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This question already has an answer here:

I'm sure this is a very simple question, but I have been confused about this topic for a while now. According to Ohm's law, resistance varies directly with voltage. This means that if resistance increases voltage increases... But obviously that's not how it really works. If I add in a resistor to a circuit, the voltage decreases. I've heard that its because a resistor reduces current which in effect reduces voltage, however I don't understand how this lines up mathematically with what I said earlier.

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marked as duplicate by Ignacio Vazquez-Abrams, PeterJ, Daniel Grillo, nidhin, Dave Tweed Jan 20 '16 at 13:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ +1 for asking a useful question that I can't believe was closed as a duplicate of that question. "V = IR but voltage drops and current is constant. Why?" is a very reasonable question. \$\endgroup\$ – person27 Feb 8 '17 at 19:49
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According to Ohm's law, resistance varies directly with voltage

You should read this the other way. Voltage varies directly with current. "R" is the constant of proportionality telling how much it varies.

If I add in a resistor to a circuit, the voltage decreases.

If you have a resistor in a circuit, with a current flowing through it, there will be a voltage dropped across the resistor (as given by Ohm's law). If the resistor is in series with some other element, and they together are powered by a constant voltage source, then the voltage dropped across the resistor means there's less voltage available for the other circuit element. It doesn't mean that the voltage of the source decreased.

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  • \$\begingroup\$ That kind of makes sense... So are you saying R is a constant and not a variable? \$\endgroup\$ – Dallin Jan 20 '16 at 6:19
  • \$\begingroup\$ Yes, the simplest situation is when R is constant and I and V are the variables. Of course you could also consider a potentiometer driven by a constant voltage or current source, or a LDR, or a thermistor if you want to think about situations with variable resistors --- but those all vary in response to an external stimulus, not to the applied voltage. You can also have resistor that varies in response to applied voltage (a diode's DC behavior is like this), but then you are talking about nonlinear circuits and Ohm's law no longer applies. \$\endgroup\$ – The Photon Jan 20 '16 at 6:26
  • \$\begingroup\$ The definition of a good "resistor" is that it has a fixed resistance. Other devices may change 'resistance' but they are generally not considered resistors. \$\endgroup\$ – Daniel Jan 20 '16 at 7:07
  • \$\begingroup\$ @Daniel, a thermistor or an NTC inrush current limiter certainly is considered a type of resistor. And in the realm of circuit theory, anything with a memoryless I-V is considered a (nonlinear) resistor. \$\endgroup\$ – The Photon Jan 20 '16 at 15:44
  • \$\begingroup\$ @ThePhoton Consider the audience, someone who doesn't understand what a resistor is. For the sake of this discussion, a good resistor doesn't change resistance with temperature or voltage or current or magnetic field, or whatever. \$\endgroup\$ – Daniel Jan 20 '16 at 20:03
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That's not Ohm's law.

Ohm's law is R = V/I -- or the ratio of voltage to current is a constant.

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I will emphasis on these points based on your question

"According to Ohm's law, resistance varies directly with voltage. This means that if resistance increases voltage increases"

Yes it is true ,i.e Voltage drop across a resistor increases,say if you connect a 1k and 10k resistor in series Voltage drop across 10k will be more when compared to 1k resistor.

Refer:http://www.allaboutcircuits.com/textbook/direct-current/chpt-5/simple-series-circuits/

" If I add in a resistor to a circuit, the voltage decreases."

Absolutely No,adding resistor in a circuit drops the voltage across itself but now across whole circuit i.e you may get less voltage at particular NODE but it but if you measure effective voltage it will be equal to the source(IDEAL condition)

Refer:http://www.allaboutcircuits.com/textbook/direct-current/chpt-6/voltage-divider-circuits/

Last but not the least

"I've heard that its because a resistor reduces current which in effect reduces voltage"

What you've heard is wrong , resistor never reduces current it just limits current i.e it slows the speed of electrons flowing in the circuit, and electric current as per wiki is An electric current is a flow of electric charge,so if you are slowing the electrons it doesn't mean you are reducing the current it simply means you are allowing limited no of electrons per second through resistor while remaining electrons are dissipated heat .

Hope this helps,

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If you have a constant current source passing through a resistor, then, yes, increasing the value of the resistor will increase the voltage drop across it.

Ohm's Law gives the resistance as the ratio of the voltage and current, as R = V/I.

None of these are necessarily constant, all three are variables. In certain circumstances, you may be able to treat one or two of them as constant, or nearly so. But Ohm's Law always gives the ratio.

Bear in mind you can measure the voltage at a resistor's terminals with a meter. You can measure the current flowing through a resistor with a meter. But the resistance, the inherent stuff going on inside the component that gives rise to the ratio, cannot be measured independently. The resistance is defined as the ratio of terminal voltage to through current, you have to make both measurements and do a sum. That is what a multimeter does when measuring resistance.

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It's simply but it is not Ohm's law. Please see the context "Kirchhoff's circuit laws". This law is also called Kirchhoff's second law, Kirchhoff's loop (or mesh) rule, and Kirchhoff's second rule.

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