# Resistors in parallel and series

1. Oct 12, 2008

### johndoe

A resistor in parallel with a voltage source or a resistor in series with a current source has no effect on the rest of the circuit?

2. Oct 12, 2008

### dlgoff

Depends on the internal resistance of the voltage source and the resistor value.

3. Oct 12, 2008

### Pumblechook

I take it these are ideal (mathematical) voltage and current sources with zero internal resistance.

4. Oct 15, 2008

### Type_R

In Parallel current is changed
In series the voltage is changed.

a resistor in parallel effects the current due to total conductance being effected.

A resistor in series has an effect because the voltage is effected due to the total resistance of the circuit being effected.

5. Oct 16, 2008

### aarabik

Sorry! my mistake!! I overlooked the fact that when you put the new resistor across the voltage source, the overall resistance falls. Now, that the source has to maintain the same voltage across this little resistor as well as the rest of the circuit, the source produces more current, which is consistent because the overall resistance is lower. And to maintain that voltage, the current in the rest of the circuit is the same as before. so, yes, there's no change in the circuit :D

Last edited: Oct 17, 2008
6. Oct 17, 2008

### rbj

if these are ideal voltage or current sources and you are analyzing a circuit and trying to reduce it to a less complex representation, then the answer is yes, that is the case and they definitely have no effect on the rest of the circuit. i cannot account for what these other guys are saying. they're either assuming a non-ideal voltage or current source (for which you need to model in the series resistance for a voltage source or a parallel resistance for the non-ideal current source), or they're mistaken.

and it has nothing to do with whether they are current-controlled or voltage-controlled or independent sources.

7. Oct 17, 2008

### Type_R

He asked if a resistor in parallel with a voltage source have effect on the circuit.........

Even with an ideal voltage source....... the current is changed as it flows through the resistor.......That's an effect.

But due to ideal voltage source(having the zero internal resistance) The ideal voltage source remains the same independent of the current flowing through it.......Current still is changed
and same with Current Source...... the voltage is changed, but you won't see it at the ideal current source.
Simple Ohms law. Total voltage or current in the circuit is changed.

I understand where you're coming from though.

Also his question was very vague.......You can put a resistor in parallel with a voltage source.....but he never stated what the "rest of the circuit" consisted of.

Last edited: Oct 17, 2008
8. Oct 17, 2008

### rbj

not on the rest of the circuit. the ideal voltage source says (a super-duper highly regulated power supply that has the whole power utility backing it up) that it will deliver the rated voltage no matter what current it has to dump out. theoretically, with an ideal voltage source, if you short circuit it, it will deliver infinite current. a resistor in shunt with the ideal voltage source will simply cause it to dump out more current. the voltage remains the same and the rest of the circuit doesn't know the difference. that resistor shunt can be removed and the rest of the circuit doesn't know the difference.

you can say the same thing about the ideal current source but exchange the words "voltage" and "current" and replace the word "shunt" (or "parallel") with "series".

i don't think that you completely understand this. this is basic linear electric circuits in the electrical engineering discipline. we use these techniques (along with the Thevenin/Norton equivalents and parallel/series formulae for components) to simplify circuit analysis.

it doesn't matter what the rest of the circuit is as long as it isn't something pathological like two ideal voltage sources (with different voltages) in parallel or two ideal current sources (with differing currents) in series. those pathological cases end up in contradiction and cannot be solved.

9. Oct 17, 2008

### Type_R

Yes I understand what an Ideal Voltage Source is... as well I should since i'm an electronics engineer major.

In Electronics we go by approximations....... Normally we just use "ideal" concepts since any further approximation is negligible. An Ideal voltage source has "Zero Internal Resistance" so it's not limited in how much power it's capable of putting out (Since current can be infinite).

Now you're saying that if you put a Resistor in Parallel( I assume since you said voltage isn't changed). The rest of the circuit is STILL effected....
Why is this? Because of the total conductance(G) which is the reciprocal of Resistance. He asked is the REST of the circuit effected?
Kirchoffs voltage law should help.
so if I have an 60v ideal voltage source....... and I put lets say two 30ohm resistors in parallel with it I have 4amps being drawn into the circuit which gets shared between the first resistor and the second resistor with two amps for both. Well 4 amps HAD to be drawn from the voltage source because of the two resistors.....While if it was only one only 2 would have to be drawn. Not only is it drawn.....but it's put back at the end of the circuit..........
So if I continue to add more and more resistors........I'm going to draw more and more current through the circuit. The total current will be increased. Each resistor might not notice..........I doubt it would even notice if the "Voltage source wasn't ideal" unless you went beyond the Power Supplys power capability. lol Which is besides the point that a resistor doesnt have a "perception" :)

Req = 1/R1+1/R2......
It=I1+I2.......
Now when you say "the rest of the circuit doesn't get effected"
I'll say yes it does get effected.
Now if you said "DO the other devices in the circuit get effected"
I'd say no. Since the Voltage source is unlimited in power and can supply each resistor their hearts desire of current.

Thevenin will even show you i'm correct.

Il= Vth/Rth+RL

increase the resistance decrease the current within the circuit

Last edited: Oct 17, 2008
10. Oct 17, 2008

### Type_R

I'm not sure how to use these threads that well.......But heres a picture to help illustrate my point

The more resistors you add the more current you'll measure with the current meter inserted into the circuit.

11. Oct 17, 2008

### rbj

but the part labeled mA is part of "the rest of the circuit." any resistor of finite conductance that you place in parallel with that ideal voltage source will do nothing to the rest of the circuit. "do nothing" is semantically the same as "have no effect."

12. Oct 17, 2008

### rbj

actually Type_R, before either of us had piped in on this, Don hit the issue on the head. the reason i am assuming that the internal resistance is zero is that is what you do with an ideal voltage source. the reason i'm assuming is this ideal voltage source is that is what you do in the discipline of linear electric circuits. we make other idealisms like conductances (or admitances) precisely add in parallel and resistances (or impedances) add in series. there's also the ideal Y-$\Delta$ conversion and this other concept of two-port networks that is an idealism of reality. the question really was asked, in my opinion, in the context of a linear electrics circuits course. and then the answer is "Yes, a resistor in parallel with a voltage source or a resistor in series with a current source has no effect on the rest of the circuit."

13. Oct 17, 2008

### Redbelly98

Staff Emeritus
Not quite. Re-read post #1, it asked if there's an effect on the rest of the circuit.

Yes, current out of an ideal voltage source will change. But no, the current through the rest of the circuit will not change. The difference in the two currents will be through the resistor that got added.

The rest of the circuit will have the same voltage across it, and current through it, as before.

For non-ideal sources, see Don's post #2.

14. Oct 17, 2008

### Type_R

Well this is turning to be a semantics issue for me.....Because the way I see "rest of the circuit" is........ "electron flowing from the negative potential all through the circuit until it reaches the positive potential" so I see that part of the circuit were I placed the amp meter as after the resistor making it "The rest of the circuit." In which now you have that same I(total) cycling through.

15. Oct 17, 2008

### Redbelly98

Staff Emeritus
I think it comes down to, what did johndoe have in mind when he started the thread? Not only what is meant by the "rest of the circuit", but also are these real or ideal sources. We could all be right about our own respective (but different) interpretations.

16. Oct 17, 2008

### rbj

i don't think your semantics of what the OP asked is correct. you have a voltage source (an ideal voltage source, in my opinion, considering the question) and you have whatever it is connected to (which is precisely what is meant by "the rest of the circuit"). now place a resistor of non-zero resistance in parallel with that (ideal) voltage source. does anything change in the rest of the circuit? there are three disjoint objects: the voltage source, the rest of the circuit, and the resistor we plan to shunt across the voltage source.

17. Oct 18, 2008

Right, the ammeter in post #10 needs to be between the two resistors, and not between the resistors and the source, to measure what's going on in the rest of the circuit. As is it's measuring what's going on at the source.

18. Oct 18, 2008

### Type_R

An electrical network is an interconnection of electrical elements such as resistors, inductors, capacitors, transmission lines, voltage sources, current sources, and switches

So the wiring is apart of the circuit..........The source is the "Voltage source"....... Because the 4 amps coming back are is the combination of the 2 amps from the resistors WHICH is why it's 4 amps in the first place. Coming back from the voltage source.

My reasoning for saying that part of the circuit should literally be considered part of the "rest of the circuit" is because of how I measure..... If I want to measure some of the circuits voltage. I'd measure from the first load to the end of some other load in the circuit. If I want to measure the rest of the whole circuit.....I'd come from beginning of the first load all the way to the end of the last load. To have a "rest of" you need a beginning point. The beginning point of the electrons journey starts from the voltage source....all the way until last load and transmission line back to the positive potential. The way I look at "rest of the circuit differs from you guys who are saying "The devices in the rest of the circuit." But it's the resisting effect which is determining the current going back towards the voltage source.

Just look at this way...
If you only had one resistor what would literally be the rest of the circuit after the resistor? The line going back to the voltage. Then the cycle begins again.

Last edited: Oct 18, 2008
19. Oct 19, 2008

### Proton Soup

meh. the answer is "it depends". depends on whether it's a real-world question or a homework question.

if it's a real-world question, then it depends on how much current your volt-source can source, or how much voltage your current-source can source. or how much current your power bus can handle.

20. Dec 7, 2008

### johndoe

Sorry for the late reply (for 3 months actually)

I first started this thread when I was working on a problem of finding thevenin equivalent circuits.

When I am finding Voc,the resistor on the far left has nothing to do with the rest of the circuit, and same for a resistor if it is in series with a current source. When I am finding Rth and turn independent votalge sources to zero, it becames a short across the far left resistor.

Anyway, all the extensive discussion is appreciated.

Last edited: Dec 8, 2008