Even though most of us have a natural, built in aversion to math, there is one formula dealing with electrical systems that we absolutely have to know about. This law was formulated by Georg Ohm back in the 1860’s and believe it or not, it is still in effect.

The best way to make use of Ohm’s law is to know it, and understand it and then use it instinctively, rather than having to think about it. Many times when analyzing an electrical circuit, it is helpful to consider Ohm’s law and how it affects circuit operation.

Ohm’s law is the single statement that draws together all the information we have studied thus far. It is deceptively simple and yet amazingly inclusive. Remember we have discussed Voltage, Current, Resistance, and Circuits. The formula for Ohm’s law comes in several forms, but we will start with:

I=E/R

Where I is current, E is voltage, and R is resistance. This formula states that in order to determine the current in a given circuit, you divide the circuit voltage by the circuit resistance.

As a quick example, if the voltage of a circuit is 12vdc, and the resistance is 2 ohms, the current would be:

I=12/2=6 amps.

If you have been following this series, you will remember one of my rules for working on electrical systems, and that is to have an idea of what to expect when you make an electrical measurement before you make it. Using Ohm’s law as described above enables you to look at a particular circuit and realize that you expect to measure 6 amps flowing in the circuit.

There are several other variations to this formula and we will talk more about them next time.

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## Glenn Gbole

A good way to remember the law is

E

I R

I into E gives you R

R into E Gives you I

I times R gives you E

and just for fun

I times E gives you P (power or watts)

## Larry Cad

Ham, good suggestion. While we haven’t yet gotten into defining or explaing the term “power”, everyone should be aware that Ohm’s law also has a variation dealing with power. There will be a future post dealing with that variation and we will discuss how power is calculated, particularly in RVs.

## Ham Radio

Nice series, Larry.

While I am familiar with electrical properties, many RV’ers could probably use a primer on battery drain. Knowing how to convert (appliances) power in watts againt ampere hours capacity would benefit a lot of boondockers.

Thanks for taking the time to write these pieces.

## Larry Cad

Interesting comments Steve. Sounds like you have an interesting job. To expand on Steve’s comments a little and to illuminate our lesson, Steve works on instruments which are part of a “circuit” and which introduce a resistance of 250 ohms into the circuit. The external part of the circuit causes a current to flow in the circuit which will vary from 4 milliamps to 20 milliamps. This is a very common value of current in instrumentation equipment. 4 milliamps can also be written as .004 amps, and 20 milliamps can also be written as .020 amps. If we look at these values in terms of our lesson and apply Ohm’s law, we get the following:

I=E/R can be rearranged by algebra to be E=I*R (trust me on this one for now)

So: E=.004 amps * 250 ohms = 1 volt, and E=.020 amps * 250 ohms = 5 volts.

So, as the circuit current varies from 4 milliamps to 20 milliamps, the voltage across the 250 ohm resistance of the instrument will vary from 1 to 5 volts.

We will talk more about these subjects as we continue with our series.

Larry

## Steve Sweet

Just a comment on a very good series. I deal with current and DC voltage on my job as an instrument technician. A good subject to mention in electrical series is polarity.

The instruments I have to work with operate on 24VDC and 4 to 20 MADC using a

250ohm resister across the + and – of the receiving instrument. That instrument would be receiving a 1 to 5 vdc input. Other instruments could operate on a current input,

4 to 20madc to create an output directly related to current input as in 3 to 15psi.If the

positive and negative leads are wrong nothing will work. With negative ground current flows from positive to negative. Of course with the current loads I work with a 10amp

multimeter is adequete. However polarity can be verified using positive lead to ground to check if wiring is properly terminated using multimeter and display of meter without measuring current. Steve