Network of resistors?

As shown in the figure ,a network of resistors of resistances R1and R2 extends to infinity toward the right.





http://session.masteringphysics.com/prob...





Q1.


Find the total resistance R1 of the infinite network. (Hint: Since the network is infinite, the resistance of the network to the right of points C and D is also equal to R_T .)


Epress your answer in terms of the variables R1 and R2


R_T= please help.

Network of resistors?
Take off the 3 resistors on the left and you still have the infinite series. Call its resistance X (easier to type than R_T). The fact that when you add back those resistors you still have resistance X gives you this equation:


2R1 + 1 / (1/R2 + 1/X) = X


Manipulating this,


2R1*(1/R2 + 1/X) + 1 = X*(1/R2 + 1/X)


2R1/R2 + 2R1/X +1 = X/R2 +1


2XR1/R2 + 2R1 -X^2/R2 = 0


Rearranging in conventional quadratic order,


-1/R2*X^2 +(2R1/R2)X +2R1 = 0


(A = -1/R2; B=2R1/R2; C=2R1)


whose roots are


[-2R1/R2 ± sqrt((2R1/R2)^2 + 8R1/R2)]/(-2/R2)


One root, [-2R1/R2 + sqrt((2R1/R2)^2 + 8R1/R2)]/(-2/R2) is negative (numerator +, denominator -).


The other root, [-2R1/R2 - sqrt((2R1/R2)^2 + 8R1/R2)]/(-2/R2), can be simplified to


[2R1/R2 + sqrt((2R1/R2)^2 + 8R1/R2)]/(2/R2) and to


[R1/R2 + sqrt((R1/R2)^2 + 2R1/R2)]/(1/R2) and to


R1 + sqrt(R1^2 + 2R1R2) (answer)


Checking:


Picking at random R1=3 and R2=7,


X = 10.14143


2R1 + 1 / (1/R2 + 1/X) = 6 + 1/(1/7 + 1/10.14143) = 10.14143 = X


Check is OK.
Reply:I worked for that one! Thank-you. Report It