Question

Figure shows the arrangement of four capacitors. The potential difference between $A$ & $B$ is $1500$ $Volt$. Find the potential difference between the plates of $2$ $\mu F$ capacitor.

• A. $1500$ $V$
• B. $900$ $V$
• C. $750$ $V$
• D. $450$ $V$

Answer 1

The correct option is B $900$ $V$

Given:

$\Delta V$ between $A$ and $B=1500v$

To find : $\Delta V$ across $2\mu f=?$

so we first find equivalent capacitance across $AB$.

so $\Rightarrow$ ( as $2\mu f$ and $3\mu F$ are in parallel )

now; $12\mu F,20\mu F$ and $5\mu F$ are in series so;

$\frac{1}{Ceq}=\frac{1}{12}+\frac{1}{20}+\frac{1}{5}=\frac{1}{3}\Rightarrow Ceq\underset{AB}{=}3\mu F$

Now using;

$Q=cv\Rightarrow Q=\left(3\mu F\right)\left(1500\right)=4500\mu C$

and $\begin{array}{c}\Delta U_{12\mu F}=\frac{Q}{C}=\frac{4500\mu C}{12\mu F}=375V\\ \Delta U_{20\mu F}=\frac{Q}{C}=\frac{4500\mu C}{20\mu F}=225V\end{array}\}\Delta U_{12\mu F}+\Delta U_{20\mu F}=600V$

hence $\Delta V_{PQ}=1500v-600v=900v\to$ total across $AB$ so $\left(B\right)$ option is correct