Explaination of voltage drop across resistors

[SyNtHeSiS]

Homework Statement

I want to know that if there is a voltage drop across a resistor, why is it that voltage increases as resistance increases (Ohm's law)?

Homework Equations

None.

The Attempt at a Solution

N/A

 

[n.karthick]

What do you mean by saying

why is it that voltage increases as resistance increases (Ohm's law)?

can you explain.

 

[SyNtHeSiS]

In Ohm's law, resistance is directly proportional to voltage, but I want to actually know why voltage increases, as resistance increases.

 

[n.karthick]

ohh are you asking similar to this :
In a circuit with two resistors R1 and R2 connected in series to a voltage source V, if R2 increases why voltage across R2 increases?

Well, the formula for voltage across R2 (say v2) is
$$V_2=\frac{R_2}{R_1+R_2} \times V$$

From this relation, you can deduce that if R2 is increased, V2 is definitely going to increase.

BTW Ohm's law says that (at constant temperature) current flowing in a conductor is proportional to voltage applied across it and the proportionality constant is 1/Resistance.

 

[SyNtHeSiS]

No I ment like I understand why current increases, as voltage increases (cause you supplying more energy in a way which makes a stronger current). But I don't see how an increase in voltage leads to an increase in resistance.

 

[jegues]

But I don't see how an increase in voltage leads to an increase in resistance

You don't get and increase in resistance by increasing voltage.

Larger resistors have larger voltage drops. Why? Before we can answer that we must better understand what exactly a resistor does.

A resistor resists the flow of current. This resistance means that some work must be done to "push" current through the resistor. Whenever work is done on charge, we have voltage. Thus, when current flows through a resistor, there is some voltage across the resistor.

The larger the resistance, the more work required to "push" the current through the resistor, the more work done the larger the voltage drop across said resistor.

 

[SyNtHeSiS]

Thanks for the explanation. And something which has been bugging me is say you had a circuit with no resistance and you wanted to calculate the current, how would this be possible since: I = V / 0 would be undefined?

 

[vela]

The only hope to obtain a finite answer for the current would be if V=0. That's why in an ideal conductor, the voltage difference between any two points is zero.

There are also effects like inductance and capacitance, which will keep the current from diverging due to a change in potential.

 

[SyNtHeSiS]

The only hope to obtain a finite answer for the current would be if V=0. That's why in an ideal conductor, the voltage difference between any two points is zero.

How would this be possible since if V=0 then I = 0/0 would still be undefined?

 

[vela]

Better to say indeterminate than undefined. You can avoid the division by zero if you write Ohm's law in its traditional form, V=IR. If R=0, the voltage has to be 0, but the current could be any value.

 

[SyNtHeSiS]

You wouldn't be able to determine I using Ohm's Law right? Since a circuit must contain a resistor to obey Ohm's Law?

 

[vela]

Right. Ohm's law doesn't tell you anything when the resistance is zero other than the voltage has to be zero as well.

 

[jegues]

Right. Ohm's law doesn't tell you anything when the resistance is zero other than the voltage has to be zero as well.

Realistically you would never have a case where the resistance is totally 0 would you? I mean, there would always be some infinitesimal amount of resistance, no?

 

[vela]

For your normal, everyday conductors, like copper wire, yes, there is always some resistance. Superconductors, however, do have 0 resistance.

 

[wwshr87]

Voltage and current are like kinetic and potential energy, there must be conservation of energy. Therefore, as voltage increases current decreases and vice versa.

 

[SyNtHeSiS]

Oh ok also say you had a circuit with only a +20V battery and one resistor connected in series with it to earth. Why is it that all 20V would be dropped across the resistor, even if R = 10 ohms or if R = 1k ohm?

 

[Zryn]

We could analyse how the Voltage is slowly reduced over very small increments of wire resistance as if we are doing a voltage divider, but relatively, the resistance of wire is most likely negligable compared to the resistance of the resistor.

For ease of calculations we assume the resistance of the wire to be zero.

In real life this is not how it always works, for example in mines where machines operate underground on 415V sources, a lot of the time generators produce ~433V because the length & resistance of the cables used drops a significant amount of voltage.