Question

In the given circuit taking potential at point $\text{B}$ as zero, find the potential of point $Q$.

• A. $15\text{V}$
• B. $12.5\text{V}$
• C. $17.5\text{V}$
• D. $7.5\text{V}$

Answer 1

The correct option is D $7.5\text{V}$

We can solve this problem, by potential method, by assigning potential to known points, we can get potential of the unknown points.

Firstly, let us find potential drop across capacitors, then we can easily apply potential method to the given circuit.

$Q_{net}=C_{eqn}{V}_{Battery}$

From the given circuit, equivalent capacitance of $C_2$ and $C_3$.

$C_{23}=\frac{C_2{C}_3}{C_2+C_3}$

$\Rightarrow C_{23}=\frac{12\times \left(4\right)}{12+4}=3\mu \text{F}$

Since, $C_{23}$ and $C_4$ are in parallel, so their equivalent capacitance,

$C_{234}=3+7=10\mu \text{F}$

And $C_{234}$ and $C_1$ are in series, so equivalent capacitance of the circuit,

$C_{eqv}=\frac{C_{234}\times \left(C_1\right)}{C_1+C_{234}}$

$\Rightarrow C_{eqv}=\frac{10\times \left(10\right)}{10+10}=5\mu \text{F}$

So, net charge flowing through the battery,

$\because Q_{net}=C_{eqv}{V}_{Battery}=(5\times {10}^{-6})20=100\mu C$

Potential drop across $C_1$ is,

$V_1=\frac{Q_1}{C_1}=\frac{100}{10}=10\text{V}$

Now, we can redraw the given circuit by assigning known potential as follows


Using Kirchhoff`s junction law, to the isolated part of the system, we get

$(V_0-10)12+(V_0-0)4=0$

$\Rightarrow 12V_0-120+4V_0=0$

$\Rightarrow 16V_0=120$

$\therefore V_0=\frac{30}{40}=\frac{15}{2}=7.5\text{V}$

Hence, the potential at point $Q$ is $7.5\text{V}$.