Question

Three capacitors of capacitances $1\mu F$, $2\mu F$ and $4\mu F$ are connected first in a series combination, and then in a parallel combination. The ratio of their equivalent capacitances will be:

• A. $2:49$
• B. $49:2$
• C. $4:49$
• D. $49:4$

Answer 1

Answer: (C)

$4:49$

Equivalent capacitance in series is reciprocal of sum of reciprocals of individual capacitances.

$\Rightarrow \frac{1}{C_S}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}$

So, the series equivalent is $\frac{4}{7}\mu F$.

In parallel, equivalent capacitance is the sum of individual capacitances.

$\Rightarrow C_P=1+2+4$

So, the parallel equivalent is $7\mu F$.

Thus, the ratio is $\frac{4}{49}$.