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Posted:

Mar 16, 2009

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active| newest | oldestYou can look at this like dragging a heavy sack across a floor: the type of floor (shiny / carpeted / sticky) affects how hard it is to move the sack, which you can call the floor's

resistance. Obviously, dragging two sacks requires twice as much energy.Electrical conductors have a property called resistance which is a measure of how easily (or not) current will flow.

Ohm's law relates the

resistanceof a material to theenergyneeded to drive electricity (current) through it.The energy term is Voltage (V)

The current (I), (in Amperes) is a measure of how much electricity is flowing (per second)

Resistance (R) is in Ohms

Voltage = current (I) x resistance (R)

From this you can see that Voltage (energy) increases with more resistance and also more current. More energy is needed if the resistance is higher, more energy is needed if more electricity is being pushed through.

or

Resistance (R) = Measured voltage (V) divided by current (I)

If the voltage is higher (more energy is needed to move electricity) the resistance will be higher. If the current increases, the resistance to it flowing must decrease.

Any good?

L

(This doesn't apply to Superconductors, which have no resistance at all)

V stands for voltage and is measured in volts

I stands for current and is measured in amps

(R = V / I) = (resistance = voltage / current) = (ohms = volts / amps)

(V = I * R) = (voltage = current * resistance) = (volts = amps * ohms)

(I = V / R) = (current = voltage / resistance) = (amps = volts / ohms)

OK I know that looks like a lot but its really not, I'm just restating the same thing in multiple terms that may make it easier for you to understand.

So basically what it's saying if if we know any two of the variables we can use math to determine the third. Now if I have a LED and a battery I want to power the led with, I know that the battery will supply 3 volts, and that my LED has a voltage drop of 2V and is rated for 20 mA and will burn out if operated above this amperage so I need to add a resistor to limit the current that may flow through the LED. Now I can say I know the voltage and I know the desired amperage so I will divide the voltage across the resistor(3 volts from the battery - the 2 volts dropped across the led =) 1V by 0.020A and use a 50 ohm resistor.