746_PartUniversity Physics Solution

746_PartUniversity Physics Solution - 25-10 Chapter 25...

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25-10 Chapter 25 25.43. IDENTIFY: The bulbs are each connected across a 120-V potential difference. SET UP: Use P = V 2 / R to solve for R and Ohm’s law ( I = V/R ) to find the current. EXECUTE: (a) R = V 2 / P = (120 V) 2 /(100 W) = 144 . (b) R = V 2 / P = (120 V) 2 /(60 W) = 240 (c) For the 100-W bulb: I = V/R = (120 V)/(144 ) = 0.833 A For the 60-W bulb: I = (120 V)/(240 ) = 0.500 A EVALUATE: The 60-W bulb has more resistance than the 100-W bulb, so it draws less current. 25.44. IDENTIFY: Across 120 V, a 75-W bulb dissipates 75 W. Use this fact to find its resistance, and then find the power the bulb dissipates across 220 V. SET UP: P = V 2 / R , so R = V 2 / P EXECUTE: Across 120 V: R = (120 V) 2 /(75 W) = 192 . Across a 220-V line, its power will be P = V 2 / R = (220 V) 2 /(192 ) = 252 W. EVALUATE: The bulb dissipates much more power across 220 V, so it would likely blow out at the higher voltage. An alternative solution to the problem is to take the ratio of the powers. 2 2 2 220 220 220 2 120 120 120 /2 2 0 /1 2 0 PV R V R V ⎛⎞ == = ⎜⎟ ⎝⎠ . This gives 2 220 220 (75 W) 120 P = = 252 W. 25.45. IDENTIFY: A “100-W” European bulb dissipates 100 W when used across 220 V. (a) SET UP: Take the ratio of the power in the US to the power in Europe, as in the alternative method for problem 25.44, using P = V 2 / R . EXECUTE: 2 2 2 US US US 2 EE E / 120 V . / 220 V RV R V = This gives 2 US 120 V (100 W) 220 V P = = 29.8 W. (b) SET UP: Use P = IV to find the current. EXECUTE: I = P/V = (29.8 W)/(120 V) = 0.248 A EVALUATE: The bulb draws considerably less power in the U.S., so it would be much dimmer than in Europe. 25.46. IDENTIFY: I = . Energy . Pt = SET UP: (9.0 V)(0.13 A) 1.17 W P EXECUTE: Energy (1.17 W)(1.5 h)(3600 s/h) 6320 J EVALUATE: The energy consumed is proportional to the voltage, to the current and to the time. 25.47. IDENTIFY and SET UP: By definition . P p LA = Use , and IEV L I J A = to rewrite this expression in terms of the specified variables. EXECUTE: (a) E is related to V and J is related to I, so use P = VI. This gives VI p LA = and so VI E JpE J LA = (b) J is related to I and ρ is related to R, so use 2 . PI R = This gives 2 . IR p LA = 22 2 2 and so LJ A L IJ A R p J AL A ρρ = (c) E is related to V and is related to R , so use 2 /. PV R = This gives 2 . V p RLA = 2 and so . L EL A E VE L R p A L = = EVALUATE: For a given material ( constant), p is proportional to or to . J E 25.48. IDENTIFY: Calculate the current in the circuit. The power output of a battery is its terminal voltage times the current through it. The power dissipated in a resistor is 2 . SET UP: The sum of the potential changes around the circuit is zero. EXECUTE: (a) 8.0 V 0.47 A 17 I Ω . Then 9 (0.47 A) (5.0 ) 1.1 W R Ω Ω = and 9 (0.47 A) (9.0 ) 2.0 W. R Ω Ω = (b) 16V (16 V)(0.47 A) (0.47 A) (1.6 ) 7.2 W. I r =− = Ω = E (c) 8V (8.0 V)(0.47 A) (0.47 A) (1.4 ) 4.1W. I r =+ = + Ω = E EVALUATE: (d) (b) (a) (c) =+ . The rate at which the 16.0 V battery delivers electrical energy to the circuit equals the rate at which it is consumed in the 8.0 V battery and the 5.0 Ω and 9.0 Ω resistors.
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Current, Resistance, and Electromotive Force 25-11 25.49. (a) IDENTIFY and SET UP: PV I = and energy = (power) × (time).
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746_PartUniversity Physics Solution - 25-10 Chapter 25...

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