Question

Three capacitors $C_1=3\mu F$, $C_2=6\mu F$ and $C_3=10\mu F$ are connected to a $50$V battery as shown in the figure.
Calculate:
(i) The equivalent capacitance of the circuit between points A and B.
(ii) The charge on $C_1$.

Answer 1

If we look at the given figure, we can see $C_1$ and $C_2$ are in series, so the resultant capacitance for this combination is given by $C_4$

$\Rightarrow \frac{1}{C_4}=\frac{1}{C_2}+\frac{1}{C_1}$

$\Rightarrow C_4=2\mu F$

now the circuit will be look like as new drawn figure. now we can see $C_4$ is in Parallel with$C_3$, So the equivalent capacitance between A and B is given by $C_{AB}$

$\Rightarrow C_{AB}=C_3+C_4\Rightarrow C_{AB}=10\mu F+2\mu F=12\mu F$

so the total charge flowing through the battery is $Q_{total}$

$\Rightarrow Q_{total=}{C}_{AB}V\Rightarrow Q_{total=}12\times 50=600\mu C$

Assume that charge flowing in chain 1 is $Q_1$ and in chain 2 is$Q_2$ and we can see both the chains are in parallel with each other then the potential difference will be also same at the ends.

$\Rightarrow Q_{total}{=Q}_2+Q_1=600\mu C...\left(1\right)$

$\Rightarrow \frac{Q_1}{C_4}=\frac{Q_2}{C_3}$

$\Rightarrow \frac{Q_1}{2}=\frac{Q_2}{10}$

$\Rightarrow Q_2={5Q}_1$

from equation $\left(1\right)$ and $\left(2\right)$

$\Rightarrow Q_1=100\mu Cand{Q}_2=500\mu C$

now the charge on $C_1$ will be $q_1$ equal to $Q_1$

because in a series of capacitors same amount of charge will flow so the charge on $C_1$ is

$\Rightarrow q_1=100\mu C$