A capacitor of 20\[\mu F\] capacity charged upto 500 V is connected in parallel with another capacitor of 10 aF which is charged upto 200 V. Their common potential is
A) 500V
B) 400V
C) 300V
D) 200V
Correct Answer:
B
Solution :
Let the charges on capacitors be \[{{q}_{1}},\,\,{{q}_{2}}\]then \[{{q}_{1}}={{C}_{1}}{{V}_{1}},\] \[{{q}_{2}}={{C}_{2}}{{V}_{2}}\] Total charge \[q={{q}_{1}}+{{q}_{2}}\] \[={{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}\] Let the equivalent potential be V and since capacitors are connected in parallel their equivalent capacitance is \[\therefore \] \[C={{C}_{1}}+{{C}_{2}}\] \[q=VC={{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}\] \[=V\,({{C}_{1}}+{{C}_{2}})\] \[\Rightarrow \] \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{1}}}\] Given, \[{{C}_{1}}=20\mu F,\] \[V=500\,v\,,\] \[{{C}_{2}}=10\mu F,\] \[{{V}_{2}}=200\,V\] \[\therefore \] \[V=\frac{20\times 500\times {{10}^{-6}}\times 200\times {{10}^{-6}}}{(20+10)\times {{10}^{-6}}}\] \[=\frac{12000\times {{10}^{-6}}}{30\times {{10}^{-6}}}=400\,V\]