Question

In the circuit shown in figure, potential difference between A and B is
A. $30V$
B. $60V$
C. $10V$
D. $90V$

Answer 1

Hint First calculate the resultant capacitance of the circuit. Then using the appropriate formula, determine its charge. Considering the charge of two capacitors connected in series are the same, calculate the voltage difference across capacitors $C$ and $3C$.
Formula used
For series connection, ${C_{eq}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}$ and for parallel connection, ${C_{eq}} = {C_1} + {C_2}$
$q = CV$ where $q$ is the charge of the capacitor, $C$ is called the capacitance and $V$ is the potential difference across the capacitor.

Complete step by step answer
A capacitor is a system of conductors and dielectric that can store electric charge. It consists of two conductors containing equal and opposite charges and has a potential difference $V$ between them.
The potential difference between the conductors is proportional to the charge on the capacitor and is given by the relation $q = CV$ where $q$ is the charge on the positive conductor and $C$ is called the capacitance.
First let us calculate the equivalent capacitance of this circuit.
So,
$3 + \left( {\dfrac{3}{4}} \right) \\\ = \dfrac{{15}}{4} \\\$
This is the capacitance of the two branches in parallel to each other where $\dfrac{3}{4}$is the equivalent capacitance of capacitors $C$ and $3C$ connected in series.
This is in series connection with the capacitance $C$
So, equivalent capacitance of the circuit is
$\dfrac{{\dfrac{{15}}{4}}}{{1 + \dfrac{{15}}{4}}} =$ $\dfrac{{\dfrac{{15}}{4}}}{{\dfrac{{19}}{4}}} = \dfrac{{15}}{{19}}$
Using the formula $q = CV$we find,
$q = \dfrac{{15}}{{19}} \times 190 = 150C$
So, the charge of the resultant capacitance is $150C$
Now, as both capacitances $1$ and $\dfrac{{15}}{4}$ are in series, they must have the same charges.
Therefore, voltage across capacitance $\dfrac{{15}}{4}$ is
$V = \dfrac{q}{C} = \dfrac{{150}}{{\dfrac{{15}}{4}}} = 40V$
From the diagram we can show that the voltage across capacitances $C$ and $3C$ is $40V$.
Again since these two capacitors are in series so they must have the same charges.
Equivalent capacitance of $C$and $3C$is
${C_{eq}} = \dfrac{3}{4}$
$q = {C_{eq}}V \\\ \Rightarrow q = \dfrac{3}{4} \times 40 = 30C \\\$
Therefore the voltage across capacitance $3C$is $\dfrac{{30}}{3} = 10V$

Thus, the correct option is C.

Note It is important to note that two or more capacitors in series will always have the same charge across their plates. As the charge is equal and constant, the voltage drop across the capacitor is determined by the value of the capacitor only.