
Now we take the log of both sides.
⎛ 2
⎞
−t / 36 ms = ln ⎜ 1−
10 ⎟
⎝
⎠
3
We multiply both sides by –36 ms.
⎛ 2
⎞
t = −36 ms ⋅ ln ⎜ 1−
10 ⎟
⎝
⎠
3
When we plug the numbers into a calculator, we get an error. We are
trying to take the log of a negative number, which causes the error. What
went wrong? The mathematical approach was correct, but the problem is
that the C never has 10 µJ of stored energy. The energy peaks at time
1
approaching infinity at wC = 0.5 µF(3V )2 = 2.25 µ J , which is less than
2
10 µJ. Thus, the answer to the question is "never".
Note, however, that the solution procedure shown above will work when
there is a time that may be solved for.