An infinite capacitor ladder circuit is formed by the infinite repetition of a single link. Each lin

Question

An infinite capacitor ladder circuit is formed by the infinite repetition of a single link. Each link consists of two identical capacitors: one in series along the top wire, and one in parallel (shunt) connecting the top and bottom wires. Each capacitor has a capacitance $C$. A constant DC voltage $V_0$ is applied across the input terminals of the ladder. The ratio of the voltage across the $n^{th}$ shunt(parallel) capacitor to the voltage across the $(n-1)^{th}$ shunt(parallel) capacitor (where the $0^{th}$ shunt is considered the input source) is independent of $n$ and can be expressed in the form $\frac{X - \sqrt{Y}}{2}$, where $X$ and $Y$ are positive integers. Find the value of $(X + Y)$.

Professor PC Thomas Classes · Answer

Let $V_n$ be the voltage across the $n^{th}$ shunt capacitor. The charge entering the $n^{th}$ section is $Q_n = C_{eq} V_n$. This entire charge passes through the $n^{th}$ series capacitor. The voltage drop across this series capacitor is $\Delta V = \frac{Q_n}{C} = V_n \frac{C_{eq}}{C}$. The voltage at the next node is $V_{n+1} = V_n - \Delta V = V_n \left(1 - \frac{C_{eq}}{C}\right)$. From the previous question, $\frac{C_{eq}}{C} = \frac{\sqrt{5}-1}{2}$. Therefore, the voltage reduction ratio is $\frac{V_{n+1}}{V_n} = 1 - \frac{\sqrt{5}-1}{2} = \frac{3 - \sqrt{5}}{2}$. Comparing this to $\frac{X - \sqrt{Y}}{2}$, we get $X = 3$ and $Y = 5$. Thus, $X + Y = 8$.