776_PartUniversity Physics Solution

776_PartUniversity Physics Solution - 26-18 Chapter 26 V...

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26-18 Chapter 26 (b) 2 V P R = and 22 (240 V) 14 4100 W V R P == = Ω . (c) At 11c / per kWH, for 1 hour the cost is (11c/kWh)(1 h)(4.1kW) 45c = / / . EVALUATE: The cost to operate the device is proportional to its power consumption. 26.51. IDENTIFY and SET UP: The heater and hair dryer are in parallel so the voltage across each is 120 V and the current through the fuse is the sum of the currents through each appliance. As the power consumed by the dryer increases the current through it increases. The maximum power setting is the highest one for which the current through the fuse is less than 20 A. EXECUTE: Find the current through the heater. so / 1500 W /120 V 12.5 A. PV I I PV = = The maximum total current allowed is 20 A, so the current through the dryer must be less than 20 A 12.5 A 7.5 A. −= The power dissipated by the dryer if the current has this value is ( )( ) 120 V 7.5 A 900 W. I = For P at this value or larger the circuit breaker trips. EVALUATE: 2 / PV R = and for the dryer V is a constant 120 V. The higher power settings correspond to a smaller resistance R and larger current through the device. 26.52. IDENTIFY: The current gets split evenly between all the parallel bulbs. SET UP: A single bulb will draw 90 W 0.75 A 120 V P I V = . EXECUTE: 20 A Number of bulbs 26.7. 0.75 A ≤= So you can attach 26 bulbs safely. EVALUATE: In parallel the voltage across each bulb is the circuit voltage. 26.53. IDENTIFY and SET UP: Ohm's law and Eq.(25.18) can be used to calculate I and P given V and R . Use Eq.(25.12) to calculate the resistance at the higher temperature. (a) EXECUTE: When the heater element is first turned on it is at room temperature and has resistance 20 . R = Ω 120 V 6.0 A 20 V I R = Ω () 2 2 120 V 720 W 20 V P R = Ω (b) Find the resistance R ( T ) of the element at the operating temperature of 280 C. ° Take 00 23.0 C and 20 . TR = Ω Eq.(25.12) gives () ( ) 1 RT R T T α =+−= ( ) ( ) 1 3 20 1 2.8 10 C 280 C 23.0 C 34.4 . Ω+ × ° °− ° = Ω 120 V 3.5 A 34.4 V I R = Ω 2 2 120 V 420 W 34.4 V P R = Ω EVALUATE: When the temperature increases, R increases and I and P decrease. The changes are substantial. 26.54. (a) IDENTIFY: Two of the resistors in series would each dissipate one-half the total, or 1.2 W, which is ok. But the series combination would have an equivalent resistance of 800 , Ω not the 400 Ω that is required. Resistors in parallel have an equivalent resistance that is less than that of the individual resistors, so a solution is two in series in parallel with another two in series. SET UP: The network can be simplified as shown in Figure 26.54a. Figure 26.54a EXECUTE: s R is the resistance equivalent to two of the 400 Ω resistors in series. s 800 . RRR =+= Ω eq R is the resistance equivalent to the two s 800 R = Ω resistors in parallel: eq eq s s s 1 1 1 2 800 ; 400 . 2 R RR R R Ω = += = = Ω EVALUATE: This combination does have the required 400 Ω equivalent resistance. It will be shown in part (b) that a total of 2.4 W can be dissipated without exceeding the power rating of each individual resistor.
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Direct-Current Circuits 26-19 IDENTIFY: Another solution is two resistors in parallel in series with two more in parallel.
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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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776_PartUniversity Physics Solution - 26-18 Chapter 26 V...

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