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# Beginner Question about Ohm's Law Thanks!

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Hi,

I'm a newbie to electronics and I'm confused about Ohm's Law especiallt about Resistors. A resistor limits current, but the voltage stays the same.  So when you make a circuit with a 9V battery and resistors  ( Just in Theory)
for example so it would be 9 Ohms x 1 A = 9V
and or  1 Ohms x 9 A = 9V

So how does a voltage divider work?
because I thought that resistors only limit current
??? ??? ???

Thanks you very much for clearing out the confusion

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A voltage divider has two resistors in series.
Some of the voltage is across one of the resistors and the remainder of the voltage is across the other resistor.

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The picture is a simple battery connected across two resistors in series, as you describe.

The first thing to understand is that the current is the same everywhere in the loop. There's nowhere for current to leave the circuit, so the current is the same at every point around it. I've labelled it I.

The second thing to know is that the total resistance of the two resistors in series is the sum of their individual resitances. That is, R1 + R2.

We can use Ohm's law to find out I. Since we know the battery's voltage, and we know the total resistance that this battery is pushing current through, we can work out the current I through those resistances:

I = Vs / (R1 + R2)

Now that we know the current I, we can work out what voltages will appear across the two resistances, again with Ohm's law:

VR1 = I * R1
VR2 = I * R2

Play with some numbers: Let's say Vs (battery voltage) is 9V, R1 = 1kOhm and R2 = 2kOhm.

I = 9V / (1000 + 2000) = 3mA

VR1 = 3mA * 1000 = 3V
VR2 = 3mA * 2000 = 6V

The sum of the two voltages across the resistors VR1 + VR2 (that you can measure to prove this empirically) will always be 9V, regardless of the resitances used.

In other words, the resistances can be chosen to divide the voltage across them into two separate voltages whose sum is 9V. The ratio of those two voltages is the ratio of the resistances. Thus, to divide your 9V by exactly 2 (to create 4.5V and 4.5V), choose two resistors of the same value.

You can use as many resistances in series as you like, to divide a voltage into any number of smaller voltages.

The maths above reduces to this:

VR1 = Vs * R1 / (R1 + R2)
VR2 = Vs * R2 / (R1 + R2)

Note that you don't need to know the current at all, but understanding the principle of the current being the same through both resistances allows you to see how the resistances each develop a share of the total voltage across them.

One last observation; if you are thinking you can now power a small 3V lamp by connecting it across the 1kOhm resistor in our example, think again. The lamp's low low resistance (just a few Ohms) in parallel with R1 will change the whole scenario.

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You also have to keep in mind the resistance of the load after the voltage passes the voltage divider network. Here are a couple of links that might help you understand a voltage divider better:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

http://www.doctronics.co.uk/voltage.htm#what

MP

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Zeppelin pointed out that the battery in my little circuit is drawn upside down. He's right - the positive side should be at the top, so that the upwards Vs arrow is in the right direction. Kirchoff must be turning in his grave. My apologies.

To hide my shameful error, I've edited the post wth a tweaked diagram.

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a fixed voltage divider has 2 resistances in series while a variable voltage divider has a resistance in series with a POT.

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Thanks very much for all the detailed info.

So if you want to vary the voltage, you use a voltage divider with a POT in series with a resistor.

If you want to vary the amperes, you just use just a POT in series in the circuit.

So about a battery, how does a battery vary the current but not the voltage or varying the voltage without varying the current. Like a car battery is only about 14 volts, but it has hundreds of amps?

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Thanks very much for all the detailed info.

So if you want to vary the voltage, you use a voltage divider with a POT in series with a resistor.

If you want to vary the amperes, you just use just a POT in series in the circuit.

So about a battery, how does a battery vary the current but not the voltage or varying the voltage without varying the current. Like a car battery is only about 14 volts, but it has hundreds of amps?

Not exactly. Your resistors are still going to limit the current. You just have two branches instead of one.

MP
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how does a battery vary the current but not the voltage or varying the voltage without varying the current. Like a car battery is only about 14 volts, but it has hundreds of amps?

How would you also design a circuit like this?

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Even though your car battery has hundreds of amps capacity, your circuit is only going to use what is needed by the device that is connected to the battery. You do not need current limiting for this purpose. However, if you get a short in the circuit or somethig else happens that starts sucking large amounts of current from the power source, you need to limit the total amount of available current or else you will have a marshmallow roast. Fuses are good protection, but if you have no current limiting, everything will be toast before the fuse does you any good.

Thus, to answer your question: A battery does NOT vary the current or the voltage. This must be achieved in your circuit design. Each circuit can be calculated down to a load, measured in ohms. In other words, the complete circuit can be looked at as a resistor.

MP

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a fixed voltage divider has 2 resistances in series while a variable voltage divider has a resistance in series with a POT.

The variable voltage divider might also be just a potentiometer and no fixed resistor.
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Hi, sorry about that , played around with the account and got deleted.

So is there a way to just adjust the current without varying the voltage and vice-versa?
like :

V    C  R
3    2  1.5
4    2  2
5    2  2.5

Thanks really much for your help!

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yes it's call a constant current source

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