1. When the length of the conductor is same and the area of cross-section reduced to half then its resistance 2. If you wish to increase the amount of current in a resistor from 240mA to 260mA by changing the 12V source, what should be the new voltage setting be? Answer = V 0.05 0.04 0.03 0.02 0.01 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 Current 3.The above graph is a showing the relationship between voltage-current. The resistance calculated from the above graph is = Voltage

1. When the length of the conductor is same and the area of cross-section reduced to half then its resistance 2. If you wish to increase the amount of current in a resistor from 240mA to 260mA by changing the 12V source, what should be the new voltage setting be? Answer = V 0.05 0.04 0.03 0.02 0.01 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 Current 3.The above graph is a showing the relationship between voltage-current. The resistance calculated from the above graph is = Voltage

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter27: Current And Resistance

Section: Chapter Questions

Problem 27.55P: Review. The healing element of an electric coffee maker operates at 120 V and carries a current of...

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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.
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.
The 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.
Integrated Concepts 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?
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.
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?
A young man owns a canister vacuum cleaner marked 535 W [at] 120 V and a Volkswagen Beetle, which he wishes to clean. He parks the car in his apartment parking lot and uses an inexpensive extension cord 15.0 m long to plug in the vacuum cleaner. You may assume the cleaner has constant resistance, (a) If the resistance of each of the two conductors in the extension cord is 0.900 , what is the actual power delivered to the cleaner? (b) If instead the power is to be at least 525 W, what must be the diameter of each of two identical copper conductors in the cord he buys? (c) Repeat part (b) assuming the power is to be at least 532 W.
A 120-V immersion heater consists of a coil of wire that is placed in a cup to boil the water. The heater can boil one cup of 20.00°C water in 180.00 seconds. You buy one to use in your dorm room, but you are worried that you will overload the circuit and nip the 15.00-A, 120-V circuit breaker, which supplies your dorm room. In your dorm room, you have four 100.00-W incandescent lamps and a 1500.00-W space heater, (a) What is the power rating of the immersion heater? (b) Will it trip the breaker when everything is turned on? (c) If it you replace the incandescent bulbs with 18.00-W LED, will the breaker nip when everything is turned on?
The IR drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain.
Three identical 60.0-W, 120-V lightbulbs are connected across a 120-V power source as shown in Figure P28.72. Assuming the resistance of each lightbulb is constant (even though in reality the resistance might increase markedly with current), find (a) the total power supplied by the power source and (b) the potential difference across each lightbulb.
Review. The healing element of an electric coffee maker operates at 120 V and carries a current of 2.00 A. Assuming the water absorbs all the energy delivered to the resistor, calculate the time interval during which the temperature of 0.500 kg of water rises from room temperature (23.00) to the boiling point.
Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00 102-m- high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00 102 N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00 102 N force to overcome air resistance and friction? See Figure 20.44.