Chapter 2, Solution 3
For silicon,   6.4 x102 -m. A   r 2 . Hence,
R
L
A

L
 r2


r2 
 L 6.4 x102 x 4 x102

 0.033953
R
 x 240
r = 184.3 mm
Chapter 2, Solution 10
2
–8A
1
4A
i2
i1
3
–6A
At node 1,
–8–i 1 –6 = 0 or i 1 = –8–6 = –14 A
At node 2,
–(–8)+i 1 +i 2 –4 = 0 or i 2 = –8–i 1 +4 = –8+14+4 = 10 A
Chapter 2, Solution 12
+ 30v –
loop 2
– 50v +
+
40v
-
+ 20v –
loop 1
+
v1
–
+ v2 –
loop 3
+
v3
–
For loop 1,
–40 –50 +20 + v 1 = 0 or v 1 = 40+50–20 = 70 V
For loop 2,
–20 +30 –v 2 = 0 or v 2 = 30–20 = 10 V
For loop 3,
–v 1 +v 2 +v 3 = 0 or v 3 = 70–10 = 60 V
Chapter 2, Solution 15
Calculate v and i x in the circuit of Fig. 2.79.
12 
+ v
10 V
+
_
+ 16 V –
–
ix
+
+
4V
_
_
Figure 2.79
For Prob. 2.15.
Solution
For loop 1, –10 + v +4 = 0, v = 6 V
For loop 2, –4 + 16 + 3i x =0, i x =
–4 A
3 ix
Chapter 2, Solution 16
Determine V o in the circuit in Fig. 2.80.
16 
14 

+
10 V
+
_
Vo
+
_
25 V
_

Figure 2.80
For Prob. 2.16.
Solution
Apply KVL,
–10 + (16+14)I + 25 = 0 or 30I = 10–25 = – or I = –15/30 = –500 mA
Also,
–10 + 16I + V o = 0 or V o = 10 – 16(–0.5) = 10+8 = 18 V
Chapter 2, Problem 27.
Calculate I o in the circuit of Fig. 2.91.
8
10V
+

Io
3
6
Figure 2.91
For Prob. 2.27.
Solution
The 3-ohm resistor is in parallel with the c-ohm resistor and can be replaced by a
[(3x6)/(3+6)] = 2-ohm resistor. Therefore,
I o = 10/(8+2) = 1 A.
Chapter 2, Problem 30.
Find R eq for the circuit in Fig. 2.94.
25 
180 
60 
R eq
60 
Figure 2.94
For Prob. 2.30.
Solution
We start by combining the 180-ohm resistor with the 60-ohm resistor which in
turn is in parallel with the 60-ohm resistor or = [60(180+60)/(60+180+60)] = 48.
Thus,
R eq = 25+48 = 73 Ω.
Chapter 2, Solution 31
Req  3  2 // 4 //1  3 
1
 3.5714
1/ 2  1/ 4  1
i 1 = 200/3.5714 = 56 A
v 1 = 0.5714xi 1 = 32 V and i 2 = 32/4 = 8 A
i 4 = 32/1 = 32 A; i 5 = 32/2 = 16 A; and i 3 = 32+16 = 48 A
Chapter 2, Solution 58
The resistance of the bulb is (120)2/60 = 240
40 
2A
+ 90 V - 0.5 A
VS
240 
+
-
1.5 A
+
120 V
80 
-
Once the 160 and 80 resistors are in parallel, they have the same voltage
120V. Hence the current through the 40 resistor is equal to 2 amps.
40(0.5 + 1.5) = 80 volts.
Thus
v s = 80 + 120 = 200 V.