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With an instrument called a multimeter. The single meter incorporates within it a volt meter, an ohm meter and an amp meter. For higher amperages a clamp on amp meter is recommended as the circuit does not have to be opened to take a reading.

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15y ago
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13y ago

Assuming a pure resistive load. Watts = Amps x Volts therefore Volts = Watts / Amps Volts = Amps x Resistance therefore Resistance = Volts / Amps Therefore, Resistance = (Watts / Amps) / Amps and simplified that means Resistance = Watts / ( Amps x Amps)

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12y ago

If it's a straight DC circuit, then by solving Ohm's law.

Resistance = Volts divided by Amps.

In an AC circuit, it may not be possible. You can still use Ohm's law, but it's technically impedance rather than resistance that you calculate.

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10y ago

amps is current

watts is power

volts is voltage

PIV

P=IV

So if you rearrange this to work out the current you have to do power divided by voltage.

Divide the watts by the volts

Watts = Volts x Amps

Amps = Watts / Volts

Volts = Watts / Amps

<<>>

The equation that you are looking for is I = W/E, Amps = Watts/Volts.

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13y ago

I=V/R

I=amp

v= volts

R= resistance

pretty simple just divide the volt by the resistance

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8y ago

The equation that you are looking for is R = E/I. Resistance = Volts/Amps.

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Q: How do you calculate amperage when given the values of watts and volts?
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Is amperage a factor in electric billing?

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