Capacitors in Parallel and in Series — Question 10

Section 25-3 · Physics pages 4070–4071

Question 10 An infinite capacitive ladder satisfies a positive fixed-point equation An infinite ladder begins with a series capacitor C, followed by a node with a shunt capacitor C to the return terminal. Beyond that node the same complete ladder repeats. All internal nodes are neutral. Let X be the input equivalent capacitance. C C repeat Schematic; labels, not drawing lengths, specify the data. (a) Derive and solve the self-consistency equation for X. (b) Calculate the first two finite approximations obtained by starting with X0=0 and iterating the network rule; identify the unphysical root.
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Question 10 - Solution An infinite capacitive ladder satisfies a positive fixed-point equation C C repeat (a) Method and calculation X = C(C + X) 2C + X ⇒X2 + CX −C2 = 0. X = √ 5 −1 2 C ≈.618034C. (b) Interpretation and further calculation X1 = C/2, X2 = C(C + C/2) 2C + C/2 = 3C/5. X−= −1 + √ 5 2 C < 0 is inadmissible for this passive positive-capacitance network. Physical check Finite passive approximations select the positive root and approach the infinite-ladder value from below.

Question 10

An infinite capacitive ladder satisfies a positive fixed-point equation

An infinite ladder begins with a series capacitor C, followed by a node with a shunt capacitor C to the return terminal. Beyond that node the same complete ladder repeats. All internal nodes are neutral. Let X be the input equivalent capacitance.

Diagram shown in the original SVG worksheet.

  1. Derive and solve the self-consistency equation for X.

  2. Calculate the first two finite approximations obtained by starting with X0=0 and iterating the network rule; identify the unphysical root.

Question 10 - Solution

An infinite capacitive ladder satisfies a positive fixed-point equation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

X=C(C+X)2C+X⇒X2+CX−C2=0.X=\frac{C(C+X)}{2C+X}\Rightarrow X^2+CX-C^2=0.X=5−12C≈.618034C.X=\frac{\sqrt5-1}{2}C\approx.618034C.

(b) Interpretation and further calculation

X1=C/2,X2=C(C+C/2)2C+C/2=3C/5.X_1=C/2,\quad X_2=\frac{C(C+C/2)}{2C+C/2}=3C/5.X−=−1+52C<0X_-=-\frac{1+\sqrt5}{2}C<0is inadmissible for this passive positive-capacitance network.

Physical check

Finite passive approximations select the positive root and approach the infinite-ladder value from below.