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An Amp (ampere) is a unit of measurement for electrical current flow. A Watt is a unit of measurement for electrical power. The relationship between current (Amps) and power (Watts) is generally not well understood.

The power (Watts) generated or transmitted by a system is directly proportional to the current (Amperes), and can be expressed in the following equation:

P = IV

Where:

· P is power in Watts

· I is current in Amperes

· V is voltage in Volts

The 'P=IV' equation is correct for all direct current (DC) circuits and for those alternating current (AC) circuits in which the power factor is 1.

When the current or the voltage is not constant, but varies with time t, the correct equation is:

p (t) = i (t) v (t)

Where:

· p (t) is the instantaneous real power in Watts

· i (t) is the instantaneous current in Amperes

· v (t) is the instantaneous voltage in Volts

P = I*V* pf

where

· P is average power in Watts

· I is RMS current in Amperes

· V is RMS voltage in Volts

· pf is the power factor of the load

When the current or the voltage is a perfect sine wave (which they rarely are),

pf = cosine (angle between the voltage and current)

so

P = I*V* cosine (angle between the voltage and current).

The power (Watts) consumed by a purely resistive system is directly proportional to the square of the current (Amperes), and can be expressed in the following equation:

P = I2R,

Where:

· P is power in Watts

· I is current in Amperes

· R is resistance in Ohms.

The power consumed can be much less than the power transmitted. For example, consider a power cable that carries 1,000 Amps at 100 Volts. The power delivered through the cable (P=IV) is 1,000 x 100, or 100,000 Watts. Let's assume the same power cable has a resistance of 0.001 Ohm (1 Milli Ohm) from one end to the other. The power consumed by that cable (P = I2R, which is the electrical power converted into heat) is 0.001 x 10002, or 1,000 Watts. That means there are only 99,000 Watts available after the current flows through the cable.

Now consider an identical power cable, but this time it carries 100 Amps at 1,000 Volts. The power delivered through the cable (P=IV) is 100 x 1000, or still 100,000 Watts. But this time the power consumed by that cable (P = I2R) is 0.001 x 1002, or 10 Watts. In this case, there are 99,990 Watts available after the smaller current flows though the same cable. This is why electrical power companies transmit high power at such high voltages - to reduce the amount of current flowing through the cables, and thus dramatically reducing the power lost to transmission.

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