SlideShare a Scribd company logo

3power

•
•
T
Talia Carbis

Power is defined as the rate at which energy is converted and is measured in watts. During energy conversion in electrical devices, some input energy is converted to heat. This heating can cause problems in devices like motors and transformers. However, heating is rarely a problem in devices like lamps. As voltage increases for a fixed resistance, current and power output will both increase due to their relationships as defined by Ohm's and Joule's laws.

1 of 5
Power •! Power is the RATE that energy is converted.(unit watts W) •! Or •! Power is a measure of the amount of energy converted in 1 second. •! The POWER RATING of a device tells the rate that energy is converted: –! 75W lamp –! 750kW motor
Power cont. •! During the conversion some input energy(electrical) is changed to heat within the device. •! The heating of the electrical device may produce a problem: –! motors –! transformers –! cables •! In other electrical devices this is rarely a problem: –! lamps
V In the above circuit, we know we have a battery voltage of 18 volts and a lamp resistance of 3 Ω. Using Ohm's Law to determine current, we get: V Now that we know the current, we can take that value and multiply it by the voltage to determine power: V Answer: the lamp is dissipating (releasing) 108 watts of power, most likely in the form of both light and heat.
Let's try taking that same circuit and increasing the battery voltage to see what happens. Intuition should tell us that the circuit current will increase as the voltage increases and the lamp resistance stays the same. Likewise, the power will increase as well: V Now, the battery voltage is 36 volts instead of 18 volts. The lamp is still providing 3 Ω of electrical resistance to the flow of electrons. The current is now: V This stands to reason: if I = E/R, and we double V while R stays the same, the current should double. Indeed, it has: we now have 12 amps of current instead of 6. Now, what about power? V Notice that the power has increased just as we might have suspected, but it increased quite a bit more than the current. Why is this? Because power is a function of voltage multiplied by current, and both voltage and current doubled from their previous values, the power will increase by a factor of 2 x 2, or 4. You can check this by dividing 432 watts by 108 watts and seeing that the ratio between them is indeed 4
Using algebra again to manipulate the formulae, we can take our original power formula and modify it for applications where we don't know both voltage and current: If we only know voltage (E) and resistance (R): An historical note: it was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the last equation (P = I2R), and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance (E=IR ; I=E/R ; and R=E/I) that they are frequently credited to Ohm.

More Related Content

3power

  • 1. Power •! Power is the RATE that energy is converted.(unit watts W) •! Or •! Power is a measure of the amount of energy converted in 1 second. •! The POWER RATING of a device tells the rate that energy is converted: –! 75W lamp –! 750kW motor
  • 2. Power cont. •! During the conversion some input energy(electrical) is changed to heat within the device. •! The heating of the electrical device may produce a problem: –! motors –! transformers –! cables •! In other electrical devices this is rarely a problem: –! lamps
  • 3. V In the above circuit, we know we have a battery voltage of 18 volts and a lamp resistance of 3 Ω. Using Ohm's Law to determine current, we get: V Now that we know the current, we can take that value and multiply it by the voltage to determine power: V Answer: the lamp is dissipating (releasing) 108 watts of power, most likely in the form of both light and heat.
  • 4. Let's try taking that same circuit and increasing the battery voltage to see what happens. Intuition should tell us that the circuit current will increase as the voltage increases and the lamp resistance stays the same. Likewise, the power will increase as well: V Now, the battery voltage is 36 volts instead of 18 volts. The lamp is still providing 3 Ω of electrical resistance to the flow of electrons. The current is now: V This stands to reason: if I = E/R, and we double V while R stays the same, the current should double. Indeed, it has: we now have 12 amps of current instead of 6. Now, what about power? V Notice that the power has increased just as we might have suspected, but it increased quite a bit more than the current. Why is this? Because power is a function of voltage multiplied by current, and both voltage and current doubled from their previous values, the power will increase by a factor of 2 x 2, or 4. You can check this by dividing 432 watts by 108 watts and seeing that the ratio between them is indeed 4
  • 5. Using algebra again to manipulate the formulae, we can take our original power formula and modify it for applications where we don't know both voltage and current: If we only know voltage (E) and resistance (R): An historical note: it was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the last equation (P = I2R), and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance (E=IR ; I=E/R ; and R=E/I) that they are frequently credited to Ohm.