voltage across resistors - parallel vs series

[stealthelectric]

Hi:
I know that the voltage across each resistor in a series circuit is different and the same across resistors in a parallel circuit, but I don't know why.
Could somebody please help me to understand?
thank you

 

[audioguru]

Ohm's Law tells you why.

 

[stealthelectric]

yeah i considered that - but it seems more logical to me, to say that if the current going through a series circuit is the same across all resistors, and voltage is proportional to the current, then surely the...
yeah im confused - why does ohms law tell me why?

 

[Electronworks]

voltage is proportional to the current

The voltage being proportional to the current is called the resistance. If the resistance changes, so will the voltage.

OK for the parallel case: the voltage along a short circuit is the same. if you short 2 resistor ends together and apply a voltage to that end, the voltage across the 2 resistors will be the same.

For the series case: picture the current being like water in a pipe. the water going in = water coming out. Current going in = current coming out. unless you have a leak somewhere (another component connected where the 2 resistors meet), then the current in each resistor will be the same

As audioguru say, this is where Ohms law comes in: for a given current, if one resistor is twice the resistance of the other resistor, it will have twice the voltage across it.

Example:

Parallel: 2 resistors in parallel, each of 1 ohm. Apply 1V across the resistors. A current of 1A flows in each resistor and a total of 2A flows from your power source

Series: Apply 1V across the two 1Ohm resistors in series and you will get 0.5A flowing. 1 Volt applied to 2Ohms causes a current of 0.5A to flow.

is the mist clearing...?

 

[Vizier87]

....or you can imagine a tube with varying diameters throughout the length with water flow.. versus multiple tubes branching out from a single tube.. Hmm the analogy doesn't really help I think.

 

[stealthelectric]

I find it counter intuitive because I expect the parallel circuit to split the voltage, as it does the current. But I guess what everyone is trying to say with the water analogy is, if the tap is opened, this is the current flowing. The voltage is just the force/push, that moves the current. The part I find counter intuitive is that if a force is being shared by three avenues, then that force is still being divided up and lessened.

I just read that the defining characteristic of a parallel circuit is that all components are connected between the same set of electrically common points. So, I guess I need to think in terms of, the voltage just being a reading of a force that doesn't actually travel. It is just a measure of the effect that the resistance is having on the current, taken at any two given points in a circuit. And as all components are connected between the same set of electrically common points in a parallel circuit, then by this definition, the voltage must therefore be equal across those points. Maybe!

 

[ljcox]

Another term for voltage is Electro Motive Force (EMF).

It is roughly equivalent to the head of water in a water tank.

The "head" is the same regardless of how many pipes you connect to the tank.

For example, if you plumb a pipe into the bottom of the tank and water flows through it at say 1 litre/minute, then you plumb another pipe of the same diameter into the bottom of the tank the flow in both pipes will be about 1 litre/minute. Thus the tank is being emptied at 2 litre/minute. If you plumb another pipe of the same diameter into the bottom of the tank, the tank will be emptied at 3 litre/minute, etc.

 

[Electronworks]

The part I find counter intuitive is that if a force is being shared by three avenues, then that force is still being divided up and lessened.

You are thinking along the right lines here. If the voltage source only had limited current to give then indeed the voltage would drop. However, we assume in this case that the voltage source can output whatever current the circuit needs.

Your last sentence hits the nail on the head. All points shorted together will share the same voltage

 

[stealthelectric]

Well. I'm glad that such a low level question can create such fabulous responses. Thanks to all of you.

Electronworks said: The voltage being proportional to the current is called the resistance.

Resistance draws current in a circuit. In a series circuit, two resistors are along the same pipeline, so V=IR, where (I) remains a constant in a series circuit. Therefore, the higher the resistance, the higher the voltage.
When a component is added in parallel it is like creating another circuit. Another pipe for the current to flow through. And as voltage is proportional to current R=V/I, the voltage then depends also on the current flowing through the specific pipelines. Because the relationship is proportional, the voltage remains the same across all resistors in a parallel circuit. This is because V=IR, if you decrease the resistance, you increase the current, therefore, keeping the voltage across the different components constant.

Have I understood you all?

 

[Jony130]

Have you read this:

 

[ljcox]

The analogies illustrated by Jony130 are very good.

The "head" of water is equivalent to the voltage of a battery or power supply.

It is the potential to do work. That's why voltage is sometimes called potential.

 

[ljcox]

Well. I'm glad that such a low level question can create such fabulous responses. Thanks to all of you.

Electronworks said: The voltage being proportional to the current is called the resistance.

Resistance draws current in a circuit. In a series circuit, two resistors are along the same pipeline, so V=IR, where (I) remains a constant in a series circuit. Therefore, the higher the resistance, the higher the voltage.
When a component is added in parallel it is like creating another circuit. Another pipe for the current to flow through. And as voltage is proportional to current R=V/I, the voltage then depends also on the current flowing through the specific pipelines. Because the relationship is proportional, the voltage remains the same across all resistors in a parallel circuit. This is because V=IR, if you decrease the resistance, you increase the current, therefore, keeping the voltage across the different components constant.

Have I understood you all?

You're still confused. Resistance is the constant of porportionality between voltage & current.

V = IR. Hence I = V/R. Also, if you know the voltage across a resistor and the current through it, you can calculate the resistance using R = V/I.

Examples:-
If you connect a 1000 Ohm resistor across a 3 Volt battery, the current will be 3/1000 = 0.003 Amp or 3 mA.

If you connect a 500 Ohm resistor across a 3 Volt battery, the current will be 3/500 = 0.006 Amp or 6 mA.

If you connect a 500 Ohm resistor across a 9 Volt battery, the current will be 9/500 = 18 mA.

 

[stealthelectric]

yes they are good illustrations. And the link is a good read for both series and parallel circuits, and more. Thanks Jony130. It seems there are many good pages here. I think I will spend some time on the theory in these pages. It's very well written.
So, the voltage is proportional to the current and Resistance is the constant of proportionality. So if there is an increase in voltage there is a proportional increase in current, and vice versa - ohm's law; provided the temperature remains constant. What if there is an increase in resistance? Total voltage in a series circuit is equal to the sum of the individual voltage drops. So the voltage will remain the same but be split proportionally between all components in the circuit. But from the equation V=IR, if 50V=5A x 10ohms, with an increase in resistance so, 5A x 20ohms= 100V. These seem to be saying the opposite thing. Which I think is one of the points that has been confusing me. This would only hold true if there were only one resistor in the series circuit. Add more than one and ohm's law cannot be directly applied. This is because the two points in the circuit we are taking the reading from are now not electrically common, as there is more than a single component between them.

 

[ljcox]

Your post above is partially correct, but you're still missing the point.

If the resistance is increased, the current would be 50/20 = 2.5 Amp.

Your calculation would be correct if the resistor was connected across a CURRENT source.

But as far as I can see, you're assuming a VOLTAGE source.

 

[ljcox]

Ohm's law V = IR can be re-arranged to make either I or R the dependent variable.

ie. I = V/R and R = V/I.

You choose which ever form you need for the situation.

It depends on what you know.

Examples:-
1. if V and R are known, then you use I = V/R.

2. if you apply a voltage V to an unknown resistor and use a meter to measure the current, then you use R = V/I to determine the resistance.

 

[stealthelectric]

Ahh! I see where I went wrong there yes. Thanks. Because they are proportional you cannot increase or decrease one without it changing the other. I'm feeling silly now.
However, I have been looking at Kirchoff's circuits and in the image attached there are a number of resistors and a couple of voltage sources some series and parallel connections, but the calculation for the voltage drop across R2, V2 is, V2=I2 x R2. Which is a standard V=IR equation.
How can this be?

 

[stealthelectric]

I think I can answer that. Because each part of a Kirchoff circuit has independent current. If you know the current in a series circuit going through a component you can just use Ohm's law.

 

[ljcox]

The voltage across R2 is not V2 since V2 = 12 V.

The voltage across R2 is I2 x R2 as you said.

ie. Vdc = I2 x R2.

You need to use Kirchoff's Voltage Law on both loops and then solve the resulting similtaneous equations.

 

[wrinkledcheese]

I'm no electronics expert, or even knowledgeable, but in series means, to me, that one resistor end is connected to the next...in a series. Whereas in parallel, one connection is split between multiple resistors and each resistor ends at the end point of the circuit, one next to the other instead of one after the other. Since the resisters are independent from each other in parallel they would have no effect on each other. In a series the first resistors effects would have happened before the next resistor in the series is reached in the circuit.


here is a simple diagram

Sample series

circuit in-[]--[]--[]-circuit out

Sample parallel

            -[]-
circuit in-<     >- circuit out
            -[]-

A little legend for my makeshift symbols

> & < circuit path split/merge
-[]- single resistor

P.S.
I am probably wrong, just my two cents.

....so since I'm wrong here is some information that more clearly states what I'm trying to say.
**broken link removed**
**broken link removed**

After reading these my statement that parallel have no effect on each other since they are on different circuit paths seems to be true, likewise for the series.

in case that's not enough information for you, here tool with nothing but knowledge....and porn.
https://www.google.ca/search?q=parallel+resistors