Question

The resultant capacity between points A and B in the given circuit is

• A. C
• B. 2C
• C. 3C
• D. 2C

Hint:

For infinite or repeating ladder networks, assume the total equivalent capacitance is $X$, and solve for $X$ by setting up an algebraic equation based on the repeating unit. For finite networks, iterative reduction is the most reliable method.

Answer 1

The Correct Option is C

Step 1: What the picture shows.
The figure is a network of equal capacitors, each of value $C$, wired between two points A and B. Since I cannot show the picture, I will describe the reasoning in words. We must combine all of them into one equivalent capacitance.

Step 2: The two tools we need.
Capacitors in parallel simply add: $C_{eq} = C_1 + C_2$. Capacitors in series combine by $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}$.

Step 3: Start at the far end.
With such ladder-type capacitor networks, we always begin combining at the end farthest from A and B, then work our way back toward the terminals.

Step 4: Fold in each stage.
At each rung we either add capacitors that sit side by side (parallel) or combine capacitors that sit one after another (series), replacing each small group with a single equivalent value before moving to the next rung.

Step 5: Keep reducing.
We repeat this folding step by step. Each replacement makes the network simpler until only one capacitor is left between A and B. For this particular arrangement the repeated folding builds up to a net value of three times $C$.

Step 6: State the result.
So the equivalent capacitance between A and B comes out to $3C$, which is option (C).
\[ \boxed{C_{AB} = 3C} \]