Physics Help Forum Resistance combinations help

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 Apr 29th 2018, 03:44 AM #1 Junior Member   Join Date: Apr 2018 Posts: 13 Resistance combinations help A little bit confusing the way this circuit was set up if you can help it would be great questions are equivalent resistance current through 18 and 12 ohms resistor power dissipation in 4.5 ohm
Apr 29th 2018, 04:06 AM   #2
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 A little bit confusing the way this circuit was set up
Never be afraid to redraw the circuit, perhaps more than once.

This would be a good start.

Can you see a series or parallel combination that would simplify it?

Also label your components. It is easier to refer to R2 than "the 25ohm resistor"

and R5 is easier than "the parallel combination of ...."

Ask again if you need more help
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 Apr 29th 2018, 04:28 AM #3 Junior Member   Join Date: Apr 2018 Posts: 13 thank you now it looks so much easier so could you pls check if these are right current through 18 ohms is just 6/18 A and 12 ohms is just 6/12 A and power dissipation is (R3)^3(36)/R1+R3 and equivalent resistance is 7.41 ohm? Last edited by Musawwir; Apr 29th 2018 at 12:24 PM.
Apr 29th 2018, 04:35 AM   #4
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 Originally Posted by Musawwir thank you now it looks so much easier so could you pls check if these are right current through 18 ohms is just 6/18 A and 12 ohms is just 6/12 A and power dissipation is (R3)^3(36)/R1+R1 and equivalent resistance is 7.41 ohm?
I don't think you have got it yet.

Yes the current through R4 is 6/18 amps.

But the current through R1 is not 6/12 amps.

Since there are several methods to solve this please indicate how you are approaching this so we can make your method work.

 Apr 29th 2018, 04:56 AM #5 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 What's the problem?
Apr 29th 2018, 12:20 PM   #6
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 Originally Posted by studiot I don't think you have got it yet. Yes the current through R4 is 6/18 amps. But the current through R1 is not 6/12 amps. Since there are several methods to solve this please indicate how you are approaching this so we can make your method work.
here is my re worked solution

Apr 29th 2018, 12:53 PM   #7
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 here is my re worked solution
Well sort of.

Here are the stages I worked through to get R7, the equivalent resistance.

Can you see how I have built it up in stages, one step at a time?

You need this for the later parts of the question.

Do you understand that each of the redrawn circuits are equivalent to the first one?
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 Apr 30th 2018, 02:43 AM #8 Junior Member   Join Date: Apr 2018 Posts: 13 yes i got it thank you
Apr 30th 2018, 03:00 AM   #9
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 Originally Posted by Musawwir yes i got it thank you
OK so the next thing to do is to calculate the total current.

Do you know how to do this?

 Apr 30th 2018, 06:30 AM #10 Junior Member   Join Date: Apr 2018 Posts: 13 yes you just take the equivalent resistance divide by the voltage

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