1
CHAPTER 4
Solutions for Exercises
E4.1
The voltage across the circuit is given by Equation 4.8:
)
/
exp(
)
(
RC
t
V
t
v
i
C
−
=
in which
V
i
is the initial voltage.
A t the time
t
1%
for which the voltage
reaches 1% of the initial value, we have
)
/
exp(
01
.
0
%
1
RC
t
−
=
Taking the natural logarithm of both sides of the equation, we obtain
RC
t
/
605
.
4
)
01
.
0
ln(
%
1
−
=
−
=
Solving and substituting values, we find
t
1%
= 4.605
RC
= 23.03 ms.
E4.2
The exponential transient shown in Figure 4.4 is given by
)
/
exp(
)
(
τ
t
V
V
t
v
s
s
C
−
−
=
Taking the derivative with respect to time, we have
)
/
exp(
)
(
τ
τ
t
V
dt
t
dv
s
C
−
=
Evaluating at
t
= 0, we find that the initial slope is
.
/
τ
S
V
Because this
matches the slope of the straight line shown in Figure 4.4, we have shown
that a line tangent to the exponential transient at the origin reaches the
final value in one time constant.
E4.3
(a) In dc steady state, the capacitances act as open circuits and the
inductances act as short circuits. Thus the steadystate (i.e.,
t
approaching infinity) equivalent circuit is:
From this circuit, we see that
A.
2
=
a
i
Then ohm’s law gives the voltage
as
V.
50
=
=
a
a
Ri
v
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(b)
The dc steadystate equivalent circuit is:
Here the two 10
Ω
resistances are in parallel with an equivalent
resistance of 1/(1/10 + 1/10) = 5
Ω
.
This equivalent resistance is in
series with the 5
Ω
resistance.
Thus the equivalent resistance seen by
the source is 10
Ω
, and
A.
2
10
/
20
1
=
=
i
Using the current division
principle, this current splits equally between the two 10
Ω
resistances,
so we have
A.
1
3
2
=
=
i
i
E4.4
(a)
ms
1
100
/
1
.
0
/
2
=
=
=
R
L
τ
(b)
Just before the switch opens, the circuit is in dc steady state with
an inductor current of
A.
5
.
1
/
1
=
R
V
s
This current continues to flow in
the inductor immediately after the switch opens so we have
A
5
.
1
)
0
(
=
+
i
.
This current must flow (upward) through
R
2
so the initial value of the
voltage is
V.
150
)
0
(
)
0
(
2
−
=
+
−
=
+
i
R
v
(c)
We see that the initial magnitude of
v
(
t
) is ten times larger than the
source voltage.
(d)
The voltage is given by
)
1000
exp(
150
)
/
exp(
)
(
1
t
t
R
L
V
t
v
s
−
−
=
−
−
=
τ
τ
Let us denote the time at which the voltage reaches half of its initial
magnitude as
t
H
.
Then we have
)
1000
exp(
5
.
0
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 Fall '05
 KELLEY/SEYLER
 Steady State, Volt

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