Ten capacitors each of 2 μF capacity were arranged in two rows such that first row contains 6 capacitors in series where as the rest were in second row. The equivalent capacitance of the network is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{5}{6}μF$
$C_{eq}$, Net capacitance when 6 capacitors are connected in series = $\frac{1}{\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)}$
$=\frac{6}{2}=\frac{1}{3}μF$
$C_{eq}'$, Net capacitance when 4 capacitors of 2μF are connected in series = $\frac{1}{\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)}$
$C_{resistant}$, when both $C_{eq}$ and $C_{eq}'$ are = $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}μF$