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There is no direct relation of electric current and power. In order to know

how much power (or energy) the current gives up, you must know what the

current is flowing through.

The easiest way to describe anything through which the current is flowing is

to measure and state its electrical resistance.

Once you know the resistance through which the current is flowing . . .

Power delivered by the current = (magnitude of the current, amperes)2 x (resistance of the path, ohms)

The power is delivered continuously. Its unit is watts.

Each watt of power means 1 joule of energy every second.

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Q: Realation of current with power and time?
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Does current or voltage actually perform the work in a circuit?

I would say both. Power (energy, or work, per unit time) can be calculated with the formula: P = IV (power = current x voltage)


If you multiply the voltage in a circuit by the current what are you going to get for your answer?

Voltage x current = power (watts)


What do you multiply to get electrical energy?

Electrical energy = (power) multiplied by (time) or (voltage across a component or circuit) times (current through it) times (time)


Energy carried by current?

For DC: power (in watts) = current (in ampere) x voltage (in volts). Energy = power x time, so energy = current x voltage x time (time in seconds). For AC, a power factor may have to be included (the cosine of the angular displacement between current and voltage). This is often near one, but it may be less.


If current in an electric stove is doubled how will the thermal energy produced per unit time change?

P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.