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Network with resistors problem

  • Thread starter mopar969
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  • #1
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Please see attachment for the problem. I used the rules for parallel and series resistors to find a total resistance in the network of 4560/59 ohms and then found a current of 59/152 amps through the 20 ohm resistor in the original problem. Am I correct?

Homework Statement





Homework Equations





The Attempt at a Solution

 

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  • #2
ideasrule
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Please see attachment for the problem. I used the rules for parallel and series resistors to find a total resistance in the network of 4560/59 ohms and then found a current of 59/152 amps through the 20 ohm resistor in the original problem. Am I correct?
No, that's not correct. 4560/59 is smaller than 240 ohms, which is not possible because the 240-ohm resistor is in series with the rest of the circuit.
 
  • #3
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I think I found my mistake I now found 3/8 amp through the 20 ohm resistor and found a total resistance of 80 ohms in the circuit. Is this correct?
 
  • #4
SammyS
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The comment made by ideasrule still applies, because 80 < 240, so your answer cannot be correct.

To get more specific help, please indicate the steps you take to get your results.
 
  • #5
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I added the 10 ohm resistors in series which makes 20 ohms and the three 60 resistors in parallel made 20 ohms. Then the 2 20 ohms resistors are in parrallel which makes 10 ohms. Then the 40, 10, 50, and 20 ohm resistors are in series and make 120 ohms which is then parralel to the 240 ohm resistor.


Maybe I mad a mistake some where let me know. Thanks.
 
  • #6
gneill
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I added the 10 ohm resistors in series which makes 20 ohms and the three 60 resistors in parallel made 20 ohms. Then the 2 20 ohms resistors are in parrallel which makes 10 ohms. Then the 40, 10, 50, and 20 ohm resistors are in series and make 120 ohms which is then parralel to the 240 ohm resistor.


Maybe I mad a mistake some where let me know. Thanks.
How do figure that the 240 Ohm resistor is in parallel with the 120 Ohms you just worked out? Both of their ends are not connected. You can trace a single path through the battery and both resistances.
 
  • #7
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Thanks I see my error. I now got that the total resistance is 360 ohms and the 20 ohm resistor has a current of 1/6 ampps through it. Is this correct?
 
  • #8
gneill
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That looks better! Correct.
 
  • #9
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Thanks for all the help much appreciated.
 

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