Question

Three capacitors, whose capacitances are 10 uF, 15 uF and 30 uF
ON
respectively, are connected in series order, then find the equivalent
capacitance of the combination.​

Answer 1

Given:

Three resistors of capacitances

$C_{1}$ = 10 μF

$C_{2}$ = 15 μF

$C_{3}$ = 30 μF

To find:

Equivalent  capacitance in series combination

Solution:

In series combination;

• $\frac{1}{C_{eq}}$ = $\frac{1}{C_{1} }$ + $\frac{1}{C_{2} }$ + $\frac{1}{C_{3} }$

              = $\frac{1}{10}$ + $\frac{1}{15}$ + $\frac{1}{30}$

              = $\frac{3+2+1}{30}$ = $\frac{6}{30}$ = $\frac{1}{5}$

• $C_{eq}$ = 5 μF