# How can a resistor affect current AND potential at the same time?

Though the resistor is always introduced as one of the most simple components it is the one which makes least sense to me.

Ohm's law defines the resistance as$$R = \frac{V}{I}$$ this means that the voltage is defined as $$V = I \cdot R$$ and the current as $$I = \frac{V}{R}$$.

So following the law a resistor must affect both voltage and current however the reality is that it only changes one size.

1. To lower the voltage
2. To lower the current

This does not make much sense to me because in my understanding voltage and current must be lowered both but in the common LED resistor example it only affects one size:

$$U = 9V\\ I = 30mA\\ R = 300Ω$$

you also find use cases where only voltage is affected. How do I interpret this?

What is the factor which determines if the resistor affects either voltage or current?

• What is the "common LED resistor example"? What are "cases where only voltage is affected."? Nov 22 '14 at 14:04
• why was this question downvoted ? It asks a questions and shows his reasonsing behind his thoughts ? Nov 22 '14 at 15:37
• May be I didn't understand well your question, but have in mind that the flow rate of an electric current would increase with increased potential difference and decrease if the resistance of the circuit is increased. Nov 23 '14 at 4:34
• It changes both... Aug 14 '18 at 1:58

There is no factor that determines if the voltage or the current is reduced. That whole concept is erroneous.

The simple statement you are looking for is:

A Resistor Defines the Relationship Between the Voltage and the Current

That is, if the current is fixed, then the resistor defines the voltage. If the voltage is fixed, then the resistor defines the current.

In all three of the Ohm's Law formulae you will have two of the three values as fixed values - values you know, through measurement, or whatever, and the third variable is the one you want to find. From there it's simple maths.

The LED example, though, throws an extra spanner in the works, since the LED isn't a linear device. So its influence on the circuit is calculated separately before Ohm's Law is applied.

You have three known values, and you want to calculate a fourth.

The known values you have are: the supply voltage (9V), the LED forward voltage (say, 2.2V as an example), and the current you want to flow through the LED (30mA).

From that you want to calculate the value of the resistor.

So you subtract the LED's forward voltage from the supply voltage, since those are both fixed voltages, and the result will be the amount of voltage that must be dropped across the resistor for the whole to total 9V. So 9V - 2.2V is 6.8V. That is a fixed voltage. The current you want is fixed too - you have decided on 30mA.

So the resistor value is then: $$R=\frac{V}{I}$$ $$\frac{6.8}{0.03} = 226.\overline{6} \Omega ≈ 227 \Omega$$ You will always have two of the three values as fixed values - either because they are set by external factors, like the power supply or battery voltage, or they are a specific value that you require or desire, when it is you who has set that value. The third value is what must be calculated to make both those fixed values hold true.

• but how do I determine if a voltage or current is fixed or not? Nov 22 '14 at 14:10
• You will know already from the circuit. If you know the voltage, and you know the resistance, then you will be trying to calculate the current. Nov 22 '14 at 14:21
• @bodokaiser - there exist "fixed current" power supplies as well as "fixed voltage" - and everything in between - most of the time though, stuff tends to be closer to the fixed voltage. Nov 22 '14 at 20:58
• Also in this example, if you have a fixed current supply (thats supplying the desired current) then the resistor value doesn't matter, and so can be 0 (i.e. you don't need a resistor). Of course, this will rarely be the case. Jun 13 '16 at 22:15
• @BWalker Most constant current supplies (or sinks) are effectively a variable resistor with feedback - the resistance is varied to keep the current constant. It's virtually doing the resistance calculation for you real-time. Jun 13 '16 at 22:19

however the reality is that it only changes one size.

Ohm's law relates the voltage across and current through a resistor. In general, a change in resistance will change both the voltage across and current through the resistor.

For example, consider the simple voltage divider circuit - a voltage source $V_S$ and two resistors $R_1$, $R_2$, connected in series.

The series current is just

$$I_S = \frac{V_S}{R_1 + R_2}$$

and the voltage across the second resistor is, by Ohm's law,

$$V_{R_2} = I_S R_2 = V_S\frac{R_2}{R_1 + R_2}$$

Now, double the resistance of the second resistor $R'_2 = 2R_2$

Both the voltage across and current through will change:

$$I_S = \frac{V_S}{R_1 + 2R_2}$$

$$V_{R'_2} = I_S R'_2 = V_S\frac{2R_2}{R_1 + 2R_2}$$

Only in the case that the voltage across is fixed by the circuit will only the current through change when the resistance is changed. An example would be a single resistor connected across a voltage source.

And, only in the case that the current through is fixed by the circuit will only the voltage across change when the resistance is changed. An example would be a single resistor connected across a current source.

In summary, Ohm's law holds for resistors but one must apply it in conjunction with other circuit laws such as KVL and KCL to fully determine the resistor voltage and current.

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied across the conductor. That means if you increase voltage the current will increase proportionally in the conductor.

For example if you have a voltage of one volt across a conductor having a resistance of 1 ohm then the current flowing through the resistor will be 1 amp. If the voltage is increased to 2 volts then the current flowing through the resistor be 2 amps. For 3 volts 3 amps and so on.

The basic thing in an electrical circuit is that voltage applied across a fixed circuit alone determines current flowing through the circuit. If you increase voltage the current will increase.

When you are connecting LED to power supply, LED has fixed resistance So voltage drop across LED is fixed. Hence voltage across LED and resistance of LED are fixed and so third term which is current will vary.

• The LED does not have a fixed resistance. It's I-V curve is exponential. In every point you can approximate the resistance, but it's only valid in that DC operating point for (very) small signal swings. Jun 4 '17 at 17:11