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How to find the voltage ?

Given is the following mixed circuit below, every resistor has the value $1$ ohm, the current flow through the last resistor is $1$A and I have to find the voltage V

enter image description here

If I first find the substitute resistor; I begin from right, $2$ most right resistors are in series and the a third one is parallel to them, so I call first $2$ from right $R_0$ then adding another $2$ from left gives $R_1$ etc, and the general formula looks like;

$\displaystyle R_{n+1}=1+\frac{1\cdot R_n}{1+R_n},\quad$with $R_0=2\Omega$

so in this case I have to compute $R_3$ and I got $\frac{34}{21}\Omega$

and using the reciprocal rule, I get the following currents:

enter image description here

So the current through the most left resistor is $21$A and the voltage is then $$V=R\cdot I=\frac{34}{21}\cdot 21 =34 V$$ Am I right?

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  • $\begingroup$ Seems correct. Interesting how the currents form the Fibonacci sequence. $\endgroup$ Apr 9, 2015 at 11:04
  • $\begingroup$ @ArpanBanerjee Thanks for your response, but the problem here is that the task has three partial exercises and in the first exercise have to find the voltage and in the second one the equivalent resistor, so is there a method to find the voltage without computing first the equivalent resistor ? $\endgroup$
    – derivative
    Apr 9, 2015 at 11:07
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    $\begingroup$ This seems like a check-my-work question, which is typically off-topic. $\endgroup$
    – Sean
    Apr 9, 2015 at 11:21

1 Answer 1

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In response to your comment, there is a way to find the voltage without finding equivalent resistance.

You can write the potentials at each junction point. Taking the points on the lower line to be zero potential, the potential gain on going up the last branch is $1$V. Then going left, add another $1$V.

Then, current through the last branch is 1A and the resistance is $2\Omega$ whereas the resistance of the second last branch is $1\Omega$, hence the current through it is 2A.

Hence, on going left from the point with potential $2$V, you are going across $3$V (3A and 1$\Omega$).

Similarly keep on building the potentials and you'll get your answer.

enter image description here

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    $\begingroup$ I believe the left most node voltage is $13 + 21 = 34\mathrm V$ $\endgroup$ Apr 9, 2015 at 11:38
  • $\begingroup$ My mistake. Edited now $\endgroup$ Apr 9, 2015 at 11:39

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