Question

calculate the equivalent capacitance between point a and b when connected in series C1 is 1uf C2 is 2uf C3 is 3uf and C4 is uf

Answer 1

◈ If n capacitors are connected in parallel,

 $\implies \rm \orange{C_{parallel}=C_1+C_2+...\ +C_n}$

◈ If n capacitors are connected in series,

$\implies \rm \green{\dfrac{1}{C_{series}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+...\ +\dfrac{1}{C_n}}$

❊ Observe the diagram in the attchment carefully,

Node A = Node Q

Node P = Node R

So, replace Node Q with with A abd Node R with P.

❖ Referring to the attchment,

In between node A and P, we have three capacitors, which are in parallel.  

➤ $\sf C_1+C_2+C_3$

➤ $1+2+3$

➤ $\sf 6 \mu F$

Now, this resultant capacitance is in series with $\sf 6 \mu F$.

➠ $\sf C_{AB}= \dfrac{6 \times 6}{6 + 6}$

➠ $\sf C_{AB}= \dfrac{36}{12}$

➠ $\sf C_{AB}=3 \mu F$

So, equivalent capacitance between the points A and B is $\sf 3 \mu F$