Question

The charge on the $6\mu F$ capacitor in the circuit shown is

Answer 1

Hint:
To solve the above circuit we can see that capacitors $12\mu F$ and $6\mu F$ are in parallel so we just need to calculate C effectively from a given circuit and, according to parallel or series combination we can distribute charges and voltage on each capacitor.

Formula used:
 For parallel combination of capacitors, its equivalent capacitor is${{C}_{1}}+{{C}_{2}}$
For series combination, its equivalent capacitor is $\dfrac{1}{C}=\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{1}}}$

Complete Step by step solution:
Capacitor $12\mu F$ and $6\mu F$ are in parallel so effective capacitance between them is given by,
${{C}_{1}}+{{C}_{2}}$, which is equivalent to 12+6 =$18\mu F$.

Now, capacitors of capacitance $18\mu F$and $9\mu F$ are in series so, the effective capacitance C in series is given by,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{C2}=\dfrac{1}{18}+\dfrac{1}{9}=\dfrac{3}{18}$

So,$C=\dfrac{18}{3}=6\mu F$

Now, the effective capacitance between A and B is $6\mu F$so,

Total charge $Q$on circuit will be given by, $Q=CV$
Here, $Q$(total charge), $C$(effective capacitance between two point) and $V$(total voltage)

So,$Q=6\times 90=540$micro coulomb.

$\therefore $When capacitors in series charges remains same on capacitors and for in parallel voltage remains same.

Capacitors $18\mu F$ and $9\mu F$ are in series so charges remain the same, but potential varies from one capacitor to other so.

We know that, $V=Q/C$, so for $18\mu F$ voltage will be $V=\dfrac{540}{18}=30V$ since charge 540micro coulomb is the same for both capacitors.

Now, when we look in further circuit our $6\mu F$ and $12\mu F$ are in parallel so voltage will remains same which is equal to $30V$ and we need to calculate charge on $6\mu F$ capacitor,
$Q=CV=6\times 30=180$micro coulomb

So,$Q=180\mu C$ option (3) is correct.

Note:
When we need to find charge or potential on a capacitor in any circuit first we need to simplify the circuit to the extent till when we can calculate effective charge charge Q between point A and B and then by applying distributive properties of charge and voltage in series as well as parallel combinations of capacitors.