# Resistors in series and parallel

1. Mar 25, 2004

### oooride

Resistors in series and parallel (fixed)

___^^^^_____
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x|___
 __^^^^___||
x||
|||
|_||
|__^^^^_______|

Can someone explain why this is a series - parallel circuit? I don't understand why when starting from the top terminal R1 and R2 reduces in series..

The reason I'm confused is say the current is starting at the top terminal, after passing through R1 (top resistor), the current comes to a junction where it is splits, and doesn't some current go through R2 (middle resistor), while some continues on and goes through R3 (bottom resistor)?

If so, in order to be in series doesn't the same amount of current need to travel through the both resistors? Meaning wouldn't the same amount have to travel through R1 and R2 in order to be in series?

I'm confused on why there is a series involved, I understand why the bottom is in parallel.

^ = resistors (starting from the top R1 = 1 ohm, R2 = 2 ohms, R3 = 3 Ohms.
x = terminals (between the terminals is 132 volts).
- = conductor.

= Just so the diagram will stay spaced out.

Last edited: Mar 25, 2004
2. Mar 25, 2004

### pattiecake

B/c if the current flows out of the top terminal: there is only one direction it can go. It MUST go through that first resistor on the top first. There is no other choice. Whenever a current HAS to go one direction, it's in series. So the circuit has the first resistor in series, where after that, the current can go in one of two directions on the left side there, where you already said-- that part is parallel.

3. Mar 25, 2004

### oooride

But the problem I'm having trouble grasping, is after say 20 A of current travels through R1 that 20 A won't travel through R2 due to the junction/split in the conductor will it?

Won't say 10 A go through R2 and 10 A go through R3? That's the reason I'm confused, because I thought that if 20 A are going through R1 the same 20 A MUST go through R2 in order for it the two resistors to be reduced in series.

Last edited: Mar 25, 2004
4. Mar 25, 2004

### pattiecake

"But the problem I'm having trouble grasping, is after say 20 A of current travels through R1 that 20 A won't travel through R2 due to the junction/split in the conductor will it?"

No- after the current travels through R1, it is divided between R2 & R3. Think of the current as water rushing through a system of pipes. If the pipes split (like here at a junction) the water will gush both ways-but more will be able to pass through where there is less resistance.

"Won't say 10 A go through R2 and 10 A go through R3? That's the reason I'm confused, because I thought that if 20 A are going through R1 the same 20 A MUST go through R2 in order for it the two resistors to be reduced in series."

You have it right here. But like I said before, when the current comes out of the first terminal at the top--it has to go though R1. This is why the first part is considered in series. As far as I know, it doesn't matter that there is a split AFTER R1. The current must flow directly between the first terminal to R1. Because of this the top part is considered to be in series.

5. Mar 26, 2004

### oooride

Okay, well it looks like I got the question wrong on the test and all the other questions that relied on those correct values.. I reduced the R1 and R2 in parallel and then that new one with R3 in parallel, which I kinda knew was wrong when seeing what the final numbers calculated were.

I kinda knew the professor was going to give us a series-parallel circuit problem, but when I was analyzing it I guess I was thinking about it way too much. I didn't think it could be that easy, but then thought well maybe that's the trick of this problem..

What does "in series" mean then?
Because it looks like that is my problem to begin with.

I mean I understand the current leaving the top terminal can only 'flow' through R1 in one direction, but I thought in series meant "one resistor after another", and if one resistor is "after another", then an X amount of current flowing through one resistor must also equal the same X amount of current flowing through the second resistor in order for it to be considered and reduced in series.

I understand that R1 and R2 here are in series (+)___^^^^____^^^^___(-)
,but I figured in the above circuit that split was the reason it couldn't be in series..

So from now on when reducing resistors/capacitors if current can move in only one direction through one resistor, then that resistor and the other one involved will be in series, regardless of the path the current (or amount of current) travels to the second resistor?

Last edited: Mar 26, 2004
6. Mar 26, 2004

### oooride

I see it now, I was approaching the problem the hard way. The reason I believe I was getting confused was I was trying to follow the current instead of just looking at the big picture.

If I would of reduced the middle and bottom resistors in parallel first then that equivalent resistance and the top resistor would be in series.

Thanks for all the help.

7. Mar 26, 2004

### pattiecake

No problem...I'm glad you figured it out! =D