In This post you will get answer of Q.
Find out the effective capacitance between points A and B as shown in the figure. Here $ce{C_1 = C_2 = 20 , mu F }$ and $ce{C_3 = C_4 = 10 , mu F}$
Solution:
Here, $C_1$ and $C_2$ are in series. Hence, their effective capacitance $C′$ is given by
$frac{1}{C’}=frac{1}{C_{1}} + frac{1}{C_{2}}$
$frac{1}{C’} =frac{1}{20}+frac{1}{20}$
$ Rightarrow :: C’ = 10 mu F $
Similarly, $frac{1}{C”} = frac{1}{C_3} + frac{1}{C_4}$
$ frac{1}{C”} = frac{1}{10} + frac{1}{10} Rightarrow C” = 5 , mu F$
Here C′ and C″ are in parallel. Hence, the equivalent capacitance between points A and
B is
$C_{AB} = C’ + C” = 10 + 5 = 15 , mu F$
Solution:
Here, $C_1$ and $C_2$ are in series. Hence, their effective capacitance $C′$ is given by
$frac{1}{C’}=frac{1}{C_{1}} + frac{1}{C_{2}}$
$frac{1}{C’} =frac{1}{20}+frac{1}{20}$
$ Rightarrow :: C’ = 10 mu F $
Similarly, $frac{1}{C”} = frac{1}{C_3} + frac{1}{C_4}$
$ frac{1}{C”} = frac{1}{10} + frac{1}{10} Rightarrow C” = 5 , mu F$
Here C′ and C″ are in parallel. Hence, the equivalent capacitance between points A and
B is
$C_{AB} = C’ + C” = 10 + 5 = 15 , mu F$