Question

Seven capacitors each of capacitance 2μF are to be connected to obtain a capacitance of $\left( \frac{10}{11} \right)$μF. Which of the following combination is possible

• A. (a) 5 in parallel 2 in series
• A. (b) 4 in parallel 3 in series
• A. (c) 3 in parallel 4 in series
• A. (d) 2 in parallel 5 in series

Answer 1

(a)

5 capacitors in parallel gives 5 x 2 μF = 10μF capacity. Further, two capacitors in series gives a capacity 1μF. When the two combinations are connected in series, they give a resultant capacitance $\left( \frac{10\text{ x 1}}{\text{10} + 1} \right)$ = $\left( \frac{10}{11} \right)$μF