Plain language definition of Ohms & Ohm's Law of electrical resistance:
What are Ohms and how are ohms or resistance measured & used in understanding electrical circuits, electrical resistance, and heat?
We also give a little of the history of the development of Ohm's law as well as links to supporting research.
This article series gives definitions of amps, volts, watts, resistance, current, ohms, electrical phases. We include basic formulas relating amps, volts, resistance, watts, and we explain what these electrical terms mean in practical applications such as for building or appliance electrical power, electrical wiring, and basic troubleshooting.
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Electrical Resistance is illustrated at left courtesy of Carson Dunlop Associates, a Toronto home inspection, education & report writing tool company [ carsondunlop.com ].
Electrical resistance is measured in Ohms Ω and is related to Watts and Volts by the simple equations we show here:
Watts = Volts 2 / Ohms
Current (Amps) = Potential (Volts) / Resistance (Ohms)
Electrical resistance can be thought of as how easily electricity flows through a material. Where resistance is high more effort is needed. A smaller-diameter electrical wire has more resistance to electrical flow than a larger-diameter wire.
A reason that the light bulb filament has high resistance is that it is very small in diameter.
Beginning with Thomas Edison, researchers discovered that if resistance in a wire is high enough the wire will get hot enough to glow (produce light) or even to start a fire (which is why the inside of an incandescent light bulb is a vacuum - to deny oxygen and thus protect the filament from simply burning up).
Ohm's Law, written about as early as 1791 and first formally published by Georg Simon Ohm (1789-1854) in 1827, states very simply the relationship between electrical current (Amps), electrical voltage (Volts) and the resistance of movement of electrical energy in a metallic conductor (Ohms).
I = V / R
tells us that the current (Amps) through a conductor (wire) between two points on a circuit is proportional to the potential difference (Voltage drop) across the two points and that the current (Amps) between the same two points is inversely proportional to the resistance between them (Ohms or Ω).
We can re-write Georg Ohm's law to describe each of amps, volts, or resistance in terms of the other parameters, as shown below.
I = the current, measured in Amps; I = V / R
and using simple algebra to re-write the Ohm's Law equation,
V = the difference in potential between the same two points, measured in Volts;
V = I x R
R = Ω or the resistance in the conductor or circuit between the same two points, measured in Ohms;
R = V / I
Also see ELECTRICAL RESISTANCE vs HEAT GENERATED
Also See JOULES HEATING LAW
Resistivity
Resistivity or electrical resistance describes how strongly a material opposes or "resists" the flow of electric current. Low resistance means easy flow. High resistance means more-limited flow of electricity.
Symbols for resistivity are the greek letter ρ (rho) or the SI unit of electrical resistivity the ohm-metre (Ω . m)
The resistance between two contacts at two points of an electrical circuit is expressed in ohms or Ω.
Resistivity is the reciprocal of conductivity.
Technically, resistivity or electrical resistance is a measure of the scattering of electrons, where when more electrons are scattered the resistance is higher and can be written as
σ = ne2ℓ / mevrms
Where
σ = electrical conductivity [S/m]
n = density of free electrons [e/m3]
e = charge of an electron (1.60 × 10−19 C)
me = mass of an electron (9.11 × 10−31 kg)
vrms = root-mean-square speed of electrons [m/s]
ℓ = mean free path length [m]
Conductivity Electrical conductivity is a measure of the degree to which a material will conduct electricity.
Similarly, heat conductivity is the rate with which heat passes through a material, or the amount of heat that flows per unit of time through a unit area with a temperature gradient of one degree per unit of distance.
Conductivity can be calculated as the ratio of current density in the material to the electric field that causes the current flow.
The conductivity between two contacts at two points of an electrical circuit is expressed by the greek letter σ (sigma) as you'll see making a cameo appearance in Glenn Elert's homely formula above. Some texts use kappa κ or gamma γ or in SI units (S/m) (Siemens per metre) to express conductivity.
Conductivity is most-easily expressed as the reciprocal of resistivity.
Technically, conductivity in metals is a statistical/thermodynamic quantity. But that's not going to help us undestand how the little electric heater works in a room thermostat heat anticipator.
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Below you will find questions and answers previously posted on this page at its page bottom reader comment box.
I have a 1967 Airstream International planted in the backyard and want to run an extension cord out to it to run basic electric. It has 5 amber running lights and a porch light on the outside that runs on DC. It is completely gutted and I have access to the single wire from each fixture. What do I need to do to hook up the outside running lights to a power strip. - Rob
Rob,
At your local electrical supplier or auto supply store or even at Radio Shack you should be able to pick up a small External AC to DC power converter (aka AC to DC Power Supply). What you need is a converter that will have high enough Amps output for the few trailer lights that you cited.
I used the term "External AC to DC Power converter or power supply because you don't want to have to buy a separate cabinet or case and do assembly. When shopping don't rule out existing computer or other electronic power supply "bricks" - just take a look at the DC Wattage output that the supply can provide. If it's big enough you'll be OK.
For just a few running lights and a porch light, most likely you won't even be drawing 10 watts, but to be on the safe side and to allow for expanded use of your power supply once you figure out how useful it is, I'd look at a unit with 35 watts or larger output.
Take a look at our DEFINITIONS of ELECTRICAL TERMS beginning at the top of this page for help with AC, DC, Amps, and Watts.
Run a weather-resistant outdoor-rated extension cord from an outdoor GFCI protected receptacle over to the Airstream and inside it where it will be weather protected. To that 120V (probably 15A) circuit, you'll plug in your AC to DC Converter. You'll then wire the DC output terminals to your Airstream lighting circuit.
If you're going to use this electrical supply system frequently you might want to put together a suitable plug connection feeding the DC powered circuit and a suitable connector on the wires that you run from the AC-DC converter.
For more permanent power conversion of an older Airstream or other mobile home or trailer, I'd look at what's offered by RV suppliers. Certainly I had no trouble retrofitting an AC to DC connector and power supply on an old slide-on pickup truck camper, thus allowing us to "plug in" to any 120V DC power source and run all of the DC devices in the camper.
(May 26, 2015) Mike L said:
Is 1 amp AC equal to 1amp DC? My golf club stores the batteries for member's walking power carts. There are about 20 battery chargers and batteries plugged into one AC circuit that is presumably a 15 amp circuit. My charger has a 2 amp output to the battery and the others are probably the same.
If all of the batteries needed charging at the same time, drawing 40 amps, then the batteries would have to share the available 15 amps resulting in a very slow charge.
Am I correct?
Technically AC & DC amps are not precisely equivalent. The DC amperage draw will be slightly less than AC.
For example if you are selecting an electrical switch capable of handling 5A at 125VAC, it should be fine to use that switch on a DC current since the actual DC amperage draw will be a bit under 5 amps.
Apparent Power (VA) = Amps x Volts regardless of AC or DC
20A at 120VAC = 2,400 Watts (apparent power)
Also the amperage rating of a switch is increased as the voltage is decreased.
Example: a 120VAC 5-Amp switch can handle about 600 Watts and according to batterystuff.com's calculator, supports a 12-Volt DC amperage of 55.
The more-accurate formula for calculating watts is
Calculate Actual Power in Watts
Amps (A) x Volts (V) x Power Factor = Watts (W)
where the power factor of an AC circuit is the ratio of actual or "real" power consumed by the circuit (W) divided by the "apparent power" consumed by that circuit (VA).
So our first formula above is really "apparent power" in Watts.
The power factor is a measure of the efficiency with which an electrical device onverts volt-amps into power. A 100% efficient electrical device would have a power-factor of 1.
But in all conventional electrical devices, some of the energy supplied by the circuit is lost in the form of heat. Thus the true power factor will be some number less than 1.
To know the actual efficiency or power factor of an electrical device we use a power meter, or a watt meter, an electrical test tool, to measure the actual power consumed in watts.
For an accurate calculation of actual energy consumed that includes the effects of AC current and power factors,
see DEFINITION of POWER FACTOR, REAL POWER
(June 18, 2015) barry said:
is there a formula i can use to calculate the cost of a water well pump, and a booster pump?
(June 18, 2015) bbrechbiel1950@gmail.com said:
hi my name is barry i had ask a question about calculating a well pump and booster pump for cost. i didn't receive an answer so i put my e-mail, by the way you do an excellent job on teaching basic electrical, keep up the good work. thanks barry 2000hrs 6-18-2015
To convert the current draw (from the pump data tagor by actual measurement) to watts and watt hours to relate the electricity usage to what appears on your electric meter and electric bill in watt hours:
Energy E in watt-hours (Wh) = Power P (in watts W) x time period t in hours (h), or
E (Wh) = P(W) x t (h)
(July 5, 2015) layne said:
In selecting inverters for a specific product I know the Volts need to be the same, but as for the amps....if your inverter produces more Amps than needed, is this a problem?
Layne
No not as I understand it; the amps rating is the maximum amount of current draw or demand that can be asked of the inverter. If your application draws less current that's ok.
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