Network with resistors problem

In summary, you 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.
  • #1
mopar969
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0
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?
 

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  • #2
mopar969 said:
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
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
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
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
mopar969 said:
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
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
That looks better! Correct.
 
  • #9
Thanks for all the help much appreciated.
 

1. What is a network with resistors problem?

A network with resistors problem involves analyzing a circuit that contains resistors connected together in a specific pattern. The goal is typically to determine the voltage, current, or resistance at different points in the circuit.

2. How do I solve a network with resistors problem?

To solve a network with resistors problem, you will need to use Ohm's Law and Kirchhoff's Laws, which describe the relationship between voltage, current, and resistance in a circuit. You will also need to use techniques such as series and parallel circuit analysis to simplify the circuit and make it easier to solve.

3. What are the most common mistakes when solving a network with resistors problem?

The most common mistakes when solving a network with resistors problem include forgetting to use the correct units, not properly applying Ohm's Law or Kirchhoff's Laws, and making errors in calculations. It is important to double check your work and use a systematic approach to avoid these mistakes.

4. Do I need to consider the internal resistance of a battery in a network with resistors problem?

Yes, the internal resistance of a battery should be considered in a network with resistors problem. This resistance can affect the overall resistance and current in the circuit, and should be included in calculations to ensure accurate results.

5. Are there any helpful tips for solving a network with resistors problem?

Some helpful tips for solving a network with resistors problem include drawing a clear and organized circuit diagram, labeling your variables and unknowns, and breaking the circuit down into simpler parts. It is also useful to check your answer by using different methods or by plugging in your values to ensure they are consistent with your calculations.

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