# Electrical resistance calculator question [closed]

Consider this circuit diagram: This series is continued till $n$ times.

Neglect the first two resistances. Start with the four resistances in the box like pattern.

How I tried to solve this:

I wanted to find a general formula for the net resistance . I tried to discover a pattern in the resistances when $$n=1,n=2,n=3$$ and so on. The resistances were $$\frac{3}{4},\frac{11}{15},\frac{41}{56}$$ when $$n=1,n=2,n=3$$ respectively and so on. I came to know that the net resistance for the first loop could be taken as $$\frac{a}{b}$$ where $$a=3$$ and $$b=4$$ The net resistance for second loop was $$\frac{a+2b}{a+3b}$$ where $$a=3$$ and $$b=4$$ Let us take this equal to $$\frac{c}{d}$$ The net resistance for the third loop was $$\frac{c+2d}{c+3d}$$ where $$c=a+2b$$ and $$d=a+3b$$ Let us take this equal to $$\frac{e}{f}$$ The net resistance for the fourth loop was $$\frac{e+2f}{e+3f}$$ where $$e=c+2d$$ and $$f=c+3d$$ Let us take this equal to $$\frac{g}{h}$$ This pattern is continuing like this. But I don't know how to find the general term for such a series. I will be grateful to you if you provide me with the solution.

• Is there sone reason you are neglecting the first two resistors? – Triatticus May 31 '18 at 7:50
• I am doing that only because I want the method for the loops. I wanted to add a picture to make the circuit easier to understand but I couldn't get any picture only with loops. So I told to neglect. We can add 2 to the formula to get the Answer if we don't want to neglect it . – MathsWiz May 31 '18 at 11:59
• Try defining the resistance from A to B as Rab and then use the parallel resistance formula for n iterations. – user45664 May 31 '18 at 16:50
• Possible duplicate of Equivalent resistance in ladder circuit – sammy gerbil May 31 '18 at 17:05
• It is an infinite series with a different pattern although the kind of question is a little bit same. That question can be solved using quadratic but I couldn't solve this one using quadratic . – MathsWiz Jun 1 '18 at 9:49