# Ohms Law

Volts, Amps, Watts and Ohms... how one unit of measurement relates to another.

## Definitions

Voltageis the electromotive force (pressure) applied to an electrical circuit measured in Volts (E)Currentis the flow of electrons in an electrical circuit measured in Amperes (I)Resistanceis the opposition to the flow of electrons in an electrical circuit measured in Ohms (R)Poweris the product of the voltage times the current in an electrical circuit measured in Watts (P)

- 1V (Volt) = 1000mV (millivolts)
- 1A (Amp) = 1000mA (milliamps)
- 1W (Watt) = 1000mW (milliwatts)
## Basics

Basically, Ohm's law states that the current in an electrical circuit is directly proportional to the applied voltage and the resistance of the circuit. The 3 most common mathematical expressions are:

Power can be expressed as P = E * I (Power = Voltage

- E = I * R (Voltage = Current
timesResistance)- I = E / R (Current = Voltage
divided byResistance)- R = E / I (Resistance = Voltage
divided byCurrent)timesCurrent), so with a little algebra we can combine these expressions and derive P = E² / R (Power = Voltage squareddivided byResistance).## More Formulae!

Now if we introduce Power (expressed in Watts), we can represent all 12 combinations via charts. If you need to know a specific parameter, the result can be found by using two other parameters.

Another way to look at those same 12 formulae Voltage = Resistance = Current = Power = (Square Root of Power) * Resistance Voltage / Current Voltage / Resistance Voltage * Current Power / Current Power / (Current squared) (Square Root of Power) / Resistance (Current squared) * Resistance Current * Resistance (Voltage squared) / Power Power / Voltage (Voltage squared) / Resistance

What all the above boils down to is that as long as you know two parameters (out of Voltage, Current, Resistance & Power), you can work out the others.## Examples

OK, how can we apply some of these formulae?## Example 1

Say we have a 12 Volt power supply and we know that it is rated for 670mA (0.67A) of output current. What power (Watts) would this be?

Looking at the 4th column in the table, we have three choices. Only one of them (the first) lists Voltage and Current, so let's use that.

12 (Volts) times 0.67 (Amps) gives us 8.04 Watts. Wasn't that simple!## Example 2

What about if we wanted to know how much voltage drop there will be in a cable? If we know the cable resistance and how much current is flowing, we can use the first column in the table above.

In this example, we'll say the cable has a total resistance of 0.3 Ohms (0.15 Ohms for each wire of a fig.8 cable) and we are drawing 2.5 Amps through it.

The third formula in the first column has the two known parameters, so let's substitute them into the equation:

2.5 (Amps) times 0.3 (Ohms) gives us 0.75 (Volts), so now you know that the voltage at the load end of the cable will be 0.75V less than the supply end.

As the relationship is proportional, a higher current (more Amps) through the same cable will result in a higher voltage drop. Conversely, using thicker cable (with a lower resistance) will reduce the voltage drop.

We can also see how much power is wasted (as heat) in the above cable example, by using the second formula from the last column:

2.5 (Amps) squared * 0.3 (Ohms) gives us 1.875 Watts, lost as heat. You can also double check this by multiplying the Voltage we got (0.75V) by the Current (2.5A) and see that you still get 1.875 Watts.

## Summary

Hopefully you're still reading and have understood at least some of the above. Once you master Ohms Law (or can at least read the charts above!), you'll have a better understanding of how and why electricity behaves the way it does. Categories: Electronics pagesThis page has been seen 890 times.

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