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Material Type: Exam; Class: Fund Electric Circuits; Subject: Electrical & Computer Engg; University: University of Utah; Term: Unknown 1989;
Typology: Exams
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N. Cotter PRACTICE FINAL EXAM SOLUTION: Prob 1
t = 0
v
After having been open for a long time, the switch is closed at t = 0.
1
2
= 12.5Ω L = 6.25 μH
a. Two capacitances are available: 250 nF and 2 nF. Specify the value of C
that will make v(t) overdamped.
b. Using the value of C found in (a), write a time-domain expression for v(t).
ans: a) C = 250 nF
b) v(t) = 13.3 (e
-0.4Mt
–1.6Mt ) V
sol'n: (a) To make the response overdamped, we must have two real characteristic
roots. We use the circuit for t > 0, consisting of C, R 1
, L, and v A
in series.
We may find the characteristic equation by looking it up in a textbook or by
setting the impedance of R 1
, C, and L in series to zero.
1
2
1
The characteristic roots for the quadratic equation are
1 ,
2
or
1 ,
2
o
2
o
We want an overdamped response, (real roots α
2
ω o
2 ).
Try each C value in turn.
C = 2 nF:
o
ω o
= 8.9M rad/s > α
2 (underdamped)
C = 250 nF:
o
ω o
= 0.8M rad/s < α
2 (overdamped)
We need C = 250 nF for an overdamped solution.
sol'n: (b) We use the exponential solution for the overdamped case:
1
s 1
t
2
s 2
t
3
Because the value of A 3
is all that is left of v(t) as t → ∞, we first find the
constant term, A 3
. (The other terms decay because the characteristic roots
always have negative real parts in a passive RLC circuit. When the switch
opens, the energy sloshing back and forth in the L and C will decay owing to
power dissipated by the series resistor R 1
As t → ∞, the circuit reaches equilibrium. C acts like an open circuit, L acts
like a short circuit or wire.
Model:
v C
i L
v C
i L
Since L acts like a wire, there is no voltage drop across it.
Thus, A 3
= v(t→∞) = 0.
We find coefficients A 1
and A 2
by matching initial conditions in the circuit.
We find initial conditions by examining the circuit at t = 0
has reached equilibrium. We find the values of i L
and v C
, the energy
variables, at t = 0
(since the energy in the
circuit cannot change instantly).
A
L
1
C
The same equation applies for t > 0, and we may differentiate to find dv(t)/dt
in terms of energy (or state) variables i L
and v C
A
L
1
C
L
1
C
The basic equations for L and C, rearranged, allow us to translate the
derivatives on the right side of this equation into non-derivatives we can
calculate numerically.
L
L
C
C
Applying these identities, we have
L
1
C
Only now that we have differentiated do we finally evaluate the derivative
we seek at t = 0:
t = 0
L
1
C
t = 0
C
Since i C
is in series with i L
, we have i C
) = i L
) = 4A.
t = 0
Now we find A 1
and A 2
1
2
2
1
t = 0
1
1
2
2
1
1
2
1
2
2
o
2
2
o
2
2
o
2
2
2
Concluding the algebra, we find the numerical values of the coefficients A 1
and A 2
1
2
Using the values of α and ω o
from above, we find the values of s 1
and s 2
s 1
s 2
Plugging into the general form of underdamped solution completes our
answer:
v(t) = 13.3 (e
-0.4Mt
–1.6Mt ) V