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Topic 5.1 Extended B – Electric power

Average Power. Average Power. Instantaneous Power. FYI: Electrical power is measured in joules / second or watts , just as in mechanics. Topic 5.1 Extended B – Electric power.

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Topic 5.1 Extended B – Electric power

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  1. Average Power Average Power Instantaneous Power FYI: Electrical power is measured in joules/second or watts, just as in mechanics. Topic 5.1 ExtendedB – Electric power Back in the day when we talked about power we defined it more or less as the rate at which work is being done. Work time P = But in electricity we know that the work W in moving a charge q through a potential difference of V is given by Thus W = qV. qV t P = If the current I and the voltage V are constant, we can write the average power as the instantaneous power like this, P = IV P = I2R V2 R P = since I = q/t. Since V = IR and I = V/R we can rewrite instantaneous power in two other ways:

  2. Topic 5.1 ExtendedB – Electric power The collisions of the electrons with the lattice structure of the resistor (or any conductor, for that matter) will cause heat to be produced. We call this thermal energy joule heat or I squared R losses. In transmission lines, joule heat is a non-desirable phenomenon. In heating elements joule heat is desirable. A heating element is rated at 1500 W when plugged into a 120 V outlet. If the heating coil breaks at one end, and you repair it by cutting off 10% of its total length, how will the power rating be affected? The resistance is proportional to the length, so the new R is equal to .9R0. = 1.11P0 = 1.11V2/R0 Since P = V2/R = V2/.9R0 then P = 1.11(1500) = 1667 W FYI: This home repair could convert a safe heater into a potentially dangerous one!

  3. Topic 5.1 ExtendedB – Electric power Electric bills (someday you, too, will be paying them, reflect payment for electricity in units of kilowatt hours. Note that watt is a unit of power, whereas hour is a unit of time. Thus a kWh is a unit of energy. 1 kWh = (1000 W)(3600 s) = 3.6106 J How much does it cost to "run" a 150 W light bulb per day if it is on all the time and the electric company charges 15¢ per kWh? Unit cancellation is useful: 1 kW 1000 W 150 W $0.15 kW h 24 h 1 day = $0.54 per day How much energy is used up in one day by the bulb? Energy = power times time:  24  3600 s = 12,960,000 J Energy = 150 W

  4. Topic 5.1 ExtendedB – Electric power In our society, about 30% of our electrical energy is used for lighting. FYI: This is the equivalent of about 100 power plants of 1 GW each! About 7% of our electrical energy is used by refrigerators. FYI: This is another 25 power plants! FYI: This is why the government is pushing for energy efficient refrigerators, air conditioners, dishwashers, water heaters, etc. Over 30,000,000,000 kWh of energy have been save so far.

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