Question

Each of the capacitor shown in the figure has a capacitance of $2\mu F$. Find the equivalent capacitance of the assembly between the points $A$ and $B$. Suppose a battery of emf $60volts$ is connected between A and B. Find the potential difference appearing on the individual capacitors.

Answer 1

$\begin{array}{c}\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\\ C_{eq}=\frac{2}{3}\\ C_{eq}=3\times \frac{2}{3}=2\mu F\end{array}$
Net charge came out of the battery
$=C_{eq}V=2\times 60=120\mu C$
Now, due to symmetry
This will be equally divided in 3 branch
q in each branch
$=\frac{q_{eq}}{3}=\frac{120}{3}\mu C=40\mu C$
so in each row, charge on each capacitor =$2\mu C$ as in series charge on each capacitor is same so potential difference
$=\frac{q}{c}=\frac{40}{2}=20V$