Download Exercise 2

Circuit Analysis Exercise 2
Exercise Ch.2 Resistive Circuits
Problem 1. (Hambley 2.3)
Problem 2. (Hambley 2.6)
Problem 3. (Hambley 2.7)
Problem 4. (Hambley 2.8)
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Circuit Analysis Exercise 2
Problem 5. (Hambley 2.12)
Problem 6. (Hambley 2.16)
Problem 7. (Hambley 2.17)
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Circuit Analysis Exercise 2
Problem 8. (Hambley 2.20)
Problem 9. (Hambley 2.21)
Problem 10. (Hambley 2.27)
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Circuit Analysis Exercise 2
Problem 11 (Hambley 2.34)
Problem 12 (Hambley 2.39)
Problem 13 (Hambley 2.47)
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Circuit Analysis Exercise 2
Problem 14. (Rizzoni 2.45)
Problem 2.45
Solution:
Known quantities:
Circuits of Figure 2.45.
Find:
Equivalent resistance and current i
Analysis:
Step1:
4| 4| 22 24Ω
Step 2:
24|8
| 6Ω
Therefore, the equivalent circuit is as shown in the figure:
Further,
46| 90
| 9 W 
The new equivalent circuit is shown below.
Thus,
Rt ot al
10Ω.
We can now find the current i by the current divider rule as
follows:
 10 
i 
 5   0 .5 A
 1 0 +9 0 
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Circuit Analysis Exercise 2
Problem 15. (Rizzoni 2.46)
Solution:
Known quantities:
Circuits of Figure 2.46.
Find:
Resistance R
Analysis:
Combining the elements to the right of the 15  resistor, we compute
Re q
4|4|6|24
| 410
Ω.
The power dissipated by the 15- resistor is
P
15Ω 
therefore,
v2
 15 W ,
15
v15= 15 V and i1 = 1 A.
Using the current divider rule:
Applying KCL, we can find iR:
i
2

15
i1 1.5 A.
10
iR i1 i2 2.5A.
2 5 2.5 R  1 5 0
Using KVL:
Therefore, R = 4.
Problem 16. (Rizzoni 2.47)
Solution:
Known quantities:
Circuits of Figure 2.47.
Find:
Equivalent resistance.
Analysis:
2Ω  2Ω  4Ω
6Ω 12Ω  4Ω
4Ω 4Ω  2Ω
4Ω 4Ω  2Ω
2Ω  2Ω  4Ω
Req  3Ω  4Ω 4Ω  5Ω
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Circuit Analysis Exercise 2
Problem 17. (Rizzoni 2.48)
Solution:
Known quantities:
Circuits of Figure 2.48.
Find:
a)
Equivalent resistance
b) Power delivered.
Analysis:
(a)
2  1  3
3 3  1.5
4  1.5  5  1 0.5
1 0.5 6  3.8 1 8
Req  3.8 1 8  7  1 0.8 1 8
(b)
14V
 1.29A
10.818Ω
P  14V1.29A  18.06W
I
Problem 18. (Rizzoni 2.49)
Solution:
Known quantities:
Circuits of Figure 2.49.
Find:
Equivalent resistance.
Analysis:
( 6R| R4)  R5  ( 1 ,|01 00
00
)  9 . 
1 1 00, resulting in
the circuit shown below.
Therefore, the equivalent resistance is
Re q 1 0| 03|| R2
|  RR1  1 0| 1 0| 0| 1 0| 0  0
 50 5
R1
.26 Ω
R
2
R
3
100 ?
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