A) 500 V
B) 400 V
C) 300 V
D) 200 V
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 \[C={{C}_{1}}+{{C}_{2}}\] \[\therefore \] \[q=VC={{C}_{1}}\,{{V}_{1}}+{{C}_{2}}\,{{V}_{2}}\] \[=V\left( {{C}_{1}}+{{C}_{2}} \right)\] \[\Rightarrow \] \[V=\frac{{{C}_{1}}\,{{V}_{1}}+{{C}_{2}}\,{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] Given, \[{{C}_{1}}=20\,\mu F\], \[V=500\,v,\] \[{{C}_{2}}=10\,\mu F\], \[{{V}_{2}}=200\,volt\] \[\therefore \] \[V=\frac{20\times 500\times {{10}^{-6}}+10\times 200\times {{10}^{-6}}}{\left( 20+10 \right)\times {{10}^{-6}}}\] \[=\frac{12000\times {{10}^{-6}}}{30\times {{10}^{-6}}}=400\,\text{volt}\] Note : Capacitors are combined in parallel when we require a large capacitance to a small potential.You need to login to perform this action.
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