Chapter 1, Solution 1
(a)
17
q = 6.482x10 x [-1.602x10
-19
18
C] = –103.84 mC
-19
(b) q = 1. 24x10 x [-1.602x10
19
-19
(c) q = 2.46x10 x [-1.602x10
20
C] = –198.65 mC
C] = –3.941 C
-19
(d) q = 1.628x10 x [-1.602x10
C] = –26.08 C
Chapter 1, Solution 11
-3
q= it = 90 x10 x 12 x 60 x 60 = 3.888 kC
E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ
Chapter 1, Solution 17
∑ p=0
∑ -205 + 60 + 45 + 30 + p = 0
3
p = 205 – 135 = 70 W
3
Thus element 3 receives 70 W.
Chapter 1, Solution 19
I = 8 –2 = 6 A
Calculating the power absorbed by each element means we
need to find vi for each element.
p
= –8x9 = –72 W
8 amp source
p
p
element with 9 volts across it
element with 3 bolts across it
p
6 volt source
= 2x9 = 18 W
= 3x6 = 18 W
= 6x6 = 36 W
One check we can use is that the sum of the power absorbed must
equal zero which is what it does.
Chapter 1, Solution 20
p
30 volt source
= 30x(–6) = –180 W
p
12 volt element
= 12x6 = 72 W
p
28 volt e.ement with 2 amps flowing through it
p
28 volt element with 1 amp flowing through it
p
the 5Io dependent source
= 28x2 = 56 W
= 28x1 = 28 W
= 5x2x(–3) = –30 W
Since the total power absorbed by all the elements in the
circuit must equal zero,
or 0 = –180+72+56+28–30+p
or
into the element with Vo
p
into the element with Vo
Since p
into the element with Vo
= 180–72–56–28+30 = 54 W
= V x3 = 54 W or V = 18 V.
o
o
Chapter 2, Solution 17
Applying KVL around the entire outside loop we get,
–24 + v + 10 + 12 = 0 or v = 2V
1
1
Applying KVL around the loop containing v , the 102
volt source, and the 12-volt source we get,
v + 10 + 12 = 0 or v = –22V
2
2
Applying KVL around the loop containing v and the
3
10-volt source we get,
–v + 10 = 0 or v = 10V
3
3
Chapter 2, Solution 21
Applying KVL,
-15 + (1+5+2)I + 2 V = 0
x
But V = 5I,
x
-15 +8I + 10I =0, I = 5/6
V = 5I = 25/6 = 4.167 V
x
Chapter 2, Solution 35
Combining the resistors that are in parallel,
30║70 =21 Ω
, 5║20 =4 Ω
i = 200/(21+4) = 8 A
v = 21i = 168 V, v = 4i = 32 V
1
o
i = v1/70= 2.4 A, i = vo /20=1.6 A
1
2
At node a, KCL must be satisfied
i =i +I
2.4 = 1.6 + I
I = 0.8
1
2
o
o
o