Total equivalent resistance of a combined circuit

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Homework Help Overview

The discussion revolves around calculating the total equivalent resistance of a combined circuit involving resistors in series and parallel configurations. The subject area includes circuit analysis and Ohm's law.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the configuration of resistors R1, R2, R3, and R4, discussing whether they are in series or parallel. There are attempts to calculate the equivalent resistance using various interpretations of the circuit setup. Questions arise regarding the application of formulas for combining resistors in parallel and series.

Discussion Status

There is ongoing clarification regarding the correct application of formulas for calculating equivalent resistance. Some participants have provided guidance on how to combine resistors, while others are questioning the accuracy of previous calculations. Multiple interpretations of the circuit configuration are being explored.

Contextual Notes

There is confusion regarding the values of R2, with differing claims about its resistance. Participants are also addressing potential errors in the calculations presented earlier in the thread.

dnk3294
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Homework Statement
Three light bulbs are connected in a circuit as shown in the diagram. Vs= 15V, R1= 30 ohm, R2= 5 ohm, R3= 10 ohm, R4= 15 ohm. Determine the total equivalent resistance (Req)
Relevant Equations
1/Req= 1/R1+1/R2+1/R3 for parallel resistors.
R= R1+R2+R3 for resistors in series.
I attempted to solve the problem by following the positive terminal of the battery and I realized that Resistor R1 has a different path than R2+R3+R4. I came to the conclusion that R1 is in parallel with R2+R3+R4. The series connection with R2,R3,R4 would be 30 ohm total by adding the three resistors. To find the equivalent resistance, I added 30 ohm to the resistance of a parallel resistor which was 1/Req= 1/30 and Req=30 ohm. The total equivalent resistance was calculated to be 60 ohm.
 
Last edited:
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dnk3294 said:
, I added 30 ohm to the resistance of a parallel resistor which was 1/Req= 1/30 and Req=30 ohm. The total equivalent resistance was calculated to be 60 ohm.
You do not seem to have applied that formula correctly. Please post the details of that calculation.
 
Sorry about the error, I thought that R2,R3,R4 were in series so the resistors in series would equal:
R2(5)+R3(10)+R4(15)= 30 Ohms.
R1 is in parallel so it would be :
1/Req= 1/30 = 30 Ohms.
The total equivalent resistance would be 30ohms +30 ohms=60 ohms.
 
Last edited:
dnk3294 said:
R2(20)
The problem statement says R2=5Ω
dnk3294 said:
R1 is in parallel so it would be :
1/Req= 1/30 = 30 Ohms.
No, you are not using the parallel resistance formula correctly.
Having merged R2, R3 and R4 into the single equivalent resistance R=R2+R3+R4 you have simplified it to R1 in parallel with R. How do you combine two resistors in parallel? The formula you quoted in post #1 is for combining 3.
 
dnk3294 said:
R2(20)+R3(10)+R4(15)= 45 Ohms.

Does R2 = 5 ohms or 20 ohms?
 
R2 is equal to 5 ohms.
 
Would the equivalent resistance be :
1/Req= 1/30+1/30 = 2/30 and the answer would be 15 ohms for the equivalent resistance due to R and R1 being in parallel?
 
dnk3294 said:
Would the equivalent resistance be :
1/Req= 1/30+1/30 = 2/30 and the answer would be 15 ohms for the equivalent resistance due to R and R1 being in parallel?
yes
 

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