Question

If three 15 uf capacitors are connected in series, the net capacitance is

Answer 1

1/C= 1/5+1/5+1/5
1/C=3/15
C=5uf

Answer 2

Answer:

$5\ \mu F$

Explanation:

When capacitors are connected is series, then their equivalent combination is calculated using the following formula:

$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+.....+\dfrac{1}{C_n}$

• $C$ = capacitance of the capacitors = $15\ \mu F$

According to the question, three capacitors of the same capacitance are connected in series. So, its equivalent capacitance is given by:

$\dfrac{1}{C_{eq}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}\\\Rightarrow \dfrac{1}{C_{eq}}= \dfrac{3}{C}\\\Rightarrow C_{eq}= \dfrac{C}{3}\\\Rightarrow C_{eq}= \dfrac{15\ \mu F}{3}\\\Rightarrow C_{eq}=5\ \mu F$

Hence, the net capacitance of the connection is $5\ \mu F$.