Find this?
The infinite circuit shown above is formed by repetition of a same link consisting of R 1 = 4 Ω and R 2 = 3 Ω .
Find the resistance of the circuit between A and B .
The answer is 6.
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2 solutions
Good solution,easy solution .
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General solution too, The same can be applied to other simple infinite circuits.
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@Nnsv Abhiram @Chew-Seong Cheong yes .it is applied for other infinite circuites. these method is used in other problems related to number theory....etc
I AM INSPIRED BY IRODOV
.SO THERE IS SPECIAL FORMULA FOR THIS ....
So anwer is 6.
@Nnsv Abhiram , you should not use all cap in text. All cap in text is equivalent to shouting in voice which is rude.
I have redone the image and wording for you.
How did you change the image?,And how did you type the ohm symbol.?
I drew the image using Print, saved it and posted it here. You can enter LaTex code by keying in \ ( [formulas] \ ) (no space between backslash \ and the brackets ()). In the LaTex backslash-brackets type \Omega Ω . \alpha α , \beta β , \gamma γ ....
@Nnsv Abhiram @Chew-Seong Cheong do anyone know the derivation of this formula?
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Obviously it is from the third line by solving the following quadratic equation for R .
R 2 − R 1 R − R 1 R 2 ⟹ R = 0 = 2 R 1 + R 1 2 + 4 R 1 R 2 = 2 R 1 ( 1 + 1 + 4 ( R 1 R 2 ) )
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@Chew-Seong Cheong Oh!! I didn't recognize it. Thanks
Let the resistance of the infinite circuit between A and B be R . Then the circuit is equivalent the circuit below where
R = R 1 + R 2 ∣ ∣ R ( R − R 1 ) ( R + R 2 ) R 2 − R 1 R − R 1 R 2 R 2 − 4 R − 1 2 ( R − 6 ) ( R + 2 ) ⟹ R = R 1 + R + R 2 R R 2 = R R 2 = 0 = 0 = 0 = 6 Ω Rearranging Putting in R 1 = 4 , R 2 = 3 Since R > 0