YumYum247
  • YumYum247
Yelp ME !!!! X(
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
YumYum247
  • YumYum247
Please check my work!!!! :")
Michele_Laino
  • Michele_Laino
ok! Please wait, I'm checking your computation...

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YumYum247
  • YumYum247
8')
Michele_Laino
  • Michele_Laino
your computation is right! :)
Michele_Laino
  • Michele_Laino
I got the same results!
YumYum247
  • YumYum247
are you sure :( because if you look at the resistance section, resistor 1 and 2 and 3 and 4 don't add up to be the same......:(
Michele_Laino
  • Michele_Laino
resistors #1 and #2 are not series linked, whereas resistors #3 and #4 are series linked furthrmore, there is no reason such that \(R_1+R_2=R_3+R_4\)
Michele_Laino
  • Michele_Laino
furthermore*
YumYum247
  • YumYum247
ok one quick question, when it says the total resistance in the circuit is 10 ohms.....what does it mean because it doesn't look like.....
Michele_Laino
  • Michele_Laino
total resistance is given by this computation: \[\Large \begin{gathered} {R_{TOTAL}} = {R_1} + \frac{1}{{\frac{1}{{{R_2}}} + \frac{1}{{{R_3} + {R_4}}}}} = \hfill \\ \hfill \\ = 6 + \frac{1}{{\frac{1}{5} + \frac{1}{{8 + 12}}}} = ...? \hfill \\ \end{gathered} \] in particular: e have to compute the sum \(R_3+R_4\) since they are series linked resistors, then I have to compute the parallel composed by the \(R_3+R_4\) resistor and \(R_2\) resistor, and finally I have to compute the series between such resistance and \(R_1\)
Michele_Laino
  • Michele_Laino
oops.. in particular: we* have...
YumYum247
  • YumYum247
what is the formula????? please!! :)
Michele_Laino
  • Michele_Laino
here is the first step: |dw:1449595214044:dw| here we have: \(R_a=R_3+R_4\)
Michele_Laino
  • Michele_Laino
second step: |dw:1449595327836:dw| here we have: \[\Large {R_b} = \frac{1}{{\frac{1}{{{R_2}}} + \frac{1}{{{R_a}}}}} = \frac{1}{{\frac{1}{{{R_2}}} + \frac{1}{{{R_3} + {R_4}}}}}\]
YumYum247
  • YumYum247
yes!!!
Michele_Laino
  • Michele_Laino
third step: |dw:1449595464577:dw| here we have: \[\Large {R_d} = {R_1} + {R_b} = {R_1} + \frac{1}{{\frac{1}{{{R_2}}} + \frac{1}{{{R_3} + {R_4}}}}}\]
YumYum247
  • YumYum247
yes yes!!! you broke them into smaller pieces!!
Michele_Laino
  • Michele_Laino
yes! Usually I compute the total resistance, step by step as the procedure above
YumYum247
  • YumYum247
Thank you so much, i wish we could just talk face to face so you don't ave to type so much...thank you and i appreciate every little effort you made into making your point across!! I LUV YOU!!! :")
Michele_Laino
  • Michele_Laino
Thanks!! :)

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