1 you must select the two bulbs of the following

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Unformatted text preview: s (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) | e-Text 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 e-Text 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 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents...
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