1
APPENDIX C
PC.1
Because the capacitor voltage is zero at
t
= 0, the charge on the
capacitor is zero at
t
= 0. Then using Equation 3.5 in the text, we have
t
dx
dx
t
i
t
q
t
t
3
3
0
)
(
)
(
0
0
=
=
+
=
∫
∫
For
s,
2
µ
=
t
we have
C
6
10
2
3
)
3
(
6
µ
=
×
×
=
−
q
PC.2
Refer to Figure PC.2 in the book.
Combining the 10
Ω
resistance and the
20
Ω
resistance we obtain a resistance of 6.667
Ω
, which is in series
with the 5
Ω
resistance.
Thus, the total resistance seen by the 15V
source is 5 + 6.667 = 11.667
Ω
.
The source current is 15/11.667 = 1.286
A.
The current divides between the 10
Ω
resistance and the 20
Ω
resistance.
Using Equation 2.27, the current through the 10
Ω
resistance is
A
8572
.
0
286
.
1
10
20
20
10
=
×
+
=
i
Finally, the power dissipated in the 10
Ω
resistance is
W
346
.
7
10
2
10
10
=
=
i
P
PC.3
The equivalent capacitance of the two capacitors in series is given by
F
4
/
1
/
1
1
2
1
µ
=
+
=
C
C
C
eq
The charge supplied by the source is
C
800
10
4
200
6
µ
=
×
×
=
=
−
V
C
q
eq
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2
PC.4
The input power to the motor is the output power divided by efficiency
W
1865
80
.
0
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 Spring '08
 Kumar
 Alternating Current, Volt, Magnetic Field, output voltage, 10 resistance

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