question_answer

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

A)  200 V  

B)  300 V   

C)  400 V  

D)  500 V

Correct Answer: A

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, \[{{C}_{1}}=10\,\mu F,{{V}_{1}}=250V,{{C}_{2}}=5\,\mu F\] \[{{V}_{2}}=100V\] \[\therefore \] \[V=\frac{(10\times {{10}^{-6}}\times 250)+(5\times {{10}^{-6}}\times 100)}{(10\times {{10}^{-6}}+5\times {{10}^{-6}})}\] \[\Rightarrow \]\[V=\frac{3000\times {{10}^{-6}}}{15\times {{10}^{-6}}}=200V\]