# Simple circuit - parallel resistors

B4ssHunter
Simple circuit -- parallel resistors

## Homework Statement

i am asked to find two or more resistors having the same voltage across their terminals , the answer is R2 and R0 , but i want to know why , note that there are no known values(they might or might not be equal i have no idea ) for R nor did he give me any values for the voltage or the intensity

## The Attempt at a Solution

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Mentor
Have you learned about how to tell which resistors might be in series or which might be in parallel? Have you learned about the features that parallel or serial connected components have with respect to potential difference and current?

B4ssHunter
yes i have , but i still cant tell why R2 and R0 have the same voltage . they told me because they are connected in parralel , but i thought the Whole two branches were connected in parallel

Staff Emeritus
When a path splits it forms parallel paths only for that length where the paths remain fully separated before re-joining.

Electron flow can split between R0 and R2, but then it joins or mixes again in the vertical wires at their ends. So that parallel path extends only for the length of R0 (and R2).

Likewise, the path again splits between R1 and R3, then joins again at either end. So R1 and R3 represent parallel paths.

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B4ssHunter
When a path splits it forms parallel paths only for that length where the paths remain fully separated before re-joining.

Electron flow can split between R0 and R2, but then it joins or mixes again in the vertical wires at their ends. So that parallel path extends only for the length of R0 (and R2).

Likewise, the path again splits between R1 and R3, then joins again at either end. So R1 and R3 represent parallel paths.

so the current from R2 and R0 rejoin completely , and they split upon the other two resistors ?
where do they rejoin exactly ? in the middle of the vertical line ? do both currents flow in the vertical wire in the middle ?
for me it seems like only a part of the current rejoins , as the current moving in R2 some of it will go through R3 and some will Rejoin by moving through the vertical wire , so they dont completely rejoin , only a part of them rejoins

Mentor
so the current from R2 and R0 rejoin completely , and they split upon the other two resistors ?
where do they rejoin exactly ? in the middle of the vertical line ? do both currents flow in the vertical wire in the middle ?
for me it seems like only a part of the current rejoins , as the current moving in R2 some of it will go through R3 and some will Rejoin by moving through the vertical wire , so they dont completely rejoin , only a part of them rejoins

Keep in mind that circuit diagrams are not scale models of a physical circuit, they are "schematic" diagrams indicating the topology (in the geometric sense) of the circuit component interconnections. On such a diagram the wires are considered to be perfect conductors, and everywhere reachable by an unbroken path of such wiring is thus at the same potential, so a given contiguous network of such wiring makes a single node.

Unless otherwise specified, a node is treated as a single dimensionless point where all the components connected to that wire join. That means you're free to redraw circuit diagrams, moving the components and wire locations, change wire lengths, straighten bends, add bends, alter their paths, etc,. so long as the topology remains the same. Any component that connects to a given contiguous path can have its connection relocated to any point on the same path without changing the circuit topology.

So for the circuit we're looking at here, the "I" shape wiring joining all the resistors is taken to be a single node at the same potential everywhere. You can treat it as though the wires from all the resistors met at a single point. So yes, we consider that all the currents "mix" and are indistinguishable at the node. UNLESS OTHERWISE INSTRUCTED!

If we were given instructions that the central vertical wire was to be taken as a separately identifiable physical path in the "real world" implementation of the circuit, and that we were required to find the current flowing through that path, then we could apply KCL at the intersections to find the individual currents.

Note well that having the separate path in the physical implementation would not change the overall node potential or the individual resistor currents in any way. So it's generally not worth worrying about how a given conducting path is drawn on paper, or how currents "mix" at a node. As far as we're concerned, all of a given node is at one potential and the sum of the currents entering the node is equal to the sum of the currents leaving it, and that's all the information required to analyze most any circuit.

B4ssHunter
Keep in mind that circuit diagrams are not scale models of a physical circuit, they are "schematic" diagrams indicating the topology (in the geometric sense) of the circuit component interconnections. On such a diagram the wires are considered to be perfect conductors, and everywhere reachable by an unbroken path of such wiring is thus at the same potential, so a given contiguous network of such wiring makes a single node.

Unless otherwise specified, a node is treated as a single dimensionless point where all the components connected to that wire join. That means you're free to redraw circuit diagrams, moving the components and wire locations, change wire lengths, straighten bends, add bends, alter their paths, etc,. so long as the topology remains the same. Any component that connects to a given contiguous path can have its connection relocated to any point on the same path without changing the circuit topology.

So for the circuit we're looking at here, the "I" shape wiring joining all the resistors is taken to be a single node at the same potential everywhere. You can treat it as though the wires from all the resistors met at a single point. So yes, we consider that all the currents "mix" and are indistinguishable at the node. UNLESS OTHERWISE INSTRUCTED!

If we were given instructions that the central vertical wire was to be taken as a separately identifiable physical path in the "real world" implementation of the circuit, and that we were required to find the current flowing through that path, then we could apply KCL at the intersections to find the individual currents.

Note well that having the separate path in the physical implementation would not change the overall node potential or the individual resistor currents in any way. So it's generally not worth worrying about how a given conducting path is drawn on paper, or how currents "mix" at a node. As far as we're concerned, all of a given node is at one potential and the sum of the currents entering the node is equal to the sum of the currents leaving it, and that's all the information required to analyze most any circuit.
so i can just think of , as if the current from the above R2 and R0 , entered the node , got mixed together , and Then get distributed all over the other two resistors ?
also , if the the middle I shaped wire has a resistance , how do i measure the current flowing through it ? i measure the potential difference across its two ends and then divide it by its resistance ?
or do i treat it as a series resistor to R3 or R1 * depending on which has the highest resistance , if R1 has higher resistance then obviously more current is going to go into R3 so i treat the resistance of the middle wire as if it is connected in series with the R3 *

Staff Emeritus
for me it seems like only a part of the current rejoins , as the current moving in R2 some of it will go through R3 and some will Rejoin by moving through the vertical wire , so they dont completely rejoin , only a part of them rejoins
Whether they rejoin completely or partially is immaterial, it means they are not fully separate paths. Therefore paths are not parallel beyond any point where they rejoin.

As gneill points out, a piece of wire is a short circuit so the paths are effectively joined at a point (termed a node) where you see a vertical connecting wire.

Mentor
so i can just think of , as if the current from the above R2 and R0 , entered the node , got mixed together , and Then get distributed all over the other two resistors ?
Sure. Currents enter a node, currents leave a node. Which particular charge carriers come from where or go to where makes no difference as long as the totals balance.
also , if the the middle I shaped wire has a resistance , how do i measure the current flowing through it ? i measure the potential difference across its two ends and then divide it by its resistance ?
Yup. If the wire has a significant resistance compared to other components then it in effect becomes a component itself and would be replaced on the diagram with a resistor symbol with an appropriate value. That would effectively split the single node into two independent nodes, and you would analyze the circuit accordingly.
or do i treat it as a series resistor to R3 or R1 * depending on which has the highest resistance , if R1 has higher resistance then obviously more current is going to go into R3 so i treat the resistance of the middle wire as if it is connected in series with the R3 *
Just replace any wire from the real-world implementation of the circuit that has significant resistance with a resistor symbol. So if the center vertical wire has some significant resistance (maybe because it happens to be several hundred meters long and consists of small diameter wire), then replace it in the diagram with a resistor of appropriate size: For wires already in series with resistors you can add resistor symbols to represent the wires or just add the resistance to the resistor it connects to.

Short wires tend to have very low resistance values (less than a milliohm per cm of length), so they won't significantly effect a circuit that has resistors in the tens to tens of thousands of Ohms.

Garden variety resistors are typically sold with tolerances of ±20%. Better tolerances can be had for increasing amounts of money -- ±10%, ±5% , ±1% are typical, although there are standards for even better tolerance parts down to ±0.1%. Anything else would be special order, and unless you're very wealthy you'd be better off redesigning your circuit to do without or add "trimmer" resistors to tweak the critical values manually.