Question

Three condenser of capacity $2\mu F,4\mu F$ and $8\mu F$ respectively, are first connected in series and then connected in parallel. The ratio of equivalent capacitances in two cases will be :

• A. 7:3
• B. 49:4
• C. 3:7
• D. 4:49

Answer 1

The correct option is D 4:49

For series, $\frac{1}{C_{series}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}$

$\therefore C_{series}=\frac{8}{7}\mu F$

For parallel, $C_{parallel}=C_1+C_2+C_3=2+4+8=14\mu F$

$\frac{C_{series}}{C_{parallel}}=\frac{8}{7\times 14}=\frac{4}{49}$