Download Chapter 02-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity!
The voltage across a 5-kΩ resistor is 16 V. Find the current through the resistor.
Chapter 2, Solution 1
v = iR i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Problem 2.
Find the hot resistance of a lightbulb rated 60 W, 120 V.
Chapter 2, Solution 2
p = v^2 /R → R = v^2 /p = 14400/60 = 240 ohms
Chapter 2, Problem 3.
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 240 Ω at room temperature, what is the cross-sectional radius of the bar?
Chapter 2, Solution 3
For silicon, ρ = 6.4 10 x^2 Ω-m. A = π r^2. Hence,
2 2 2 2
L L L x x x R r A r R x
− = = ⎯⎯→ = = =
r = 0.1843 m
Chapter 2, Problem 4.
(a) Calculate current i in Fig. 2.68 when the switch is in position 1. (b) Find the current when the switch is in position 2.
Chapter 2, Solution 4
(a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA
For the network graph in Fig. 2.69, find the number of nodes, branches, and loops.
Chapter 2, Solution 5
n = 9; l = 7; b = n + l – 1 = 15
Chapter 2, Problem 6.
In the network graph shown in Fig. 2.70, determine the number of branches and nodes.
Chapter 2, Solution 6
n = 12; l = 8; b = n + l –1 = 19
Chapter 2, Problem 7.
Determine the number of branches and nodes in the circuit of Fig. 2.71.
Figure 2.71 For Prob. 2.7.
12 V _ 8 Ω 5 Ω 2 A
Chapter 2, Solution 7
6 branches and 4 nodes.
In the circuit in Fig. 2.67 decrease in R 3 leads to a decrease of: (a) current through R 3 (b) voltage through R 3 (c) voltage across R 1 (d) power dissipated in R 2 (e) none of the above
Chapter 2, Solution 10
I 1
I 2
4A
3A
-2A
At node 1, 4 + 3 = i 1 i 1 = 7A At node 3, 3 + i 2 = -2 i 2 = -5A
Chapter 2, Problem 11.
In the circuit of Fig. 2.75, calculate V 1 and V 2.
V 1
_
V 2
_
5 V
_
1 V – + 2 V –
Figure 2.75 For Prob. 2.11. Chapter 2, Solution 11
− V 1 (^) + 1 + 5 = 0 ⎯⎯→ V 1 =6 V − 5 + 2 + V 2 (^) = 0 ⎯⎯→ V 2 =3 V
In the circuit in Fig. 2.76, obtain v 1, v 2 , and v 3.
Chapter 2, Solution 12
V 1
+ V 2 -
V 3
– 25V + 10V^ -
+ 15V -
20V
LOOP
LOOP
LOOP
For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35v For loop 2, -10 +15 -v 2 = 0 v 2 = 5v For loop 3, -V 1 + V 2 + V 3 = 0 v 3 = 30v
Given the circuit in Fig. 2.78, use KVL to find the branch voltages V 1 to V 4.
V 2
V 4
V 1
V 3
3 V
4 V 5 V
2 V
Figure 2.
Chapter 2, Solution 14
3V V 1 I 4 V 2
- I 3 - + 2V - +
- + V 3 - + +
4V
I 2 - I 1
V 4 5V
For mesh 1,
− V 4 + 2 + 5 = 0 ⎯ →⎯ V 4 = 7 V
11
8
For mesh 2,
- 4 + V 3 + V 4 = 0 ⎯ →⎯ V 3 = − 4 − 7 = − V
For mesh 3,
− 3 + V 1 − V 3 = 0 ⎯ →⎯ V 1 = V 3 + 3 = − V
For mesh 4,
− V 1 − V 2 − 2 = 0 ⎯ →⎯ V 2 = − V 1 − 2 = 6 V
Thus, V 1 = − 8 V , V 2 = 6 V , V 3 = − 11 V , V 4 = 7 V
