Charles Sully's Electrical Primer, Installment 2 PDF Print E-mail
This discussion will show how a simple circuit works. It will involve a couple of three element equations which will help you to understand the operation of the circuits used in your hobby, your home and your car. If you have this knowledge and a volt ohm meter, you are set to trouble shoot and repair circuits which malfunction. This simple knowledge has been lots of fun to me and I look forward to passing it along to you here and at the field.
These discussions will include batteries, chargers, glow plug circuits, starters, and car battery analysis. The symbols involved are E, V, I, R, P. The symbol E stands for electromotive force which is the pure driving force of the circuit.  Inside our batteries lies the E force which can be measured by the volt meter connected across the two open circuit terminals. The voltmeter draws negligible current and will display the true internal potential of the battery without regard to the internal resistance. The E force into our homes is 240 volts from a transformer which is center tapped to give two 120 volt sources. The symbol V is the voltage measured across the circuit load such as a glow plug or light bulb. Each element of the circuit will have a voltage across it. If we have 12 Christmas tree lights in series in a circuit then there will be 10 volts across each bulb since the 120 volt supply will be divided equally across each bulb.
If we measure the voltage across a battery which is supplying current to a circuit, we will measure less voltage than the E which is inside because there will be an internal voltage loss due to the battery's resistance. This discussion will be clear after we see ohm's law. 
The symbol I is the current flow in amps through the circuit. Symbol I will be flowing in the opposite direction to the electron flow and is called a conventional current. You can think of I as consisting of tiny positive charges which are emitted from the positive side of the battery and are drawn into the negative side.
Let's follow a single positive charge around the circuit. Starting at the negative battery terminal, the particle feels a rise in energy as it moves through the battery. It follows the conductor to the load of our glow plug element and is pushed through it causing heat which takes away it's gained energy. It then follows the conductor back to the negative battery terminal.
When the particle passes through a resistance, there is a voltage loss. The polarity of the loss goes from plus to minus as the particle passes through the load. The symbol R is used to represent the various resistive losses in the circuit and is expressed in Ohms. In our low voltage hobby world, resistance is the enemy and must be eliminated in every way possible. Resistance is present in our batteries, wires and connectors. Our batteries come with low internal resistance and as they are used and abused, the resistance increases to a point that they are no longer able to supply the needed current. When current passes through the battery's internal resistance, we get a voltage loss called an IR drop which leaves less external voltage to do the job.
To prove that this is happening, let someone put a voltmeter across your field battery while you are running the starter to turn the model's engine. The 12 volts will drop due to the voltage lost across the battery's internal resistance and since high current is flowing, there will be even less voltage available at your starter motor's terminals due to the voltage lost in the supply wire and connectors.
The symbol P is for power supplied or used in watts. There are 746 watts in one horsepower which is a useful tool in powering our electric models. The popular Lipo 3 cell 2100 mah battery supplies 11.1 volts and can easily supply 20 amps. Since power is defined as voltage multiplied by current, the power output would be 11.1 times 20 which is 222 watts. If 222 is divided by 746 we have about 0.3 hp. A peppy model will use 100 watts per pound.
The attached circuit, shows a 12 volt battery supplying a 6 ohm resistance. There are no losses considered in the battery or wires. Ohm's law is given as E=IR and can be expressed in three ways as shown. We find that the circuit carries 2 amps. When the current flows through the load, then the IR drop is 2 times 6 or 12 volts. the loss in the load is what we gained at the source.
Joule's power law is P= EI which shows that the battery is supplying 12 times 2 or 24 watts. The load is consuming 24 watts which is equal to VI. Power across a resistor is also expressed as I squared times R which would be 2 times 2 times 6 or 24 watts.  If you want to test your new skills why not add one ohm in the circuit for the battery's internal resistance and refigure the current and voltages.

 

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