Question

The figure shows a network of seven capacitors. If the charge on the 5μF capacitor is 10μC, the potential difference between the points A and C is given as $\dfrac{x}{3}$.

Answer 1

We should have knowledge about kirchhoff's law of capacitance for solving this circuit. Q=CV, we will deal with equivalent capacitance across two points. Charge conservation method should be kept in mind. We will find ${{Q}_{eq}}$, ${{C}_{eq}}$ and ${{V}_{AB}}$ and by using formula we will solve the question.

Formula used:
${{Q}_{eq}}={{C}_{eq}}{{V}_{AB}}$

Complete answer:

Firstly we will find potential across 5 capacitor
Let the potential across the $5\mu f$ capacitor be ${{v}_{5}}$.
Potential across $5\mu f$ is ${{v}_{5}}=\dfrac{{{q}_{5}}}{{{c}_{5}}}$
$=\dfrac{10}{5}=2$
As we can see that$\left( 3\mu f,4\mu f,5\mu f \right)$ are in parallel in below given diagram

So now we will find charge on $3\mu f$ and $4\mu f$ capacitor voltage across both the capacitance will be the same as both are in parallel i.e =2volts.
Charge on $3\mu f$ is ${{Q}_{3}}=3\times 2=6\mu C$
Charge on $4\mu f$ is ${{Q}_{4}}4\times 2=8\mu C$
Charge on $5\mu f$ is ${{Q}_{5}}=5\times 2=10\mu C$
Now we can see that ${{C}_{2}}$ is in series with $\left( 3\mu f,4\mu f,5\mu f \right)$so charge on ${{C}_{2}}$ and group $\left( 3\mu f,4\mu f,5\mu f \right)$are same.
So we can find the charge on ${{C}_{2}}$ is ${{Q}_{2}}=10+6+8=24\mu C$
And potential across it ${{V}_{2}}=\dfrac{24}{4}=6V$
Thus we will now find potential across point A and B
${{V}_{AB}}={{V}_{5}}+{{V}_{2}}=2+6=8V$
Now equivalent capacitance of lower branch of circuit is
${{C}_{eq}}=\dfrac{3\left( 4+2 \right)}{3+4+2}=2\mu f$ And;
Total equivalent charge${{Q}_{eq}}={{C}_{eq}}{{V}_{AB}}=2\left( 8 \right)=16\mu C$
$\therefore {{V}_{AC}}=\dfrac{{{Q}_{eq}}}{3}=\dfrac{16}{3}$
Thus on comparing it with $\dfrac{x}{3}$ we obtain x=16.
On solving the circuit by basic charge conservation method we obtain that x=16.

Note:
We have to apply charge conservation carefully properties of capacitor in series and parallel should be kept in mind
Formula to find the charge on capacitor is
q=CV
Where,
q=charge on capacitor
V= voltage applied across capacitor
C= Capacitance of capacitor
When voltage is applied across a parallel plate capacitor it produces positive charge on one plate and negative charge on another plate. Charge across the capacitor is directly proportional to voltage.