4 resistors in series and parallel

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SUMMARY

The discussion focuses on calculating the equivalent resistance of four resistors, each with a resistance of 40Ω, when connected in series and parallel. The correct formula for series resistance is R=R1+R2+R3+R4, yielding a total of 160Ω. However, the initial attempt at calculating parallel resistance was incorrect; the proper formula is R=1/(1/R1 + 1/R2 + 1/R3 + 1/R4), which results in an equivalent resistance of 10Ω. The user corrected their misunderstanding of the parallel resistor formula after community feedback.

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  • Understanding of Ohm's Law
  • Familiarity with series and parallel resistor configurations
  • Knowledge of equivalent resistance calculations
  • Basic algebra skills for manipulating equations
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  • Study the derivation of the parallel resistor formula R=1/(1/R1 + 1/R2 + 1/R3 + 1/R4)
  • Practice calculating equivalent resistance with different resistor values in series and parallel
  • Explore the impact of resistor tolerance on equivalent resistance
  • Learn about more complex resistor networks and their analysis techniques
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abruski
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Homework Statement



I have 4 resistors R1=R2=R3=R4=40\Omega

I need to find the Equivalent Resistance when they are connected in series and when they are connected in parallel.

The problem is my answer for the parallel differs from the answer given in the book.

Homework Equations



Series: R=R1+R2+R3+R4
Parallel: R=\frac{R_{1}*R_{2}*R_{3}*R_{4}}{R_{1}+R_{2}+R_{3}+R_{4}}

The Attempt at a Solution



The first is easy: 4*40=160

Parallel: R=\frac{40^{4}}{40*4}=16000\Omega

in the book the answer is: 10\Omega
 
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You're almost there, however the second equation is wrong I'm afraid, get that checked up on (the parallel one)
 
abruski said:
Series: R=R1+R2+R3+R4
OK.
Parallel: R=\frac{R_{1}*R_{2}*R_{3}*R_{4}}{R_{1}+R_{2}+R_{3}+R_{4}}
:bugeye: Where did you get this equation? (Look up resistors in parallel.)
 
The parallel resistor formula I took from the same lesson of the same book. Very weird. Anyway, thanks guys :)
 
I take it the formula from that lesson only had two parallel resistors? In that case \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \Rightarrow R=\frac{R_1R_2}{R_1+R_2}. Now add a third resistor to the first equation and see that the generalization you made is not correct.
 
I saw the wrongness of my ways now. Thank you
 

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