Which process will double the power dissipated by a resistor? a doubling the current while making the potential difference half as big b doubling the current and doubling the potential difference c doubling the current and making the resistance half as big d doubling the current while doubling the resistance

Which process will double the power dissipated by a resistor? a doubling the current while making the potential difference half as big b doubling the current and doubling the potential difference c doubling the current and making the resistance half as big d doubling the current while doubling the resistance

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 14CQ: The power dissipated in a resistor is given by P = V2/R, which means power decreases if resistance...

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Similar questions

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.
A 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?
The potential difference across the filament of a light-bulb is maintained at a constant value while equilibrium temperature is being reached. The steady-state current in the bulb is only one-tenth of the current drawn by the bulb when it is first turned on. If the temperature coefficient of resistivity for the bulb at 20.0C is 0.004 50 (C)-1 and the resistance increases linearly with increasing temperature, what is the final operating temperature of the filament?
A rechargeable battery has an emf of 13.2 V and an internal resistance of 0.850 . It is charged by a 14.7-V power supply for a time interval of 1.80 h. Alter charging, the battery returns to its original state as it delivers a constant current to a load resistor over 7.30 h. Find the efficiency of the battery as an energy storage device. (The efficiency here is defined as the energy delivered to the load during discharge divided by the energy delivered by the 14.7-V power supply during the charging process.)
A portable coffee heater supplies a potential difference of 12.0 V across a Nichrome heating clement with a resistance of 2.00 . (a) Calculate the power consumed by the heater. (b) How many minutes would it take to heat 1.00 kg of coffee from 20.0C to 50.0C with this heater? Coffee has a specific heat of 4 184J/(kg C). Neglect any energy losses to the environment.
A battery with an internal resistance of 10.0 produces an open circuit voltage of 12.0 V. A variable load resistance with a range from 0 to 30.0 is connected across the battery. (Note: A battery has a resistance that depends on the condition of its chemicals and that increases as the battery ages. This internal resistance can be represented in a simple circuit diagram as a resistor in series with the battery.) (a) Graph the power dissipated in the load resistor as a function of the load resistance. (b) With your graph, demonstrate the following important theorem: The power delivered to a load is a maximum if the load resistance equals the internal resistance of the source.
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.
While taking photographs in Death Valley on a day when the temperature is 58.0C, Bill Hiker finds that a certain voltage applied to a copper wire produces a current of 1.000 A. Bill then travels to Antarctica and applies the same voltage to the same wire. What current does he register there if the temperature is 88.0C? Assume that no change occurs in the wires shape and size.
When the current in the portion of the circuit shown in Figure P32.65 is 2.00 A anti increases at a rate of 0.500 A/s, the measured voltage is Vab = 9.00 V. When the current is 2.00 A and decreases at the rate of 0.500 A/s. the measured voltage is Vab = 5.00 V. Calculate the values of (a) I. and (b) R

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a	doubling the current while making the potential difference half as big
b	doubling the current and doubling the potential difference
c	doubling the current and making the resistance half as big
d	doubling the current while doubling the resistance