question_answer

A capacitor of 20 \[\mu \]F charged to 500 V is connected in parallel with another capacitor of 10 \[\mu \]F charged to 200 V. The common potential will be:

A) 400 V                                    

B) 300 V

C) 200 V                                    

D) 100 V

Correct Answer: A

Solution :

The capacitance (C) of a conductor is defined as the ratio of charge (q) given to the rise in the potential (V) of the conductor\[C=\frac{q}{V}\]Equivalent capacitance   of the capacitors joined in parallel is \[C={{C}_{1}}+{{C}_{2}}\] The common potential is given by \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] Putting the numerical values, we get \[V=\frac{20\times {{10}^{-6}}\times 500+10\times {{10}^{-6}}\times 200}{(20+10)\times {{10}^{-6}}}\] \[V=\frac{12000}{30}=400\,\,V\]