The equivalent resistance R of Resistors R1 and R2 connected in parallel in an electrical circuit is 1 R1 -. Use differentials to determine by approximately what percentage the R2 given by %3D R resistance R change if R¡ is increased from 100 ohms to 105 ohms and R2 = 25 ohms is decreased by 1.5 ohms.

The equivalent resistance R of Resistors R1 and R2 connected in parallel in an electrical circuit is 1 R1 -. Use differentials to determine by approximately what percentage the R2 given by %3D R resistance R change if R¡ is increased from 100 ohms to 105 ohms and R2 = 25 ohms is decreased by 1.5 ohms.

Inquiry into Physics

8th Edition

ISBN:9781337515863

Author:Ostdiek

Publisher:Ostdiek

Chapter7: Electricity

Section: Chapter Questions

Problem 13P

See similar textbooks

Similar questions

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) 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.)
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.
Show that the power dissipated by the load resistor is maximum when the resistance of the load resistor is equal to the internal resistance of the battery.
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.
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.
Integrated 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.
Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has a=0.06/C ) when it isat 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 37C (normal bodytemperature)?
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?
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.