The capacitance of an infinite circuit formed by the repetition of the same link consisting of two identical capacitors, each with capacitance \(C\), as shown in the figure is:
Show Hint
In infinite electrical networks:
The remaining circuit after removing the first block has the same equivalent value.
This self-similarity leads to a quadratic equation for the equivalent resistance or capacitance.
Concept:
For an infinite repeating network, the equivalent capacitance of the entire circuit remains the same even after removing the first repeating section.
This self-similarity allows us to form an equation for the equivalent capacitance.
Let the equivalent capacitance between points \(A\) and \(B\) be \(C_{\text{eq}}\).
Step 1: {Represent the infinite network using self-similarity.}
After the first section, the remaining infinite network still has equivalent capacitance \(C_{\text{eq}}\).
The first section consists of:
One capacitor \(C\) in series along the top branch
A vertical capacitor \(C\) connected to the remaining network
Step 2: {Combine the vertical capacitor with the remaining network.}
The vertical capacitor \(C\) is in parallel with the equivalent capacitance of the remaining infinite network.
Thus,
\[
C' = C + C_{\text{eq}}
\]
Step 3: {Find equivalent capacitance of the series combination.}
The series combination of \(C\) and \(C'\) is
\[
C_{\text{eq}} = \frac{C \cdot C'}{C + C'}
\]
Substitute \(C' = C + C_{\text{eq}}\):
\[
C_{\text{eq}} =
\frac{C(C + C_{\text{eq}})}{2C + C_{\text{eq}}}
\]
Step 4: {Solve the equation.}
\[
C_{\text{eq}}(2C + C_{\text{eq}}) = C(C + C_{\text{eq}})
\]
\[
2CC_{\text{eq}} + C_{\text{eq}}^2 = C^2 + CC_{\text{eq}}
\]
\[
C_{\text{eq}}^2 + CC_{\text{eq}} - C^2 = 0
\]
Dividing by \(C^2\):
\[
\left(\frac{C_{\text{eq}}}{C}\right)^2
+
\left(\frac{C_{\text{eq}}}{C}\right)
-
1 = 0
\]
Step 5: {Solve the quadratic equation.}
\[
\frac{C_{\text{eq}}}{C} =
\frac{-1 + \sqrt{5}}{2}
\]
Thus,
\[
C_{\text{eq}} =
\frac{(\sqrt{5}-1)C}{2}
\]
Hence, the correct option is (B).
