How to calculate Resistors in a circuit ?

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To calculate resistors in a circuit, it's essential to differentiate between series and parallel connections. For resistors in series, simply add their resistances together. For parallel resistors, use the formula 1/R_total = 1/R1 + 1/R2 + 1/R3 to find the equivalent resistance. After determining the equivalent resistance of parallel resistors, it can be added to the series resistors to find the total resistance. Understanding these concepts is crucial for accurate circuit analysis.
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Homework Statement
i tried to solve for this question as far as i know to calculate the resistors you only need to add all of them but my answer of 12 was wrong
Relevant Equations
r1+r2+r3.....
here is the question below with all the possible answers, if anyone can help it would be appreciated
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You only add resistors that are connected in series. What do you do with resistors that are connected in parallel?
 
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Orodruin said:
You only add resistors that are connected in series. What do you do with resistors that are connected in parallel?
for parallel i use the formula 1/r1+1/r2+1/r3
but this one is a complex of both parallel and a series
 
Start by replacing the parallel resistors by an equivalent single resistor.
 
Orodruin said:
Start by replacing the parallel resistors by an equivalent single resistor.
1/4 + 1/4 = 2 that's the total of the parallel resistors but from there i don't know how to add it to the sum of the series resistors to get the total answer
 
Vantom said:
1/4 + 1/4 = 2 that's the total of the parallel resistors
This is not correct. There are three parallel resistors, not two.

Edit: Also, it should obviously be 1/(1/4 + 1/4) = 2.
 
Orodruin said:
This is not correct. There are three parallel resistors, not two.

Edit: Also, it should obviously be 1/(1/4 + 1/4) = 2.
its the same answer (2) as mine
but the problem is i don't know how to add it to the other ohms in series so i can get the total resistance
 
Vantom said:
its the same answer (2) as mine
My point is that you quoted the computation wrong. Also, the answer is wrong because there are three resistors in parallel, not two. Also, there are not two 4 Ω resistors that are connected in parallel.

So let us go back to the basics: Which resistors in your figure are connected in parallel? Which resistors are connected in series?
 
4 Ω and the other 4 Ω are in parallel while 2 Ω and 2 Ω are in series
 
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Vantom said:
4 Ω and the other 4 Ω are in parallel while 2 Ω and 2 Ω are in series
No. This is incorrect. Look up the definition of series and parallel.
 
  • #11
Orodruin said:
No. This is incorrect. Look up the definition of series and parallel.
so 2 Ω and 2 Ω and 4 Ω are in parallel and the 4 Ω alone is in series
i have to sum the parallel ohms to replace them with a single resistor of 8 right ?
 
  • #12
Vantom said:
so 2 Ω and 2 Ω and 4 Ω are in parallel and the 4 Ω alone is in series
i have to sum the parallel ohms to replace them with a single resistor of 8 right ?
No. It is correct that the 2-2-4 combination is in parallel. However, you did not use the correct relation to get the equivalent resistance to replace that entire combination.
 
  • #13
Orodruin said:
No. It is correct that the 2-2-4 combination is in parallel. However, you did not use the correct relation to get the equivalent resistance to replace that entire combination.
inverse 1/2 + 1/2 + 1/4 = 0.8
then i add 0.8 to the single 4 Ω and the sum = 4.8

i was only confused by the parallel and series situation as the circuit is rotated
thank you
 
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