- Home
- Standard 12
- Physics
1. Electric Charges and Fields
normal
A capacitor of capacitance $C_0$ is charged to a potential $V_0$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S$, the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by
A
$\frac{{{C_0}\left( {{V_0} - V} \right)}}{{{V_0}}}$
B
$\frac{{{C_0}\left( {V - {V_0}} \right)}}{{{V_0}}}$
C
$\frac{{{C_0}\left( {V + {V_0}} \right)}}{V}$
D
$\frac{{{C_0}\left( {{V_0} - V} \right)}}{V}$
Solution
Acconding to law of conservation of charge
$\mathrm{C}_{0} \mathrm{V}_{0}=\mathrm{C}_{0} \mathrm{V}+\mathrm{CV}$
$\therefore {\rm{C}} = \frac{{{{\rm{C}}_0}\left( {{{\rm{V}}_0} – {\rm{V}}} \right)}}{{\rm{V}}}$
Standard 12
Physics
Similar Questions
normal
