Question

Three uncharged capacitors of capacitance $C_1=1\mu \text{F},C_2=2\mu \text{F}$ and $C_3=3\mu \text{F}$ are connected as shown in figure. If potential at shown points are $V_A=10\text{V};V_B=25\text{V}$ and $V_D=20\text{V}$. The potential at point $O$ will be

• A. $20\text{V}$
• B. $10\text{V}$
• C. $30\text{V}$
• D. $40\text{V}$

Answer 1

The correct option is A $20\text{V}$

Let the charge flows towards point $O$ from point $\text{B}$ and $\text{D}$, because $V_B>V_D>V_A$


Applying KCL junction law at $O$,

$\Rightarrow q_3+q_2=q_1$

$\Rightarrow C_3(V_D-V_O)+C_2(V_B-V_O)=C_1(V_O-V_A)$

Substituting given value of potentials, we get

$\Rightarrow 3(20-V_O)+2(25-V_O)=1(V_O-10)$

$\Rightarrow 60-3V_O+50-2V_O=V_O-10$

$\therefore V_O=\frac{120}{6}=20\text{V}$

Hence, option $\left(a\right)$ is correct.

Why this question? It intends to test your understanding of potential difference and charge flow. In this problem we can infer that positive charge will flow from high potential to low potential $V_B>V_D>V_B$, thus directions of flow will be towards point $\text{A}$. Caution: The potential difference across a capacitor is difference between potential of +ve plate (high potential) and -ve plate (low potential).