Current Electricity Charging Discharging Of Capacitors Question 703
Question: 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)
Options:
A) $ \frac{R _{1}-R _{2}+\sqrt{R _{1}^{2}+R _{2}^{2}+6R _{1}R _{2}}}{2} $
B) $ \frac{R _{1}+R _{2}+\sqrt{R _{1}^{2}+R _{2}^{2}+6R _{1}R _{2}}}{2} $
C) $ \frac{R _{1}-R _{2}-\sqrt{R _{1}^{2}+R _{2}^{2}+6R _{1}R _{2}}}{2} $
D) None of these
Show Answer
Answer:
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}=kR _{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}kR _{AB}}{R _{1}+kR _{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}+6R _{1}R _{2}}}{2} $
BETA
