Is there a non recursive formula for mixed resistors?

In equivalent resistors in multiple parallel resistors, like this simulate this circuit – Schematic created using CircuitLab

I need a formula for the equivalent resistance for the whole circuit, but in a way that, this formula can be generalized for more than 5 nets.

I've done this:

2E is the equivalent resistor for the last net, because they're in serie. Then, the next is in parallel: 2ED/(2E+D) and so on... but it looks really awful when finally get to the first net.

Every time this expression keeps growing and growing, I would like a simple formula for this: By using some power of 2 instead of the R's, I get those simplifications: But, nothing simple to make a generalization for further terms. NOTE: $Req(i)$ represents a circuit with $i$ nets on it.

• For questions that look very much like homework problems, it's appreciated if you at least show you've put some work into solving it yourself. – Dan Laks Oct 22 '14 at 17:30
• and using a solid colour background doesn't aid the readability of a circuit. – JIm Dearden Oct 22 '14 at 17:34
• @DanLaks doesn't read like homework to me - it's obvious hkviktor knows the right formula and method - just not if there's a "better" or "tidier" way of doing it. – Majenko Oct 22 '14 at 17:34
• Maybe the maths SE might help you reduce your long-hand formula into a more compact version? The methods used wouldn't be specific to electronics. – Majenko Oct 22 '14 at 17:37
• My image had a white background, but here appears to have a green one, @JImDearden. – hkviktor Oct 22 '14 at 17:44

It's to do with the golden ratio I think: - or... For N = 1,

x = $\dfrac{13}{8}+\dfrac{-1^{2} \cdot 3!}{3!\cdot 1!\cdot 4^{5}}$ =

x = $\dfrac{13}{8}+\dfrac{3!}{3!\cdot 4^{5}}$ = 1.6259765625

Anyway the link is here and I can't vouch for it being perfect.

• But that only works if all the resistors have the same value, I guess. – Matt L. Oct 22 '14 at 21:24
• Oh, now I see that the OP has updated the question, and indeed, now all R's are the same ... – Matt L. Oct 22 '14 at 21:25
• Maybe, It's certainly possible for me to make a math mistake. (I screwed up a pressure conversion making my first flow impedance and made it x10 too big..) – George Herold Oct 22 '14 at 23:21
• I suppose that by $N$ you mean the value of one resistor? What do you do when $N\neq1$? – user17592 Nov 22 '14 at 13:43

I get a series that looks like (Assuming equal R's)

1.) R(1+1/1)

2.) R(1+2/3)

3.) R(1+5/8)

4.) R(1+13/21)

5.) R(1+34/55)

I can write down the rest of the terms... but I'm not very good at finding a closed form solution.

Edit Re: ratio, So do a few sections and check me. Given number N(i) = 1+a/b the next number is,

N(i+1) = 1+(a+b)/(2a+b)

and N(1) = 1+1/1.. (a=1, b=1)

That's still a recursive relation.

• Actually, $$R_i=\frac{R_{i+1}}{2}$$, for each $$0\leq i\leq n$$ – hkviktor Oct 22 '14 at 19:44
• Hmm, I don't follow... the terms continue..89/144, 233/377, 610/987, 1597/2584.... if you calcutlate the ratio it's 0.61803... – George Herold Oct 22 '14 at 20:08
• Ah the golden ratio!! en.wikipedia.org/wiki/Golden_ratio or... $\dfrac{1}{x} = 1+x$ – Andy aka Oct 22 '14 at 20:13

I'm becoming quite sure that it's not possible to give a non-recursive formula for this. Diverger's answer shows a way to represent the equivalent resistance as a continued fraction:

$$R_{T_n} = R + \frac{\vdots}{\frac{1}{R}+\frac{1}{R+\frac{1}{\frac{1}{R}+\frac{1}{R+\frac{1}{ \frac{1}{R} + \frac{1}{R} }}}}}$$

However, to make this a general function, that works for any $n$, without dots, we would need a generalized continued fraction, like this:

$$A_n = b_n A_{n-1} + a_n A_{n-2}, \qquad B_n = b_n B_{n-1} + a_n B_{n-2}$$

However, this uses recursion ($A_n$ is defined in terms of $A_{n-1}$ and $A_{n-2}$, and analogically the same with $B_n$), and you didn't want recursion.

Andy's answer using the golden ration seems to work only when R=1. Otherwise the formula doesn't hold.

The previous answers are all incorrect since the answer is 0 Ω from inspection. Note that A and B are directly connected. All the resistors are just red herrings to see if you fall for doing all the math and then coming up with the wrong answer anyway. Looks like it worked.

• Heh, good point. But I actually think that's my mistake, I put the A and B nodes - and on the wrong place, apparently. Maybe this is indeed the intention of the question, or the resistance between the bottom left and the top right has to be calculated. – user17592 Nov 22 '14 at 14:48

If you label your resistors from right to left as $r_1, r_2, ... , r_n$ as in the following circuit: simulate this circuit – Schematic created using CircuitLab

then the equivalent resistance $R_n$ for $n$ resistors is

$$\begin{eqnarray*} R_n & = & r_n + \frac{r_nR_{n-1}}{r_n+R_{n-1}} \\ & = & r_n (1 + \frac{R_{n-1}}{r_n+R_{n-1}}) \end{eqnarray*}$$

but it doesn't seem very easy to solve this in terms of $n$ and the $r_n$ values. You could easily write a program/spreadsheet/etc to do these calculations, but you'd have to try some place a bit more math-heavy to find a closed-form answer.

• How did you get that formula? I mean, this formula is for the $n$ net of the circuit or for the equivalent resistor of a circuit with $n$ nets? – hkviktor Oct 22 '14 at 18:36
• I took the right side of the circuit as one resistor $R_{n-1}$ and combined it with the new pair of resistors (in series and parallel, like you did in your question). – Greg d'Eon Oct 22 '14 at 18:38
• I didn't see the second half of your comment - $R_n$ is the resistance of the ladder with $n$ rungs. (ie: you calculated $R_3$ but with E, D, and C instead of $r_1, r_2, r_3$) – Greg d'Eon Oct 22 '14 at 19:08
• OP is looking for a non-recursive formula (see the title), but you define $R_n$ in terms of $R_{n-1}$. – user17592 Nov 22 '14 at 11:35 simulate this circuit – Schematic created using CircuitLab

Start Thévenin equation at left end. Every stage contain a series resistor and a resistor pull-down. The series resistor will be in series with the previous stage's thévenin equivalent resistor.

$$R_{T0}=0\\ R_{T1} = (R_{T0}+R_{1}) || R_{2}\\ R_{T2} = (R_{T1}+R_{3}) || R_{4}\\ R_{T3} = (R_{T2}+R_{5}) || R_{6}\\ ...\\ R_{Tn} = (R_{Tn-1}+R_{2n-1})||R_{2n}$$

If all $R$s are equal. Then

$$R_{Tn}=(R_{Tn-1}+R)||R$$

Use Python or other languages to make a small app can calculate this, but it's a "recursive formula".

Let's think more. If your expand the equation above, it should be like a "Continued fraction".

Note: Continued fraction doesn't mean "infinite", if your count of resistor is fixed, then the expression will be finite.

$$R_{Tn} = \frac{\vdots}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{ \frac{1}{R} + \frac{1}{R} }+R}+\frac{1}{R}}+R}+\frac{1}{R}} + R$$

Per http://en.wikipedia.org/wiki/Closed-form_expression, Continued fraction is not a "closed-form expresson" but a "analytic expression".

• OP asked for a non-recursive formula (see the title), but you're defining $R_{T_n}$ in terms of $R_{T_{n-1}}$. Yes the last one doesn't do that, but it has dots... – user17592 Nov 22 '14 at 11:37
• The last isn't infinite! The last is the most left of the circuit. If i write from bottom, the dots should be at the top!!! – diverger Nov 22 '14 at 12:05
• True, but because of the dots, when you write it out, it's not a general function because it's different for different values of n. – user17592 Nov 22 '14 at 12:15
• Yes, so i think that's why it's not a "closed-form expression", just for discussion and sharing. +/-1 is nothing. – diverger Nov 22 '14 at 12:55
• There's a difference between finite continued fractions (this one, for example) and infinite continued fractions. According to the closed-form wiki, especially "Unlike the broader analytic expressions, the closed-form expressions do not include infinite series or continued fractions", finite series of continued fractions are members of the set of closed-form expressions. But in any case, this is not really a general function. – user17592 Nov 22 '14 at 13:00