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(a)
cost = $0.60 (standard size) R 2 = 90 ohms
R 3 = 100 ohms cost = $0.90 (standard size)
(c)
cost = $0.75 (nonstandard size)
The system should be designed for a minimum cost such that I = 1.2 A ± 5% .
(b) I
+
70 W
Power
Supply Rx Ry −
Figure 2.1
Since we need two of the three bulbs, there are only three possibilities.
(a)
Use R 1 and R 2 . R = R 1 R 2 = 80 90 = 42.35 ohms
I = 1.2 ± 5% = 1.2 ± 0.06 = 1.26, 1.14 amps
(1.26) 2 (42.35) = 67.23 W
p = I2R = 2
(1.14) (42.35) = 55.04 W
cos t = $0.60 + $0.90 = $1.50
Use R 1 and R 3 .  v v (b)  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents R = R 1 R 3 = 80 100 = 44.44 ohms
I = 1.2 ± 5% = 1.2 ± 0.06 = 1.26, 1.14 amps
(1.26) 2 (44.44) = 70.55 W
2
p=I R=
2
(1.14) (44.44) = 57.75 W
cos t = $0.60 + $0.75 = $1.35
Use R 2 and R 3 . (c) R = R 2 R 3 = 90 100 = 47.37 ohms
I = 1.2 ± 5% = 1.2 ± 0.06 = 1.26, 1.14 amps (1.26) 2 (47.37) = 75.2 W
2
p=I R=
2
(1.14) (47.37) = 61.56 W
cos t = $0.90 + $0.75 = $1.65
Note that case (b) represents the lowest cost, however both (b) and (c) have a power that exceeds
the 70 W power that can be supplied. Therefore, the correct design uses case (a), i.e. R 1 and R 2 . Given the circuit in Figure 2.1, Problem 2.20 R2
IR1
IS IR2
Rsh R1 Figure 2.1
find I R 2 for I S = 10 mA, R sh = 5 Ω, and R1
R1
R1
R1 (a)
(b)
(c)  v v (d)  = 10 kΩ
= 1000 Ω
= 50 Ω
=5 Ω and
and
and
and eText Main Menu R2
R2
R2
R2 = 10 kΩ
= 1000 Ω
= 50 Ω
=5 Ω  Textbook Table of Contents  Problem Solving Workbook Contents Using current division, R1
IR2 = IS R 1 + (R 2 + R sh ) 10 × 10 3
100 −3
IR2 = (10 × 10 ) = 2.001 × 10 4 = 4.998 mA
3
3 (10 × 10 ) + (10 × 10 ) + 5 (a) 1 × 10 3
10 −3
IR2 = (10 × 10 ) = 2.005 × 10 3 = 4.987 mA
3
3 (1 × 10 ) + (1 × 10 ) + 5 50 0.5 −3
IR2 = (10 × 10 ) = 105 = 4.762 mA 50 + 50 + 5 5
(10 × 10 −3 ) = 0.05 = 3.333 mA
IR2 = 15 5 + 5 + 5 (b)
(c)
(d) In summary,
I R 2 = 4.998 mA
(a)
(b) I R 2 = 4.987 mA (c) I R 2 = 4.762 mA (d) I R 2 = 3.333 mA These answers can be compared to the case where R sh = 0 to see just how much the current
through R 2 is affected by the internal resistance of the real ammeter. In each case, it can be
shown that  v v R1 IR2 = I = (0.5)(10 × 10 3 ) = 5 mA
R1 + R 2 S  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents...
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 Spring '12
 bkav

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