Q. A capacitor of $ 10\text{ }\mu F $ charged upto 250 V is connected in parallel with another capacitor of 5aF charged upto 100 V. The common potential is
ManipalManipal 2013Electrostatic Potential and Capacitance
Solution:
The equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances. Let the charge on the capacitors be
$ {{q}_{1}} $ and $ {{q}_{2}} $
Then total charge
$ Q=Q+{{Q}_{2}} $ $ CV={{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}} $
Since, capacitors are connected in parallel equivalent capacitance is
$ C={{C}_{1}}+{{C}_{2}} $
$ V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}} $
Given, $ {{C}_{1}}=10\,\mu F,{{V}_{1}}=250V,{{C}_{2}}=5\,\mu F $
$ {{V}_{2}}=100V $
$ \therefore $ $ V=\frac{(10\times {{10}^{-6}}\times 250)+(5\times {{10}^{-6}}\times 100)}{(10\times {{10}^{-6}}+5\times {{10}^{-6}})} $
$ \Rightarrow $ $ V=\frac{3000\times {{10}^{-6}}}{15\times {{10}^{-6}}}=200V $
