Question

Circuit of three capacitors is shown in the figure and initially switch is at position 1. Now it is shifted to position 2 . then :

• A. Initially charge on capacitor $C_1$ is $\frac{5}{4}\mu C$
• B. Charge flow to the battery when switch is shift from position 1 to position 2 is 2.5 $\mu$ C
• C. Potential energy of system of capacitors remains unchanged
• D. Heat produce in redistribution of charge is 12.5 $\mu$ J

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Initially charge on capacitor $C_1$ is $\frac{5}{4}\mu C$
B Charge flow to the battery when switch is shift from position 1 to position 2 is 2.5 $\mu$ C
C Potential energy of system of capacitors remains unchanged
D Heat produce in redistribution of charge is 12.5 $\mu$ J

Using the laws for parallel combination and series combination of capacitors, the effective capacitance of the system is $3/4\mu F$. The charge provided by battery is thus $15/4\mu C$.
In the first case as seen in the diagram the potential is distributed evenly in series and in proportion to the charges on capacitors when in parallel. Thus we get
potential across $C_1=5/4V\Rightarrow charge=5/4\mu C$.
charge on $C_2=15/4\mu C$.
In the second case,
$C_1$ and $C_2$ get interchanged.
charge on $C_1=15/4\mu C$
charge on $C_2=5/4\mu C$
Therefore, charge passed through the battery$=15/4-5/4=2.5\mu C$
Work done by battery $=qV=12.5\mu J$
This work is dissipated as heat since the potential energy of the capacitor system remains the same.