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Current through each resistor

by LakeMountD
Tags: current, resistor
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LakeMountD
#1
Feb15-05, 12:55 AM
P: 59
if you have a circuit where there are resistors in parallel as well as in series how do you calculate the current through each individual resistor? I have found examples that explain it when they are all in series or all in parallel but i am very confused otherwise.
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learningphysics
#2
Feb15-05, 01:09 AM
HW Helper
P: 4,124
Quote Quote by LakeMountD
if you have a circuit where there are resistors in parallel as well as in series how do you calculate the current through each individual resistor? I have found examples that explain it when they are all in series or all in parallel but i am very confused otherwise.
Can you upload a picture? Or just sketch using the keyboard?

Are you using a physics or engineering textbook?
Janus
#3
Feb15-05, 07:52 AM
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Try reducing the circuit to its simpliest form like you did in your last question. Use this to calculate the total current of the circuit. Then reverse the reduction process one step at a time. When you add a parallel branch just divide the current that would pass through the 'equivalent' resistor between the branches as you would for a straight parallel circuit.

LakeMountD
#4
Feb15-05, 10:15 PM
P: 59
Current through each resistor

http://img210.exs.cx/img210/2812/resistor6xg.gif

here is the circuit.. i took the quiz today and couldnt figure it out.. i just dont understand how to do it.. the damn book is worthless since the example they show only has like two resistors in series..
krab
#5
Feb15-05, 10:47 PM
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Notice that R1 and R2 are in series. So replace it with the equivalent Req1=R1+R2. Now notice this resistor is in parallel with R3. So replace it with Req2, where 1/Req2=1/Req1+1/R3. And so on.
learningphysics
#6
Feb15-05, 11:54 PM
HW Helper
P: 4,124
Quote Quote by LakeMountD
http://img210.exs.cx/img210/2812/resistor6xg.gif

here is the circuit.. i took the quiz today and couldnt figure it out.. i just dont understand how to do it.. the damn book is worthless since the example they show only has like two resistors in series..
So the question is to figure out Req1, Req2 etc...?

Read over krab's post. Look at each diagram... look at what changes from one diagram to the next. Either you have two resistors in series being replaced by the equivalent resistance, or two resistors in parallel are being replaced by an equivalent resistance.
LakeMountD
#7
Feb16-05, 01:19 PM
P: 59
Quote Quote by learningphysics
So the question is to figure out Req1, Req2 etc...?

Read over krab's post. Look at each diagram... look at what changes from one diagram to the next. Either you have two resistors in series being replaced by the equivalent resistance, or two resistors in parallel are being replaced by an equivalent resistance.
no i understand how to get that but i have to find the current through each individual resistor.. i know you have to find the voltage drop and stuff like that to do it but i cant seem to figure it out.
stunner5000pt
#8
Feb16-05, 03:04 PM
P: 1,439
Let's just put it simple

If you had a battery and three resistors connected in seriesi with each other and the battery then the CURRENT across each of the them is the SAME but the POTENTIAL DIFFERENCE across each of them is DIFFERENT
In this case for three resistors of resistances R1, R2, and R3

[tex] V_{n} = iR_{n} [/tex] for each resistor Rn observe the valu for i is the SAM for all the resistors ALL IN SERIES

If you had a battery and the three resistors are connected in parallel with each other and the battery then the CURRETN across each resistor is DIFFERENT and the POTENTIAL DIFFERENCE across each of them is the SAME

In this case [tex] V = i_{n} R_{n} [/tex] for each resistor Rn there is a different current in going through it
learningphysics
#9
Feb16-05, 10:28 PM
HW Helper
P: 4,124
Quote Quote by LakeMountD
no i understand how to get that but i have to find the current through each individual resistor.. i know you have to find the voltage drop and stuff like that to do it but i cant seem to figure it out.
Start at the right hand side, and work leftwards...
First figure out the current through R6 (the second figure from the right).

I6=V/(R6+R4eq)

Now get the current through R5. I recommend using the current divider rule (current through R6 splits and goes through R5 and Req3).

I5=I6*[Req3/(Req3+R5)]

Also, read over stunner's post. I'd like to reiterate one thing he said.

When you get the current through a particular equivalent resistance, which is made up from resistors in series.... then the current through each of the resistors in series, is the current through the equivalent resistance.
For example: Req3=Req2+R4

So when you calculate the current through Req3, you've automatically calculated the current through Req2 and R4. The current through R4, is the same as the current through Req3.

Let me know if this helps...


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