question_answer

The circuit diagram shown in figure consists of a very large (infinite) number of elements. The resistances of the resistors in each subsequent element differ by a factor of k from the resistances of the resistors in the previous elements. Determine the resistance \[{{R}_{AB}}\] between points A and B if the resistances of the first element are \[{{R}_{1}}\] and\[{{R}_{2}}\]. (k=1/2)

A) \[\frac{{{R}_{1}}-{{R}_{2}}+\sqrt{R_{1}^{2}+R_{2}^{2}+6{{R}_{1}}{{R}_{2}}}}{2}\]

B) \[\frac{{{R}_{1}}+{{R}_{2}}+\sqrt{R_{1}^{2}+R_{2}^{2}+6{{R}_{1}}{{R}_{2}}}}{2}\]

C) \[\frac{{{R}_{1}}-{{R}_{2}}-\sqrt{R_{1}^{2}+R_{2}^{2}+6{{R}_{1}}{{R}_{2}}}}{2}\]                              

D) None of these

Correct Answer: A

Solution :

[a] It follows from symmetry considerations that if we remove the first element from the circuit, the resistance of the remaining circuit between points C and D will be \[{{R}_{CD}}=k{{R}_{AB}}.\] Therefore, the equivalent circuit of the infinite chain will have the form shown in figure. Applying to this circuit the formulas for the           resistance of series and parallel resistors, we obtain \[{{R}_{AB}}=\frac{{{R}_{1}}+{{R}_{2}}k{{R}_{AB}}}{{{R}_{1}}+k{{R}_{AB}}}\] Solving the quadratic equation for \[{{R}_{AB}},\] we obtain (in particular, for k= 1/2) \[{{R}_{AB}}=\frac{{{R}_{1}}-{{R}_{2}}+\sqrt{R_{1}^{2}+/R_{2}^{2}+6{{R}_{1}}{{R}_{2}}}}{2}\]