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Question

Yelp ME !!!! X(

yumyum247 · Answer

attachment

yumyum247 · Answer

Please check my work!!!! :")

michele_laino · Answer

ok! Please wait, I'm checking your computation...

yumyum247 · Answer

8')

michele_laino · Answer

your computation is right! :)

michele_laino · Answer

I got the same results!

yumyum247 · Answer

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 · Answer

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 · Answer

furthermore*

yumyum247 · Answer

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 · Answer

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 · Answer

oops.. in particular: we* have...

yumyum247 · Answer

what is the formula????? please!! :)

michele_laino · Answer

here is the first step:
|dw:1449595214044:dw|
here we have: $R_a=R_3+R_4$