# Ohm law

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#### sfx81

##### New Member
Hi all. I am newbie to electronics and this forum.
I am trying to get my head round ohm's law ( V = IR ).
I got 2V power supply with current set to 0.04 A. I connected both ends power supply to a resistor of 470 Ohms. Now when I connect multimeter (set to volt), to the ends of resistor I get reading of 2.0 Volts.
From Wiki :
"A resistor is a two-terminal electronic component that produces a voltage across its terminals that is proportional to the electric current through it in accordance with Ohm's law"

so how come I get 2 volts ?
Any help will be appreciated
Thanks
Kazz

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#### crutschow

##### Well-Known Member
The power supply is putting out 2V as you stated. You are measuring the power supply voltage.

According to ohms law the current through the resistor is I = V/R = 2/470 = .00426A which is well below the .04A rating of your supply.

So why would you expect other than 2V to be read by the voltmeter?

#### Preher TV

##### Member
because the 470 ohm resistor drops the 2.0 volts, and the current is roughly .00425A, plug those number into I=E/R notice how they work out, the power supply set to .04A only means that you can draw that much max but that is not the current across the 470 ohm resistor, because that would mean the voltage would be over 18 V, that is not possible as a resistor is a passive device, and the source is only 2 volts. Using ohms law you can see that if you put another 470 ohm resistor in series with the 470 ohm resistor in your picture the total resistance would now be 940 ohms so the current using I=E/R would be .00212A or 2.12mA so the voltage drop across each resistor using E=I*R would be .9964 or about 1V so each resistor would drop half the 2 volt source, this is a simple voltage divider......starting to click? If not we can help till you get it electronics can frustrate sometimes..

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#### sfx81

##### New Member
What does this relation tells then
V = IR => 0.02 * 470 = 9.40 v
I mean where is this 9.40v ?
Thanks

#### Preher TV

##### Member
it's 470*.002 not .02 then you get .94 or 1V that is half the 2 volt source ,if 2 equal resistances are in series they will both drop half the source votage, in your picture if you had 2 470 ohm resistors in series instead of the 1 reistor, you would measure 1 volt across each resistor which equals the 2V source.

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#### crutschow

##### Well-Known Member
If your power supply was a current source whose output was 0.02 amps then the voltage across the resistor would be 9.40V. But your power supply is a voltage source with a current limit of 0.02 (or 0.04)A. It only supplies that current when the resistance is low enough (2/.02 = 100 ohms).

If you want to make the supply look like a current source then you set the voltage of the supply to its maximum value and adjust the current limit to the desired current. Then the supply will adjust the voltage to deliver that current to any resistance (up to the maximum voltage of the supply).

#### sfx81

##### New Member
If your power supply was a current source whose output was 0.02 amps then the voltage across the resistor would be 9.40V
I am bit more confused now. ( Correct me if wrong ) On all the batteries and power supplies, they act as voltage source and a current source. Like in ordinary AA cell its 1.5v and 1300mA = 1.3A. So in my case its voltage = 2.0V and current = 0.02A. Do you mean I can use seperate voltage and current sources ? or Can I use current only without Voltage and vice versa ? So I get from this is that if you only run current (with out voltage) through resistor, you get voltage across its ends i-e 9.4 volts ???

Thanks

#### MikeMl

##### Well-Known Member
You are confusing yourself because your supply just happens to have a current limit adjustment. Forget that it has one, or just set it to 1,000,000A. Now set your supply to 2.0V. Connect your 470Ω resistor. Now how much current flows from the supply into the resistor? (Hint: The current limit has nothing to do with it unless the current limit setting is less than 2/470= 4.25mA!)

#### crutschow

##### Well-Known Member
You need to look at a definition of current sources and voltage sources. It's true that voltage sources also act as a source of current but they are not technically a "current source".

The definition of a "voltage source" is a device that generates a fixed output voltage and provides whatever current the load requires, based upon ohms law for the load resistance and the supply voltage setting. In theory a voltage source has zero ohms output impedance (in practice it's very low).

The definition of a "current source" is a device that generates a fixed output current and generates whatever voltage the load requires, again based upon ohms law for the load resistance and current setting. In theory a current source has an infinite output impedance (in practice it can be very high, achieved by electronic feedback).

Most typical power sources are voltage sources. Some power supplies have a current limit setting but they still act as voltage sources.

You can't have voltage across a resistance without current, nor current through a resistance without voltage. The relation between the three is ohm's law. If you know the value for two of these, you can always calculate the third.

Current without voltage can only be achieved through a superconductor which has no resistance.

#### birdman0_o

##### Active Member
Current without voltage can only be achieved through a superconductor which has no resistance.
Wait what? What does that mean #### sfx81

##### New Member
Ok here's what I now understand.
power supplies can act as voltage source or current source. You need to fix one and adjust the other. So when I set my voltage to max 21V and current to 0.02A , then voltage across resistor (470 Ohms) is v = IR = 0.02 * 470 = 9.4v. Similarly if I adjust my current to a max of 5Amp, and voltage to 2.0V then I get current across a resistor I = V/R = 2.0 / 470 = 0.004Amps.

Thanks every one for your time, and I appreciate that.
Kazz

#### crutschow

##### Well-Known Member
Wait what? What does that mean Normally current always flows through some resistance which creates a voltage drop across that resistance. Thus to maintain current flow you need a voltage source to generate that voltage.

But a superconductor has no resistance, thus a current can flow without any voltage drop. For example, they make superconducting electro-magnets that are wound with a closed loop of superconducting wire to generate the magnetic field. Once they get the current flowing in the wire, it will stay flowing indefinitely without any power source, as long as the wire is maintained at a temperature below the superconducting point.

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#### crutschow

##### Well-Known Member
Ok here's what I now understand.
power supplies can act as voltage source or current source. You need to fix one and adjust the other. So when I set my voltage to max 21V and current to 0.02A , then voltage across resistor (470 Ohms) is v = IR = 0.02 * 470 = 9.4v. Similarly if I adjust my current to a max of 5Amp, and voltage to 2.0V then I get current across a resistor I = V/R = 2.0 / 470 = 0.004Amps.
Kazz
You got it exactly.

#### EN0

##### Member
This is what I do:

Find the circuit current, I=E/R - I=2V/470Ω = 4.25 or 4.3mA

Next, find the voltage drop across the resistor, where E=IxR so E = 4.3mA x 470Ω = 2.021 or 2V

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