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) \[{{R}_{AB}}=\frac{{{R}_{1}}-{{R}_{2}}+\sqrt{R_{1}^{2}+/R_{2}^{2}+6{{R}_{1}}{{R}_{2}}}}{2}\]You need to login to perform this action.
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