Question

the resultant capacitance of four capacitors connected in series is ------------ the smallest individual capacitance
a)less than b)greater than c)equals to d) four times

Answer 1

Answer:

a)less than

Explanation:

The resultant capacitance of N capacitors connected in series is given by

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+....+\frac{1}{C_N}$

In this problem, since we have 4 capacitances in series, the resultant capacitance is

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$

Each of the single term on the right-side of the equation is a positive term: this means that the term $\frac{1}{C}$ will be larger than the individual terms $\frac{1}{C_i}$.

Therefore, this means that the reciprocal, C, will be smaller than each of the individual capacitances $C_i$.

In particular, if the 4 capacitances are identical,

$C_1=C_2=C_3=C_4=C_0$

So the resultant capacitance will be:

$\frac{1}{C}=4\frac{1}{C_0}\\\rightarrow C=\frac{C_0}{4}$

So the correct answer is

a) less than