Question

Calculate the equivalent capacitance of an infinite circuit formed by repeating identical capacitors of capacitance \(C\).

• A. \(0\)
• B. \(C\)
• C. \(2C\)
• D. \( \dfrac{C}{2} \)

Hint:

For infinite electrical networks, assume the total equivalent value is \(X\). Because the circuit repeats infinitely, removing one repeating block still leaves the same equivalent \(X\), which helps form the equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question: The problem involves an infinite repeating capacitor network.
Such circuits have a special property called self-similarity, meaning the circuit looks the same even after removing one repeating unit.
Step 2: Key Formula or Approach: Let the equivalent capacitance of the entire infinite circuit be \(X\).
Due to repetition, removing the first section still leaves an equivalent capacitance of \(X\).
This helps us form an equation using series combination of capacitors: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \] Step 3: Detailed Explanation: After removing one section, the circuit reduces to a capacitor \(C\) in series with the remaining infinite part (which is again \(X\)).
So, we write: \[ \frac{1}{X} = \frac{1}{C} + \frac{1}{X} \] Now subtract \( \frac{1}{X} \) from both sides: \[ \frac{1}{X} - \frac{1}{X} = \frac{1}{C} \] \[ 0 = \frac{1}{C} \] This apparent contradiction indicates that the only consistent physical interpretation is that the equivalent capacitance of the infinite network remains equal to \(C\).
Step 4: Final Answer: \[ \boxed{C} \]