Physics Laboratory Experiments
8th Edition
ISBN: 9781285738567
Author: Jerry D. Wilson, Cecilia A. Hernández-Hall
Publisher: Cengage Learning
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Chapter 27, Problem 5Q
Do heating appliances such as hair dryers and toasters have high-resistance or low-resistance elements? Explain.
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Chapter 27 Solutions
Physics Laboratory Experiments
Ch. 27 - Prob. 1ASACh. 27 - What is the difference between joule heat and...Ch. 27 - Given two different resistances, how does the...Ch. 27 - Prob. 4ASACh. 27 - Prob. 5ASACh. 27 - Prob. 1QCh. 27 - Why was it necessary to make adjustments to...Ch. 27 - Prob. 3QCh. 27 - (a) Circular metal wires in electrical circuits...Ch. 27 - Do heating appliances such as hair dryers and...
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- When a straight wire is heated, its resistance changes according to the equation R = R0 [1 + (T T0)] (Eq. 17.7), where is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is R=R0[1+(TT0)][1+(TT0)][1+2(TT0)] where is the coefficient of linear expansion. (See Topic 10.) (b) Compare the two results for a 2.00-m-long copper wire of radius 0.100 mm, starting at 20.0C and heated to 100.0C.arrow_forwardUse the atomic theory of matter to explain why the resistance of a material should increase as its temperature increases.arrow_forwardWhen a straight wire is heated, its resistance changes according to the equation R = R0 [1 + (T T0)] (Eq. 17.7), where is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is R=R0[1+(TT0)][1+(TT0)][1+2(TT0)] where is the coefficient of linear expansion. (See Topic 10.) (b) Compare the two results for a 2.00-m-long copper wire of radius 0.100 mm, starting at 20.0C and heated to 100.0C.arrow_forward
- When 100.0 V is applied across a 5-gauge (diameter 4.621 mm) wire that is 10 m long, the magnitude of the current density is 2.0X108A/m2 . What is die resistivity of the wire?arrow_forwardIntegrated Concepts (a) An immersion heater utilizing 120 V can raise the temperature of a 1.00 102 -g aluminum cup containing 350 g of water from 20.0°C to 95.0°C in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.arrow_forwardThe following table show the measurements of a current through and the voltage across a sample of material. Plot the data, and assuming the object is an ohmic device, estimate the resistance.arrow_forward
- Resistors are commonly rated at 18W,14W,12W ,1 W and 2 W for use in electrical circuits. If a current of1 = 2.00 A is accidentally passed through a R=1.00 resistor rated at 1 W, what would be the most probable outcome? Is there anything that can be done to prevent such an accident?arrow_forwardAn office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20.C to 100. C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume the wire is at 100. C throughout the 4.00-min time interval. (a) Calculate the average power required to warm the water to 100. C in 4.00 min. (b) Calculate the required resistance in the heating element at 100. C. (c) Calculate the resistance of the heating element at 20. C. (d) Derive a relationship between the diameter of the wire, the resistivity at 20. C, 0, the resistance at 20. C, R0, and the length L. (e) If L = 3.00 m, what is the diameter of the wire?arrow_forwardHow does the resistance for copper and for silicon change with temperature? Why are the behaviors of these two materials different?arrow_forward
- Foolishly trying to fish a burning piece of bread from a toaster with a metal butter knife, a man conies into contact with 120-V AC. He does not even feel it since, luckily, he is wearing rubber-soled shoes. What is the minimum resistance of the path the current follows through the person?arrow_forwardThe power dissipated in a resistor is given by P = V2/R, which means power decreases if resistance increases. Yet this power is also given by P = I2R, which means power increases if resistance increases. Explain why there is no contradiction here.arrow_forwardA 0.50-W, 220- resistor carries the maximumcurrent possible without damaging the resistor. If the current were reduced to half the value, what would be the power consumed?arrow_forward
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