Resistance In Series/Parallel circuits

  • #1
Mark Rice
37
0

Homework Statement


Hi, I need to find the total resistance of a circuit (attached file). I'm pretty sure it's really and straight forward but I'm a chemical engineer and this is coursework so just wanted to make sure I was doing it right.

Homework Equations

The Attempt at a Solution


Do I just simplify by adding the following to each other in series to give three resistor value (R2+R3) (R4+R5+R6) (R7+R8) then work it out using these in the parallel question? ie Rcircuit= R1 + R9 + [(1/Rtotal)=1/(R2+R3) + 1/(R4+R5+R6)] + 1/(R7+R8)]I totally appreciate they way I have just typed this was not very clear so I'm happy to explain myself more!

*Edited*
 

Attachments

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  • #2
Mark Rice said:
Do I just simplify by adding the following to each other in series to give three resistor value (R2+R3) (R4+R5+R6) (R7+R8)
That cannot be right because R4 is not parallel to R7 and R8. But you have the right approach - collapse combinations in stages, starting with the simplest ones.
 
  • #3
haruspex said:
That cannot be right because R4 is not parallel to R7 and R8. But you have the right approach - collapse combinations in stages, starting with the simplest ones.

So do I add R1 + R4 + R9 in series. Then do the parallel calculation [(1/Rtotal)=1/(R2+R3) + 1/(R5+R6)] + 1/(R7+R8)]? Sorry not done physics in 3 years so I'm a bit rusty!
 
  • #4
Mark Rice said:
So do I add R1 + R4 + R9 in series. Then do the parallel calculation [(1/Rtotal)=1/(R2+R3) + 1/(R5+R6)] + 1/(R7+R8)]? Sorry not done physics in 3 years so I'm a bit rusty!

Is this what I do anyone?
 
  • #5
Mark Rice said:
So do I add R1 + R4 + R9 in series.
Not R4. There's a connection with other resistors in between R1 and R4.
In general, you look for:
- two resistors in sequence with no other connection between them; if found, combine them, adding resistances
- two resistors joined to each other at both ends; if found, combine them using the parallel resistance rule
- repeat as necessary
Yes, you can combine 2 with 3, 5 with 6, and 7 with 8. What next?
 

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