Part AA carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00 °C, the resistance of the carbon resistor is 217.6 . What is thetemperature on a spring day when the resistance is 215.4 ?Take the temperature coefficient of resistivity for carbon to be a = -5.00×10-4 C-1.▸ View Available Hint(s)Ο ΑΣΦSubmit? Part AAt room temperature, what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm ) that is needed to cause a 3.50 A current to flow?Express your answer with the appropriate units.◎°μαVmE=0.0178SubmitPrevious Answers Request Answer?× Incorrect; Try Again; 5 attempts remainingPart BWhat field would be needed if the wire were made of silver instead?Express your answer with the appropriate units.E = 1.69×10-2-2 VmSubmitPrevious AnswersCorrect>

Part A: (conductivity of copper at room temperature:…

Answered: Part A A carbon resistor is to be used…

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter20: Electric Current, Resistance, And Ohm's Law

Section: Chapter Questions

Problem 12CQ: Explain why R=R0(1 + a?T) for the temperature variation of the resistance R of an object is not as...

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Similar questions

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.
A 20.00-V battery is used to supply current to a 10-k resistor. Assume the voltage drop across any wires used for connections is negligible, (a) What is the current through the resistor? (b) What is the power dissipated by the resistor? (c) What is the power input from the battery; assuming all the electrical power is dissipated by the resistor? (d) What happens to the energy dissipated by the resistor?
Unreasonable Results (a) What current is needed to transmit 1.00 102 MW of power at 10.0kV? (b) Find the resistance of 1.00 km of wire that would cause a 0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent?
Unreasonable Results (a) To what temperature must you raise a resistor made of constantan to double its resistance, assuming a constant temperature coefficient of resistivity? (b) To cut it in half? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable, or which premises are inconsistent?
(a) Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has a= - 0.0600/°C ) when it is at the same temperature as the patient. What is a patient's temperature if the thermistor’s resistance at that temperature is 82.0% of its value at 37.0°C (normal body temperature)? (b) The negative value for a may not be maintained for very low temperatures. Discuss why and whether this is the case here. (Hint: Resistance can’t become negative.)
Review. When a straight wire is warmed, its resistance is given by R = R0,[1 + a(T T0)] according to Equation 27.20, where a is the temperature coefficient of resistivity. This expression needs to be modified if we include the change in dimensions of the wire due to thermal expansion. For a copper wire of radius 0.100 0 mm and length 2.000 m, find its resistance at 100.0C, including the effects of both thermal expansion and temperature variation of resistivity. Assume the coefficients are known to four significant figures.
Explain why R=R0(1 + a?T) for the temperature variation of the resistance R of an object is not as accurate as P=P0(1 + a?T) which gives the temperature variation of resistivity P.
A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in °C/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300 kcal/kg • °C, assuming no heat escapes?
An all-electric car (not a hybrid) is designed to run from a bank of 12.0-V batteries with total energy storage of age of 2.00 107J. If the electric motor draws 8.00 kW as the car moves at a steady speed of 20.0 m/s. (a) what is the current delivered to the motor? (b) How far can the car travel before it is out of juice"?
Using your understanding of the nature of internal energy and temperature, explain why we might expect the resistance of a solid to increase if its temperature increases.
(a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V potential as in the figure below. What is the resistance of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use this equation for power: P=I2 RP = .)
Currents of approximately 0.06 A can be potentially fatal. Currents in that range can make the heart fibrillate (beat in an uncontrolled manner). The resistance of a dry human body can be approximately 100 k . (a) What voltage can cause 0.06 A through a dry human body? (b) When a human body is wet, the resistance can fall to 100 . What voltage can cause harm to a wet body?

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