Electric Current - Splitting & Resistance Explained

Click For Summary

Homework Help Overview

The discussion revolves around the behavior of electric current as it splits between two branches in a circuit, specifically focusing on the impact of resistance on current distribution. The original poster is uncertain about how to determine the current in each branch when resistors are present.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore Kirchhoff's rules and Ohm's Law to understand current distribution. Questions arise about how current splits between resistors of different configurations and whether the total resistance affects this split.

Discussion Status

There is an ongoing exploration of Kirchhoff's laws, with some participants suggesting that these principles are necessary for solving the problem. Multiple interpretations of current flow and resistance are being discussed, with no explicit consensus reached yet.

Contextual Notes

Participants are working under the assumption that the total current splits at a junction and then rejoins, with specific values of resistance provided. There is a mention of a total resistance value, but the exact configuration and implications are still being clarified.

jimmy42
Messages
50
Reaction score
0
I'm unsure how to know how much current splits into which wire. So, let's say one wire with 10ma splits into two. On one of the wires there are two resistors each with 10k ohm and on the other wire three resistors each with 10k ohm. Does the current just split in two? Does the amount of resistance affect this?

Thanks.
 
Physics news on Phys.org
hi jimmy42! :wink:

apply Kirchhoff's rules …

the sum of the voltage drops going round the loop must be zero, so the ratio of the currents must be … ? :smile:
 
OK, I read that law but didn't see how to apply it here.

So, is it just as simple as the wire with two resistors will have 4mA and the other 6mA?
 
maybe … and maybe not :smile:

if you're not guessing, then show us your Kirchhoff's equation to prove this! :wink:
 
I didn't know there was such an equation? I didn't use that but just ratios.

All I know is Ohms Law. so V=iR. Do I need that here?

I know the total resistance to be 12 k Ohm. The wires rejoin and then go onto other stuff.
 
jimmy42 said:
All I know is Ohms Law. so V=iR.

oh i see … i assumed you'd know Kirchhoff's rules …

sorry, but you do need them here …

KCL (the current law, or junction law) says total current in = total current out at any junction;

KVL (the voltage law, or loop law) says total iR round any loop = 0 (or = the voltage of any battery in the loop, if there is one) …

in this case, if you mark the current with arrows, and go round the loop adding iR for each resistor (counting i as negative if you're going the opposite way to the arrow through any particular resistor), you should get the correct result :wink:
 
What if the current flows in the same way for all the resistors? So the wire splits and then rejoins.
 
jimmy42 said:
What if the current flows in the same way for all the resistors? So the wire splits and then rejoins.

that's not possible (unless all the reisistors are on one branch) …

the resistors on one branch have to have the current going through the "wrong way" (in the loop) …

draw it and you'll see :smile:
 
It seems to go the same way?




-----R1-------R2-----
1------- -------1
----R3----R4---R5----


Those 1s will loop back to each other and attach to a battery, so how can it go in opposite directions?
 
  • #10
but you said …
jimmy42 said:
So, let's say one wire with 10ma splits into two. On one of the wires there are two resistors each with 10k ohm and on the other wire three resistors each with 10k ohm. Does the current just split in two? Does the amount of resistance affect this?

… i assumed you meant that the current joined up again after the resistors, and carried on round the circuit to the other terminal of the battery :confused:
 
  • #11
Yes that is what happens.

The drawing did not turn out so well. What is the equation? Maybe I can work it out from that.
 
  • #12
round the loop one way …

i1R + i1R + i1R - i2R - i2R = 0

round the loop the other way …

-i1R - i1R - i1R + i2R + i2R = 0 :smile:
 

Similar threads

Replies
44
Views
5K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
14
Views
3K
Replies
1
Views
3K
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K