Question

A. What will be the potential difference between the points A and B in the diagram when the switch S will be open?
B. Which of the points will be at the higher potential?
C. What will be the final potential of the point B when the switch S is closed?
D. How much will be the charge on each capacitor variation when S will be closed?

Answer 1

Hint: In the steady state, no current will be flowing through the circuit when S will be opened. The charge flowing can be found by taking the product of the capacitance and the potential at this point. Therefore the potential at A will be $18V$, and that at B will be zero. This will help you in answering this question.

Complete answer:
A. in the steady state, no current will be flowing through the circuit when S will be opened. Therefore the potential at A will be $18V$, and that at B will be zero. Therefore we can write that,
${{V}_{A}}-{{V}_{B}}=18-0=18V$
B. it is perfectly clear that the potential at A will be the higher one.
C. When S will be closed, at last in the steady state, the current $I$ will flow as represented in the diagram. That is we can write that,
$I=\dfrac{18}{6+3}=2A$
The voltage will be six times this value of current. That is,
${{V}_{1}}=6I=6\times 2=12V$
The voltage at the point B can be shown as,
${{V}_{B}}=18-{{V}_{1}}$
Substituting the value in it will be given as,
${{V}_{B}}=18-12=6V$
D. the charge flowing can be found by taking the product of the capacitance and the potential at this point. That is we can write that,
${{q}_{1}}={{C}_{1}}{{V}_{1}}$
Substituting the values in the equation can be shown as,
${{q}_{1}}={{C}_{1}}{{V}_{1}}=6\times 12=72\mu C$
The sum of the potential will be,
${{V}_{1}}+{{V}_{2}}=18$
From this we can write that,
${{V}_{2}}=6V$
Therefore the charge in this will be given as,
${{q}_{2}}={{C}_{2}}{{V}_{2}}$
Substituting the values in the equation will give,
${{q}_{2}}=3\times 6=18\mu C$
Before closing the switch S, the potential on each of the capacitor will be $18V$ and the charge will be
$\begin{align}
  & {{q}_{10}}=18\times 6=108\mu C \\
 & {{q}_{20}}=3\times 18=54\mu C \\
\end{align}$
Hence the variation in the charge will be given as,
$\begin{align}
  & {{q}_{1}}-{{q}_{10}}=-36\mu C \\
 & {{q}_{2}}-{{q}_{20}}=-36\mu C \\
\end{align}$
Therefore the answer for the question has been found.

Note:
The capacitance can be defined as the ratio of the amount of the electric charge stored on a conductor to the change in electric potential. There are two types of capacitance. They are self-capacitance and mutual capacitance. Any material which can be electrically charged which is showing the self-capacitance.