Question

Find the charge on each capacitance, in the given circuit? (in $\mu \text{C})$

• A. $100,40,40,60$
• B. $100,30,40,60$
• C. $70,90,120,60$
• D. $100,90,120,60$

Answer 1

The correct option is A $100,40,40,60$

As we can observe from the given figure, $C_2$ and $C_3$ are in series , their effective $C^{\prime }$ is in parallel to $C_4$.

$\because C^{\prime }=\frac{12\times 6}{12+6}=\frac{72}{18}=4\mu \text{F}$

$C^{\prime }$ is in parallel with $C_4$

$\therefore C^{\prime \prime }=(4+6)=10\mu \text{F}$


As we can observe $C^{\prime \prime }$ and $C_1$ in series:

$C_{eqv}=\frac{C^{\prime \prime }{C}_1}{C^{\prime \prime }+C_1}=\frac{10\times 10}{10+10}=5\mu \text{F}$

Net charge flows from the Battery is :

$Q=C_{eqv}{V}_{Battery}=5\times 20=100\mu \text{C}$

As we know, in parallel branches charge gets divided with respect to its effective capacitance.

Therefore, we can redraw the given circuit as follows,


$Q_1=\left(\frac{C^{\prime }}{C^{\prime }+C_4}\right)Q=\frac{4}{4+6}\times 100=40\mu \text{C}$

$Q_2=\left(\frac{C_4}{C^{\prime }+C_4}\right)Q=\frac{6}{4+6}\times 100=60\mu \text{C}$

Hence, option (a) is the correct answer.