Question

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:

• A. Zero
• B. \( \dfrac{(\sqrt{5}-1)C}{2} \)
• C. \( \dfrac{(\sqrt{5}+1)C}{2} \)
• D. Infinite

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the equivalent capacitance of an infinite ladder network of capacitors.
The key to solving this type of problem is to recognize the repeating, self-similar nature of the circuit. We can assume that adding one more identical section to the infinite network does not change its total capacitance.
Step 2: Key Formula or Approach:
Let the equivalent capacitance of the entire infinite network between points A and B be \(C_{\text{eq}}\).
Because the circuit is infinite, the capacitance of the network to the right of the first repeating section is also \(C_{\text{eq}}\).
The approach involves setting up an equation for \(C_{\text{eq}}\) based on this self-similarity. We will use the formulas for capacitors in series and parallel:
- Parallel Combination: \( C_p = C_1 + C_2 \)
- Series Combination: \( C_s = \dfrac{C_1 C_2}{C_1 + C_2} \)
Step 3: Detailed Explanation:
The circuit can be simplified by replacing the entire network to the right of the first section with its equivalent capacitance, \(C_{\text{eq}}\).
This \(C_{\text{eq}}\) is in parallel with the vertical capacitor \(C\). Let's call their combined capacitance \(C'\).
\[ C' = C + C_{\text{eq}} \] Now, this combination \(C'\) is in series with the first horizontal capacitor \(C\). The equivalent capacitance of this entire arrangement is the total equivalent capacitance of the network, \(C_{\text{eq}}\).
\[ C_{\text{eq}} = \frac{C \cdot C'}{C + C'} \] Substitute the expression for \(C'\) into this equation:
\[ C_{\text{eq}} = \frac{C (C + C_{\text{eq}})}{C + (C + C_{\text{eq}})} = \frac{C^2 + C \cdot C_{\text{eq}}}{2C + C_{\text{eq}}} \] Now, we solve this equation for \(C_{\text{eq}}\):
\[ C_{\text{eq}} (2C + C_{\text{eq}}) = C^2 + C \cdot C_{\text{eq}} \] \[ 2C \cdot C_{\text{eq}} + C_{\text{eq}}^2 = C^2 + C \cdot C_{\text{eq}} \] Rearranging the terms gives a quadratic equation in \(C_{\text{eq}}\):
\[ C_{\text{eq}}^2 + C \cdot C_{\text{eq}} - C^2 = 0 \] We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(x = C_{\text{eq}}\), \(a=1\), \(b=C\), and \(c=-C^2\).
\[ C_{\text{eq}} = \frac{-C \pm \sqrt{C^2 - 4(1)(-C^2)}}{2(1)} = \frac{-C \pm \sqrt{C^2 + 4C^2}}{2} = \frac{-C \pm \sqrt{5C^2}}{2} = \frac{-C \pm C\sqrt{5}}{2} \] Since capacitance must be a positive value, we take the positive root:
\[ C_{\text{eq}} = \frac{-C + C\sqrt{5}}{2} = C \frac{(\sqrt{5} - 1)}{2} \] Step 4: Final Answer:
The equivalent capacitance of the infinite circuit is \( \dfrac{(\sqrt{5}-1)C}{2} \). This corresponds to option (B).