What is the formula for electric power?

Answer 1

the formulas for calculating electrical energy is

E (dissipated in a resistor R)=(I^2)*R*t=VI=(V^2)/R

where V is voltage across resistor ,I is current,t is time.

or if there is a function of time the equation becomes

E= integral from 0 to t (i(t)^2*R) dt

it is the integral of the power dissipated for the time in question.

Answer 2

m=(MxIxT)/(NxF)

Where m is the theoretical yield (current efficiency), M is the molar mass (weight of displaced element in grams), I is the amperes, T is the time in seconds, N is the oxidation state (amount of displacable electrons per atom) and F is Faraday's constant (96487 Coulombs)

Answer 3

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Electrical systems are more

efficient when a higher voltage

is used to reduce current.

In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation above, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb. You know that P = 100 W, and V = 6 V. So you can rearrange the equation to solve

for I and substitute in the numbers. I = P/V = 100 W / 6 V = 16.66 amps What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power? 100 W / 12 V = 8.33 amps So this system produces the same power, but with half the current. There is an advantage that comes from using less current to make the same amount of power. The resistance in electrical wires consumes power, and the power consumed increases as the current going through the wires increases. You can see how this happens by doing a little rearranging of the two equations. What you need is an equation for power in terms of resistance and current. Let's rearrange the first equation: I = V / R can be restated as V = I R Now you can substitute the equation for V into the other equation: P = V I substituting for V we get P = IR I, or P = I2R What this equation tells you is that the power consumed by the wires increases if the resistance of the wires increases (for instance, if the wires get smaller or are made of a less conductive material). But it increases dramatically if the current going through the wires increases. So using a higher voltage to reduce the current can make electrical systems more efficient. The efficiency of electric motors also improves at higher voltages. This improvement in efficiency is what is driving the automobile industry to adopt a higher voltage standard. Carmakers are moving toward a 42-volt electrical system from the current 12-volt electrical systems. The electrical demand on cars has been steadily increasing since the first cars were made. The first cars didn't even have electrical headlights; they used oil lanterns. Today cars have thousands of electrical circuits, and future cars will demand even more power. The change to 42 volts will help cars meet the greater electrical demand placed on them without having to increase the size of wires and generators to handle the greater current.

Answer 4

Power in Watts = Current in Amps x Voltage, P=IV

Power in Watts = Current in Amps squared x Resistance in Ohms, P=I2R

Answer 5

( P= VxI )

Power = Voltage * Current

Answer 6

Thermal Eff = (mechanical heat produced/electrical heat produced) x 100%

Answer 7

the formular for efficiency is output power divided by the input power. That is (Pout/Pin)

Answer 8

Energy = power x time. If power is in watts, and time in seconds, energy will be in joules. If power is in kilowatts, and time in hours, energy will be in kilowatt-hours.

Answer 9

power= current x voltage

p= I x V