What does a resistor do?

Atreyo Bhattacharjee

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Feb 7, 2017
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I've read about this online, but nothing really answers my question, or does it in terms I don't understand. So, does a resistor decrease current (amps) or voltage (volts)? Also, if something is rated for lets say 5 volts, and I run 10 volts with half the current, will it damage the component (lets just say an LED)? And how can I calculate how much current/voltage will come out of a resistor based of the resistors resistance (Ohms) and how much current/voltage I put in?
 
Solution
Resistors create voltage drops, not current.
Measure the circuit from ground to one side of a resistor and you may read 12v.
Test the other side and it may be 11v.

The formula is I (Amps) = Voltage / R (Ohms).
So the voltage drop across a resistor is Amps x Ohms.

If you run a 5V 1.0A LED at 10V 0.5A, you will probably damage it.
But the Wattage (V x A) will still be the same (5W).

It is best to think of electricity like water in a pipe.
Voltage is Water pressure. Pounds per square inch.
Current is how much water is flowing. Gallons per second
Resistance is how thin is the pipe (wire) you are using.

If you try putting a fire hose of water through a garden hose it will burst.
If you try putting 10v through a 5v circuit it will heat...
Ohm's law is I = V / R, or R = V / I or V = R * I. Use it to make your calculations. In regards to component ratings, if it's rated for 5V max, no you can't run it at 10V and half the current. You would still be putting 10V on the device, when it's rated for 5V.
 

JoeMomma

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Nov 17, 2010
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Resistors create voltage drops, not current.
Measure the circuit from ground to one side of a resistor and you may read 12v.
Test the other side and it may be 11v.

The formula is I (Amps) = Voltage / R (Ohms).
So the voltage drop across a resistor is Amps x Ohms.

If you run a 5V 1.0A LED at 10V 0.5A, you will probably damage it.
But the Wattage (V x A) will still be the same (5W).

It is best to think of electricity like water in a pipe.
Voltage is Water pressure. Pounds per square inch.
Current is how much water is flowing. Gallons per second
Resistance is how thin is the pipe (wire) you are using.

If you try putting a fire hose of water through a garden hose it will burst.
If you try putting 10v through a 5v circuit it will heat up, burn and fail.
 
Solution

TJ Hooker

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Ambassador
By and large current is a function of voltage. If a component usually draws 1 A at 5 volts, you probably can't get it to only draw 0.5 A at 10 V. You can set the voltage, but the component will draw whatever it's going to draw based on that voltage and the component characteristics.

A resistor does both of the things you mentioned. It is a component that resists the flow of electricity, and whenever there is currently flowing through it there's will be a voltage drop across it.
 
Yeah - @JoeMomma gave you a good answer there.

Just to add one thing though, you can't directly "choose" your current (=amperage/amps). You were considering running "10 volts with half the current", it doesn't work like that. You don't decide how much current flows through a circuit, the voltage and resistance of the circuit determine the current.

Let's take your example of the LED: that LED will have a specific resistance. In other words, raising the voltage has no impact on the resistance . So, according to our I = V/R equation, if you double the voltage the end result will be double the current (amperage) flowing through your LED, which will likely burn it out. You can't choose to "halve" the current; if you have fixed resistance and double the voltage, the resulting circuit will have double the current flowing through it (unless or until something breaks)... that's the way electronics work.

To offset it however, you could double the resistance in the circuit. If you added a second resistor to the circuit (either a second identical LED in series, or some other resistor which matched the resistance of the LED and could cope with the current), then it would all work fine. If you double the voltage, but double the resistance to match it, the level of current will stay the same and everything will keep working fine. What's happening to the voltage in that case is that each resistor will produce a 5V drop in voltage. The first resistor in the circuit will see the voltage drop from 10V to 5V, (with 5V drop across that resistor - so the LED is happily within its spec), while the second resistor will see the voltage drop from 5V to ground (0V).
 

TJ Hooker

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Ambassador
@rhysiam just as an FYI, diodes (i.e. LEDs) are nonlinear circuit elements Current through them is an exponential function of voltage, so doubling voltage would typically result in much more than double the current. Although this detail doesn't really affect the point you were trying to make.
 

Didn't know that, thanks! So if we were talking about incandescent light bulbs or "normal" resistors in general, my description above would be accurate, but specifically not for any diodes.

Hopefully my clarification was still helpful despite the confusion there. Thanks.