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1
CHAPTER 4
Exercises
E4.1
The voltage across the circuit is given by Equation 4.8:
)
/
exp(
)
(
RC
t
V
v
i
C
−
=
in which
is the initial voltage.
At the time
1%
for which the voltage
reaches 1% of the initial value, we have
)
/
exp(
01
.
0
%
1
−
=
Taking the natural logarithm of both sides of the equation, we obtain
/
605
.
4
)
01
.
0
ln(
%
1
−
=
−
=
Solving and substituting values, we find
1%
= 4.605
RC
= 23.03 ms.
E4.2
The exponential transient shown in Figure 4.4 is given by
)
/
exp(
)
(
τ
s
−
−
=
Taking the derivative with respect to time, we have
)
/
exp(
)
(
dt
dv
−
=
Evaluating at
= 0, we find that the initial slope is
.
/
S
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.,
approaching infinity) equivalent circuit is:
From this circuit, we see that
A.
2
=
a
Then ohm’s law gives the voltage
as
V.
50
=
=
Ri
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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
=
=
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
=
V
s
This current continues to flow in
the inductor immediately after the switch opens so we have
A
5
.
1
)
0
(
=
+
.
This current must flow (upward) through
2
so the initial value of the
voltage is
V.
150
)
0
(
)
0
(
2
−
=
+
−
=
+
v
(c)
We see that the initial magnitude of
(
t
) is ten times larger than the
source voltage.
(d)
The voltage is given by
)
1000
exp(
150
)
/
exp(
)
(
1
−
−
=
−
−
=
Let us denote the time at which the voltage reaches half of its initial
magnitude as
H
.
Then we have
)
1000
exp(
5
.
0
−
=
Solving and substituting values we obtain
ms
6931
.
0
)
2
ln(
10
)
5
.
0
ln(
10
3
3
=
=
−
=
−
−
3
E4.5
First we write a KCL equation for
.
0
≥
t
∫
=
+
+
dx
x
v
L
R
0
2
0
)
(
1
)
(
Taking the derivative of each term of this equation with respect to time
and multiplying each term by
, we obtain:
0
)
(
)
(
=
+
dt
dv
The solution to this equation is of the form:
)
/
exp(
)
(
τ
K
−
=
in which
s
2
.
0
/
=
=
is the time constant and
is a constant that
must be chosen to fit the initial conditions in the circuit.
Since the initial
(
= 0+) inductor current is specified to be zero, the initial current in the
resistor must be 2 A and the initial voltage is 20 V:
=
=
+
20
)
0
(
Thus, we have
)
/
exp(
20
)
(
−
=
)
/
exp(
2
/
i
−
=
=
[]
)
/
exp(
2
2
)
/
exp(
20
2
1
)
(
1
)
(
0
0
τ
τ
τ
−
−
=
−
−
=
=
∫
E4.6
Prior to
= 0, the circuit is in DC steady state and the equivalent circuit
is
Thus we have
(0) = 1 A.
However the current through the inductor
cannot change instantaneously so we also have
(0+) = 1 A.
With the
switch open, we can write the KVL equation:
100
)
(
200
)
(
=
+
di
The solution to this equation is of the form
)
/
exp(
)
(
2
1
−
+
=
in which the time constant is
ms.
5
200
/
1
=
=
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This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Haris
 Volt

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