Question

In the given circuit the charge passing through the battery is $48\mu C$ The potential difference of capacitor C is:

• A. $6\text{V}$
• B. $8\text{V}$
• C. $3\text{V}$
• D. $4.5\text{V}$

Answer 1

The correct option is A $6\text{V}$

Here, $3\mu F,3\mu Fand2\mu F$ and $Cand4\mu F$ are in parallel;

So, equivalent capacitance will be, $=3+3+2=8\mu F$ and $C+2\mu F$

and $8\mu Fand(C+2\mu F)$ are in series;

$C_{eq}=\frac{(C+4\mu F)\times 8\mu F}{C+4\mu F+8\mu F}$

The circuit can be redrawn as;

Since, $q=C_{eq}V$

Here, total charge be, $=48\mu C$

$48\mu C=$

$12\times \frac{(C+4\mu F)\times 8\mu F}{C+4\mu F+8\mu F}$

$4=\frac{8C+32}{C+12}$

$4C+48=8C+32$

$4C=16$

$C=4$

As $Cand4\mu F$ are in parallel, so potential will be same across these capacitors,

Potential across C, $V=\frac{Q}{C+4}=\frac{48}{4+4}=6V$

Hence, the correct answer is (A).