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Chapter 3
3.
In this
RC
circuit,
R
= 100 Ω,
C
=10
−
12
F, the initial charge of the capacitor is
Q
=
q
(0) = 0
coulombs, and the applied constant voltage is
V
= 5 volts. Thus we can use a general equation
for the charge
q
(
t
) of the capacitor derived in Example 2. Substitution of given data yields
q
(
t
)=
CV
+[
Q
−
CV
]
e
−
t/RC
=10
−
12
(5)
±
1
−
e
−
t/
(
100
·
10
−
12
)
²
=5
·
10
−
12
±
1
−
e
−
10
10
t
²
and so
E
C
(
t
)=
q
(
t
)
C
=5
±
1
−
e
−
10
10
t
²
.
Solving the equation
E
C
(
t
) = 3, we get
5
±
1
−
e
−
10
10
t
²
=3
⇒
e
−
10
10
t
=0
.
4
⇒
t
=
−
ln 0
.
4
10
10
≈
9
.
2
×
10
−
11
(sec)
.
Therefore, it will take about 9
.
2
×
10
−
11
seconds for the voltage to reach 3 volts at the receiving
gate.
5.
Let
V
(
t
) denote the voltage across an element, and let
I
(
t
) be the current through this element.
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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