Question

Figure shows five capacitors connected across a $12\text{V}$ power supply. What is the charge on the $2\mu \text{F}$ capacitor ?

• A. $6\mu \text{C}$
• B. $10\mu \text{C}$
• C. $12\mu \text{C}$
• D. $8\mu \text{C}$

Answer 1

The correct option is D $8\mu \text{C}$

Let, $V_A=x;V_D=y$

Since, we know that potential drop across the conducting wire is zero means potential throughout the wire remains same.

$\therefore V_A=V_C=x$

And, $V_D=V_B=y$

$\Rightarrow$ We can clearly see that potential difference across all three capacitors $1\mu \text{F},2\mu \text{F}$ and $3\mu \text{F}$ are same means all are in parallel and their effective capacitance is

$C_{eq}=(1+2+3)\mu \text{F}=6\mu \text{F}$

Therefore, the circuit can be reduced as shown below.

Thus we have three capacitors in series each of capacitance $6\mu \text{F}$ across $12\text{V}$ power supply. So the potential drop across each is $12/3=4\text{V}$.

This directly implies that voltage across $2\mu \text{F}$ capacitor is $4\text{V}$.

$\therefore Q=\left(2\mu \text{F}\right)\times \left(4\text{V}\right)=8\mu \text{C}.$

Hence, option $\left(b\right)$ is the correct answer.