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4 resistors in series and parallel

  • Thread starter abruski
  • Start date
  • #1
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Homework Statement



I have 4 resistors R1=R2=R3=R4=40[tex]\Omega[/tex]

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=[tex]\frac{R_{1}*R_{2}*R_{3}*R_{4}}{R_{1}+R_{2}+R_{3}+R_{4}}[/tex]


The Attempt at a Solution



The first is easy: 4*40=160

Parallel: R=[tex]\frac{40^{4}}{40*4}[/tex]=16000[tex]\Omega[/tex]

in the book the answer is: 10[tex]\Omega[/tex]
 

Answers and Replies

  • #2
370
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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)
 
  • #3
Doc Al
Mentor
44,882
1,129
Series: R=R1+R2+R3+R4
OK.
Parallel: R=[tex]\frac{R_{1}*R_{2}*R_{3}*R_{4}}{R_{1}+R_{2}+R_{3}+R_{4}}[/tex]
:bugeye: Where did you get this equation? (Look up resistors in parallel.)
 
  • #4
9
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The parallel resistor formula I took from the same lesson of the same book. Very weird. Anyway, thanx guys :)
 
  • #5
Cyosis
Homework Helper
1,495
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I take it the formula from that lesson only had two parallel resistors? In that case [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \Rightarrow R=\frac{R_1R_2}{R_1+R_2}[/tex]. Now add a third resistor to the first equation and see that the generalization you made is not correct.
 
  • #6
9
0
I saw the wrongness of my ways now. Thank you
 

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