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I've read that R3 is parallel to R2 and R4 which are serially connected. Is it true? But I do not know what will be with R6? The question is how to present this circuit by simple way with typical parallel and serial connections?

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closed as off-topic by Kyle Kanos, Qmechanic Feb 21 '17 at 6:10

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  • $\begingroup$ R2 and R4 are actually not in series with eachother because R6 connects to the node inbetween them. You can't say they're in series if something else connects to the node between them. $\endgroup$ – Cort Ammon Feb 21 '17 at 1:33
  • $\begingroup$ To simplify the circuit, start by noticing the link labeled V5. Then you'll see which resistors really are in parallel and that makes it easy to redraw the circuit in a simpler form. $\endgroup$ – hdhondt Feb 21 '17 at 1:38
  • $\begingroup$ @CortAmmon You mean that of they are not in series , then they are parallel? $\endgroup$ – student Feb 21 '17 at 2:07
  • $\begingroup$ R resistors can be in parallel , series or neither. Series and parallel are just two special cases $\endgroup$ – Cort Ammon Feb 21 '17 at 2:11
  • $\begingroup$ It is a confusing circuit. You have to look at these for a while to see what is connected to what. Starting on the - side, R4 and R6 are in parallel with one end of each connected to -. The other ends of R4 and R6 connect in series with R2, which connects to the + side of the source. Now you can see that R3 is connected to + and - and is in parallel with the rest of the circuit of R2 R4 and R6. $\endgroup$ – C. Towne Springer Feb 21 '17 at 6:31
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You can take out $R_{6}$ because there is a short circuit. This will leave you only with $R_{3}$ in parallel with $R_{2}$ and $R_{4}$. So you can substitute all that for:

$R_{t} = \frac{1}{\frac{1}{R_{3}}+\frac{1}{R_{2}+R_{4}}}$

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  • $\begingroup$ Sorry, but R6 is not shorted. And, please do not give (wrong) answers to homework questions. You should give help to let the asker figure it out for themselves. $\endgroup$ – hdhondt Feb 21 '17 at 21:50

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