Basics of Electricity
8
If you pump 1 gal/h of water into an aquarium for five hours it will contain 1
gal/h
5 h = 5 gallons. Similarly a battery charging at 1500 mA for one hour
will contain 1.5 Amp
one hour = 1.5 Ah of charge. Since we learned above
that there are 3600 Coulombs per Ah, the physicist would say the battery
contains 1.5 Ah
3600 C/Ah = 5400 Coulombs of charge. Amp-hr versus C.
They are different ways of saying the same thing.
2) Discharging
If you know a battery and an aquarium are both “full” you just read their
“capacity” (volume) specification to find the stored “charge.” The aquarium
says 20 gallons and the battery says 2000 mAh. The aquarium would take 4
hours to empty at a “current” of 5 gal/h and the battery would take 1 hour to
empty at a current of 2000 mA.
4.3
Voltage
Something forces current to flow in a wire. That something is called
voltage
and is
measured in
Volts
(abbreviated V). The
pressure
your hand feels pushing water out of a
pipe is the force analogous to voltage.
You can’t get a current without a voltage. The two are related by the concept of resistance.
4.4
Resistance
Resistance is measured in
Ohms
(abbreviated with the Greek symbol Omega
). It is the
property that limits the current, which flows in a wire for a given voltage applied across the
wire.
The very important “Ohm’s Law” summarizes this relationship.
Amps = Volts
Ohms
It says the amount of current depends on the voltage applied divided by the amount of
resistance. Push harder for a given amount of resistance and you get more flow.
Reducing the diameter of a water pipe increases its resistance and therefore reduces the
water flowing from it for a given pressure (voltage). Similarly doubling the resistance of a
wire halves the current flowing through it for a given voltage across it.
A useful thing to remember is that 1 Volt divided by 1 Ohm = 1 Amp.
It says that One Amp of current will flow in wire having a resistance of one Ohm if one
Volt of voltage is applied across it.
So given relatively fixed battery voltages, you need low resistance wires to allow high
currents.
Resistance is a general term for a rather complex phenomenon. There are different kinds of
resistance. For our purposes we will mean Direct Current or DC resistance. Alternating
Current or AC resistance is beyond the scope of this tutorial and is not widely discussed in
RC electrical systems.
Here’s an example problem using resistance.
How much voltage drop is there to a motor drawing 50 Amps if the motor is one
foot from the battery? Assume14 Ga. wire has a resistance of approximately
0.0025 Ohms per foot.
So we have two feet of wire total with 50 Amps flowing through them. Ohms
law tells us that to get 50 Amps through a wire of 2
0.0025 Ohms requires:
Volts = Amps
Ohms, = 50
(2
0.0025) = 0.25 V.
So those wires use up 0.25 V in battery voltage. Our motor would receive 7.15
Volts from a 7.4 Volt battery. Apparently, this motor consumes exactly 50
Amps at exactly 7.15 V.
"Watt's Up" & "Doc Wattson" Watt Meter and Power Analyzer User's Manual
RC Electronics, Inc.
