Question

Three capacitors, whose capacitances are 4 µF, 6 µF and 12 µF respectively, are connected in series order, then find the equivalent capacitance of the combination.​

Answer 1

Answer:

2

Explanation:

1/1/4+1/6+1/12

3+2+1/12

1/1/2

2

Answer 2

Answer:

The equivalent capacitance of the series combination of capacitors is $2\mu F$.

Explanation:

Given three capacitors connected in series.

The capacitances of capacitors are $C_{1} =4 \mu F$, $C_{2} = 6 \mu F$ and $C_{3} = 12 \mu F$

When two capacitors, $C_{1}$ and $C_{2}$ are connected in series, then the equivalent capacitance of the combination of capacitors is given by

$\frac{1}{C} =\frac{1}{C_{1}}+ \frac{1}{C_{2}}$.

Thus, the equivalent capacitance of the given capacitors in series is

$\frac{1}{C} =\frac{1}{C_{1}}+ \frac{1}{C_{2}}+ \frac{1}{C_{3}}$

$=\frac{1}{4}+ \frac{1}{6}+ \frac{1}{12}$

$= \frac{3+2+1}{12}$

$= \frac{6}{12}= \frac{1}{2}$

Since $\frac{1}{C} =\frac{1}{2}$, therefore $C=2\mu F$

Hence, the equivalent capacitance of the series combination of capacitors is $2\mu F$.