How can I simplify this mixed resistors circuit

  • #1
Member advised to use the homework template for posts in the homework sections of PF.
Hi
I am to calculate Rab of the circuit and I have no idea on how to approach this problem. It would have been all nice and eaasy if it weren't for the vertival 30 ohm resistor. Because of this I'm in a dark spot. I will appreciate any help.
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  • #2
cnh1995
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You need to fill up the three-part template and show your attempt. It is mandatory here in the HH section.
 
  • #3
Ofc you are right. Firstly I thought of calculating 30 and 20 ohms resistors as a parrarel connection and to do so with 2x 10 resistors. How wrong I had been I thought. Then I looked a it again and tried to calculate a parrarel conncection of a vertical resistor which is 30ohm with the one on the bottom right side(30ohm). Well, I couldn't do that either because there is 20 ohm resistor on the upper right side. Having no idea what to do next I saw "Y" connection of the 3 resistors(2 cases actually) 30,30,10 or 30,20,10(one vertical 30ohm with two on the bottom or upper side resistors. I didn't even start changi it to delta configuration because there must be another simpler way(I actually don't even know if this approach is correct and whether one can solve it that way. It would require to redraw the circuit using delta configuration). That was my thinking process
 
  • #4
cnh1995
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I didn't even start changi it to delta configuration because there must be another simpler way(I actually don't even know if this approach is correct and whether one can solve it that way. It would require to redraw the circuit using delta configuration).
How about using delta to star conversion instead? Can you spot a delta connection in the circuit?
 
  • #5
Oh yes there is delta connection. On the left(10 10 30) side. Accordingly I chose to go with DELTA to STAR connection. The results are as follows:
Ra = (10×10)/(10+10+30) → Ra = 2 ohm
Rb = (10*30)/(10+10+30) → Rb = 6 ohm
Rc = (10*30)/(10+10+30) → Rc = 6 ohm
With these figures I calculate 2x series connection of resistors [30 +6(Rc) and 20 + 6(Rb)], parrarel connection of both comes next and final serie-connection [2(Ra) + simplified Rb,Rc,30,20 resistors calculated above.
All of that equals 17.6 ohm as a resistance of the whole circuit.
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  • #6
cnh1995
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All of that equals 17.6 ohm as a resistance of the whole circuit.
Your simplified circuit is correct but the equivalent resistance comes out to be 17.1 ohm.
 
  • #7
Of corse yoy are right. I eneter wrong number :D
I'm most grateful for your help. Nevertheless I wonder if there is another way to solve this(not using voltage source calculating currents etc.). It is strange because my lecturer gave my group this particular scheme to think about and try to solve. We have had neither delta nor star resistors connections during lectures so that makes me think there should be another, simpler way to solve this. Thanks again
 
  • #8
cnh1995
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There is no other way. (If the 20 ohm resistance were 30 ohms instead, the circuit would be a balanced Wheatstone bridge, meaning you can omit the middle 30 ohm resistor).
 
  • #9
That was actually the first circuit. The one I had problem with was its upgrade :D
 
  • #10
The Electrician
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Of corse yoy are right. I eneter wrong number :D
I'm most grateful for your help. Nevertheless I wonder if there is another way to solve this(not using voltage source calculating currents etc.). It is strange because my lecturer gave my group this particular scheme to think about and try to solve. We have had neither delta nor star resistors connections during lectures so that makes me think there should be another, simpler way to solve this. Thanks again
Here's another way: http://www.hallikainen.com/rw/theory/theory6.html
 
  • #11
Much appreciated THANKS !
 

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