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N. Cotter PRACTICE FINAL EXAM SOLUTION: Prob 1
- (50 points)
C
R
t = 0
A
V = 100 V
R
L
v
After having been open for a long time, the switch is closed at t = 0.
R
1
= 12.5Ω R
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.
z = R
1
sC
+ sL = s
2
R
1
L
s +
LC
The characteristic roots for the quadratic equation are
s
1 ,
R
2 L
R
2 L
2
LC
or
s
1 ,
2
o
2
R
2 L
o
LC
We want an overdamped response, (real roots α
2
ω o
2 ).
R
2 L
(2) 6.25 μH
M rad/s = 1 M rad/s
Try each C value in turn.
C = 2 nF:
o
6.25 μH ⋅ 2 nF
12.5 m ⋅ 1 μ
1 M
rad/s
ω o
= 8.9M rad/s > α
2 (underdamped)
C = 250 nF:
o
6.25 μH ⋅ 250 nF
1562.5 m ⋅ 1 μ
1 M
rad/s
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:
v ( t ) = A
1
e
s 1
t
+ A
2
e
s 2
t
+ A
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:
R
v C
i L
V
A
= 100 V
v C
(∞) = 100 V
i L
(∞) = 0 A
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
- and use the same values at t = 0
(since the energy in the
circuit cannot change instantly).
v (
) = v
A
− i
L
) R
1
− v
C
) = 100 V − 4 A⋅12.5Ω − 50 V = 0 V
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
v ( t ) = v
A
− i
L
( t ) R
1
− v
C
( t )
dv ( t )
dt
di
L
( t )
dt
R
1
dv
C
( t )
dt
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.
di
L
( t )
dt
L
v
L
( t )
dv
C
( t )
dt
C
i
C
( t )
Applying these identities, we have
dv ( t )
dt
L
v
L
( t ) R
1
C
i
C
( t ).
Only now that we have differentiated do we finally evaluate the derivative
we seek at t = 0:
dv ( t )
dt
t = 0
L
v
L
) R
1
C
i
C
dv ( t )
dt
t = 0
6.25μH
⋅ 0 V⋅12.5Ω −
250 nF
i
C
Since i C
is in series with i L
, we have i C
) = i L
) = 4A.
dv ( t )
dt
t = 0
4 A
250 nF
= − 16 MV/s
Now we find A 1
and A 2
v (
) = 0 = A
1
+ A
2
⇒ A
2
= − A
1
dv ( t )
dt
t = 0
= − 16 MV/s = A
1
s
1
+ A
2
s
2
= A
1
( s
1
− s
2
s
1
− s
2
2
o
2
2
o
2
2
o
2
= 2 ( 1 M )
2
− (0.8 M )
2
= ( 2 ) 0.6 M = 1.2 M
Concluding the algebra, we find the numerical values of the coefficients A 1
and A 2
A
1
16 M v/s
1.2 M
= 13.3 v/s
A
2
= −13.3 v/s
Using the values of α and ω o
from above, we find the values of s 1
and s 2
s 1
= –1 M + 0.6 M = –0.4 M
s 2
= –1 M – 0.6 M = –1.6 M
Plugging into the general form of underdamped solution completes our
answer:
v(t) = 13.3 (e
-0.4Mt
–1.6Mt ) V
