Question

Find the sum of final charges on ${C_2}$ and ${C_3}$ if a capacitor ${C_1} = 1.0\mu F$ is charged up to a voltage $V = 60V$ by connecting it to a battery $B$ through switch (1) and now, ${C_1}$ is disconnected from a battery and connected to a circuit consisting of two uncharged capacitors ${C_2} = 3.0\mu F$ and ${C_3} = 6.0\mu F$ through a switch (2) as shown in the figure.

A. $40\mu C$
B. $36\mu C$
C. $20\mu C$
D. $54\mu C$

Answer 1

Hint First of all, calculate the equivalent capacitance between ${C_2}$ and ${C_3}$. Then, calculate the charge passing from ${C_1}$ by using the formula of charge, $Q = CV$ (where, $Q$ is the charge, $C$ is the capacitance and $V$ is the potential difference across the capacitor).
Next, evaluate the potential difference between equivalent capacitors and ${C_1}$.
Then, calculate the charge across the equivalent capacitance.



Complete step-by-step solution:Let $C'$ be the equivalent capacitance of ${C_2}$ and ${C_3}$.
As shown in figure, ${C_2}$ and ${C_3}$ are in series so, to calculate equivalent resistance, we have to use

$
  \dfrac{1}{{C'}} = \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}} \\
   \Rightarrow \dfrac{1}{{C'}} = \dfrac{{{C_2} + {C_3}}}{{{C_2}{C_3}}} \\
  \therefore C' = \dfrac{{{C_2}{C_3}}}{{{C_2} + {C_3}}} \\
 $
Now, putting the values of ${C_2}$ and ${C_3}$ in the above expression –
$
  C' = \dfrac{{3 \times 6}}{{3 + 6}} \\
  C' = \dfrac{{18}}{9} = 2\mu F \\
 $
Now, calculating the charge across the capacitor, ${C_1} = 1.0\mu F$
$\because Q = CV$
According to the question, the capacitance, ${C_1}$ is charged up to a voltage $V = 60V$
$
  \therefore {Q_1} = 1 \times 60 \\
  {Q_1} = 60\mu C \\
 $
The potential difference between ${C_1}$ and $C'$ combined.
$V' = \dfrac{{{Q_1}}}{{{C_1} + C'}}$
Putting the values of capacitance and charge of ${C_1}$ in the above expression –
$
  V' = \dfrac{{60}}{{1 + 2}} \\
   \Rightarrow V' = \dfrac{{60}}{3} \\
  \therefore V' = 20V \\
 $
Now, calculating the charge of ${C_2}$ and ${C_3}$ system –
${Q_{23}} = C'V'$
Putting the values of $C'$ and $V'$ on the above expression –
${Q_{23}} = 2 \times 20 = 40\mu C$
Therefore, the charge on capacitors ${C_2}$ and ${C_3}$ is $40\mu C$.
Hence, option (A) is the correct answer.



Note:- The device which is used to store electrical energy is called a capacitor. The capacity of the capacitor to store the electricity is called Capacitance. The S.I unit of capacitance is Farad which is denoted by $F.$
The dimensional formula of capacitor is ${M^{ - 1}}{L^{ - 2}}{I^2}{T^4}.$