761_PartUniversity Physics Solution

# 761_PartUniversity Physics Solution - Direct-Current...

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Direct-Current Circuits 26-3 EXECUTE: Using 2 /, PV R = () 22 111 2 22 28.0 V 28.0 V / 490 W, / 327 W, and 1.60 2.40 PVR == = = ΩΩ 2 2 333 28.0 V / 163 W 4.80 = Ω EVALUATE: The total power dissipated is out 1 2 3 980 W. PP P P =++= This is the same as the power in 2.80 V 35.0 A 980 W PI = E delivered by the battery. (f) 2 /. = The resistors in parallel each have the same voltage, so the power P is largest for the one with the least resistance. 26.9. IDENTIFY: For a series network, the current is the same in each resistor and the sum of voltages for each resistor equals the battery voltage. The equivalent resistance is eq 1 2 3 RR R R = ++ . 2 PIR = . SET UP: Let 1 1.60 R , 2 2.40 R , 3 4.80 R = Ω . EXECUTE: (a) eq 1.60 2.40 4.80 8.80 R = Ω+ Ω= Ω (b) eq 28.0 V 3.18 A 8.80 V I R = Ω (c) 3.18 A I = , the same as for each resistor. (d) 11 (3.18 A)(1.60 ) 5.09 V VI R Ω = . (3.18 A)(2.40 ) 7.63 V R Ω = . 33 (3.18 A)(4.80 ) 15.3 V R Ω = . Note that 123 28.0 V VVV ++= . (e) (3.18 A) (1.60 ) 16.2 W R Ω = . (3.18 A) (2.40 ) 24.3 W R Ω = . (3.18 A) (4.80 ) 48.5 W R Ω = . Since 2 = and the current is the same for each resistor, the resistor with the greatest R dissipates the greatest power. EVALUATE: When resistors are connected in parallel, the resistor with the smallest R dissipates the greatest power. 26.10. (a) IDENTIFY: The current, and hence the power, depends on the potential difference across the resistor. SET UP: / 2 = EXECUTE: (a) (5.0 W)(15,000 ) 274 V VP R Ω = (b) / (120 V) /(9,000 ) 1.6 W Ω = SET UP: (c) If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe. Therefore the maximum power in the larger resistor must be 2.00 W. Use 2 = to find the maximum current through the series combination and use Ohm’s law to find the potential difference across the combination. EXECUTE: 2 = gives I = P / R = (2.00 W)/(150 ) = 0.0133 A. The same current flows through both resistors, and their equivalent resistance is 250 . Ohm’s law gives V = IR = (0.0133 A)(250 ) = 3.33 V. Therefore P 150 = 2.00 W and 100 R 2 = = (0.0133 A) 2 (100 ) = 0.0177 W. EVALUATE: If the resistors in a series combination all have the same power rating, it is the largest resistance that limits the amount of current. 26.11. IDENTIFY: For resistors in parallel, the voltages are the same and the currents add. eq 1 2 R =+ so 12 eq , R = + For resistors in series, the currents are the same and the voltages add. eq 1 2 R = + . SET UP: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits shown in Figure 26.11. EXECUTE: eq 5.00 R . In Figure 26.11c, 60.0 V 12.0 A 5.00 I Ω . This is the current through each of the resistors in Figure 26.11b. 12 12 (12.0 A)(2.00 ) 24.0 V R Ω = . 34 34 (12.0 A)(3.00 ) 36.0 V R Ω = . Note that 12 34 60.0 V VV += . 12 V is the voltage across 1 R and across 2 R , so 12 1 1 24.0 V 8.00 A 3.00 V I R = Ω and 12 2 2 24.0 V 4.00 A 6.00 V I R = Ω . 34 V is the voltage across 3 R and across 4 R , so 34 3 3 36.0 V 3.00 A 12.0 V I R = Ω and 34 4 4 36.0 V 9.00 A 4.00 V I R = Ω .

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26-4 Chapter 26 EVALUATE: Note that 12 34 II II +=+ . Figure 26.11 26.12. IDENTIFY: Replace the series combinations of resistors by their equivalents. In the resulting parallel network the battery voltage is the voltage across each resistor.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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761_PartUniversity Physics Solution - Direct-Current...

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