Isaac Physics 'A Power Problem', Problem

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domm1
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Here's the link to the question:
https://isaacphysics.org/questions/a...b-b4d62c994132

I was confused at first so looked at the hints and saw they'd split the circuit up into different parts labelled A, B and C. There was another post I saw made on this question, and for parts A, B and C they got values of 3Ω, 20Ω, and 6Ω respectively. When calculating them myself I got the correct values for A and B, however for C I could not see how they'd arrived at 6Ω. Could someone please just quickly explain this to me please?
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Joinedup
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I'm looking at hint 3, I get the overall resistance of the whole circuit (A,B and C combined) as 6.0 Ohms .. so maybe you got the wrong end of the stick or someone partitioned up the circuit differently or something and you were probably doing alright...
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RogerOxon
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(Original post by Joinedup)
I'm looking at hint 3, I get the overall resistance of the whole circuit (A,B and C combined) as 6.0 Ohms .. so maybe you got the wrong end of the stick or someone partitioned up the circuit differently or something and you were probably doing alright...
I don't get that. Can you post your working?

You have 3 Ohms plus the last section, of 20, 20 and 15 in parallel. Combine the 20s to get 10, then the 15 in parallel. That's going to be between 5 and 10 Ohms (I know the exact answer, but don't want to give it yet).
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Joinedup
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(Original post by RogerOxon)
I don't get that. Can you post your working?

You have 3 Ohms plus the last section, of 20, 20 and 15 in parallel. Combine the 20s to get 10, then the 15 in parallel. That's going to be between 5 and 10 Ohms (I know the exact answer, but don't want to give it yet).
Oh sorry, I've had a long day
A=3.0 Ohms
B=20 Ohms
C=6.0 Ohms

note that C encompasses B so the resistance of C is the parallel combination of the resistance you got for B with the 15Ohm and 20Ohm that aren't in B

C plus the 3 Ohms from A
gives 9 Ohms total
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RogerOxon
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(Original post by Joinedup)
Oh sorry, I've had a long day
A=3.0 Ohms
B=20 Ohms
C=6.0 Ohms

note that C encompasses B so the resistance of C is the parallel combination of the resistance you got for B with the 15Ohm and 20Ohm that aren't in B

C plus the 3 Ohms from A
gives 9 Ohms total
I agree with the 6 and 9 Ohm answers. Now to see where the maximum power is.
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Joinedup
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(Original post by RogerOxon)
I agree with the 6 and 9 Ohm answers. Now to see where the maximum power is.
Im not OP btw - I'm alright with it really.
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domm1
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(Original post by Joinedup)
Oh sorry, I've had a long day
A=3.0 Ohms
B=20 Ohms
C=6.0 Ohms

note that C encompasses B so the resistance of C is the parallel combination of the resistance you got for B with the 15Ohm and 20Ohm that aren't in B

C plus the 3 Ohms from A
gives 9 Ohms total
I now see how you've gotten 6Ω for C, however I don't exactly understand why you take the total resistance for B and add it in parallel with the other parallel resistors in C; would you care to explain it to me please?

Edit: Or RogerOxon would you be able to explain? You both understand it well so a response from either will do me well.
Last edited by domm1; 4 weeks ago
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RogerOxon
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(Original post by domm1)
I now see how you've gotten 6Ω for C, however I don't exactly understand why you take the total resistance for B and add it in parallel with the other parallel resistors in C; would you care to explain it to me please?

Edit: Or RogerOxon would you be able to explain? You both understand it well so a response from either will do me well.
C contains three parallel paths, of which B is one. We can replace any sub-circuit with one of equivalent resistance. In order to calculate the effective resistance of the parallel paths, we need a single resistance for each. As you did with the two parallel resistors in A, we do the same for the two parallel 20 Ohm resistors in B. We can replace these with a 10 Ohm resistor, which then combines, in series, with the 10 Ohm resistor, to give B an effective resistance of 20 Ohm.
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domm1
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(Original post by RogerOxon)
C contains three parallel paths, of which B is one. We can replace any sub-circuit with one of equivalent resistance. In order to calculate the effective resistance of the parallel paths, we need a single resistance for each. As you did with the two parallel resistors in A, we do the same for the two parallel 20 Ohm resistors in B. We can replace these with a 10 Ohm resistor, which then combines, in series, with the 10 Ohm resistor, to give B an effective resistance of 20 Ohm.
Thank you for the explanation, I understand it now.
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