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(A) For the equivalent capacitance to be independent of the number of sections,the ladder must behave as an infinite ladder. Let the equivalent capaci

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$4$

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(A) For the equivalent capacitance to be independent of the number of sections,the ladder must behave as an infinite ladder. Let the equivalent capacitance between points $A$ and $B$ be $C_{eq} = C$.If we add one more section to the infinite ladder,the equivalent capacitance remains $C$. The circuit consists of a $2\,\mu F$ capacitor in parallel with the series combination of a $4\,\mu F$ capacitor and the rest of the infinite ladder (which also has an equivalent capacitance $C$).Thus,the equation is:$C = 2 + \frac{4 	imes C}{4 + C}$Multiplying by $(4 + C)$:$C(4 + C) = 2(4 + C) + 4C$$C^2 + 4C = 8 + 2C + 4C$$C^2 - 2C - 8 = 0$Solving the quadratic equation:$(C - 4)(C + 2) = 0$Since capacitance cannot be negative,we have $C = 4\,\mu F$.

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