**Electricity 3**

To understand electricity, one needs to be familiar with the common measurements and what they are.

**Volts: The unit of measure of electromotive force, E in Ohms Law**

1 Microvolt = .000001 volt = 1 uv. = one 1/millionth of a volt

1 Millivolt = .001 volt = 1 mv. = one 1/thousandth of a volt

1 Volt =1 volt = 1 v.

1 Kilovolt = 1000 volts = 1 kv.

**Amperes: The unit of measure of electric current, I in Ohms Law**

1 Ampere = 1 amp. = 1a.

1 Milliampere = .001 amp. = 1 ma. = one 1/thousandth of an ampere

1 Microampere= .000001 amp. = 1 ua. = one 1/millionth of an ampere

**Ohms: The unit measure of resistance, R in Ohms Law**

1 Ohm = 1 ohm [omega symbol, not on keyboard]

1 Kilohm = 1 k ohms = 1000 ohms

1 Megohm = 1 meg. or 1 m. = 1 Million Ohms

**Watt: The unit of measure of electrical Power**

1 Microwatt = .000001 watt = 1 uw. = one 1/millionth of a watt

1 Milliwatt = .001 watt = 1 mw = one 1/thousandth of a watt

1 Watt = 1 watt = 1 w.

1 Kilowatt = 1000 watts =1 kw = 1 thousand watts

1 Megawatt = 1000000 watts = 1 megw = 1 million watts

**Horsepower: 746 Watts = the amount of energy in 1 horsepower**

There are only 2 formulas that you need to know and understand, in order to solve most common electrical problems on small power equipment.

The first is

Ohms Law:E = IR. This says thatVoltage = Current X Resistance. Conversely, you can FindIwithI = E/RandR = E/I.The second is the

Powerformula:P = EI. This says that Power inWatts = Voltage X Current(Note: current is often called amperage). Conversely,E = P/IorI = P/EandP = E^{2}/RandP = I^{2}R.Now, let's apply these formulas to a couple of tractor examples to see how they work.

Your tractor has two 12 volt headlights rated at 36 watts each, connected in parallel. Your tractor has a 6 amp charging system. You would like to add two more lights, can you do this with the charging system you have? There are two ways to solve this.Example 1:

1st Method, using the Power formula:P, or Power = 12 volt system, soE[12 Volts] xI[Current 6 amps.] = 72 Watts. This is the capability of the charging system. Two lamps in parallel with 36 watts each = 72 watts. No extra power is available for more lights. Connecting more light would result in the battery not staying charged and likely "smoke" the rectifier or stator in the charging system.

2nd Method, using Ohm's Law: Your lights are rated at 36 Watts. Current available,I= P[36W]/E[12V], so Current, orI= 3 amps each. Two lamps, at 3 amps each, equals 6 amps, the total capacity of the charging system.

Example 2:Your mower has an electric clutch with 4 ohms resistance and the two lights above. You are looking at a replacement engine. What size alternator does it need? To solve this we need to use Ohms Law to find how much current the clutch draws. You have a 12 volt system, so:

I, Current= E[12 Volts]/ R[4 ohms] = 3 AmpsYou would need a charging system that would provide current to charge the battery after starting, plus 3 Amps for the clutch, plus 6 Amps for the lights or more than a 9 Amp charging system. In this case an engine with a 6 amp alternator would not be a deal even if it was cheap, as it would need to be converted to a different charging system.

Next, I will go to some basic circuits, and explain how resistance, voltage and current interact in a circuit. Once you understand these principles, how you would use a voltmeter to locate troubleshoot wiring problems in you tractor, or actually any electric circuit.

First we need to know some common symbols and what they are. Then later we can use them to draw a diagram and analyze how it works.

These symbols are used to create schematic diagrams, these show electrically how a circuit works. Some industrial and other diagrams use some variation of these symbols. In the next article [4] we will look at series and parallel circuits and how they work, using some of these symbols.

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