A) \[1500\,\,V\]
B) \[1000\,\,V\]
C) \[500\,\,V\]
D) \[400\,\,V\]
Correct Answer: C
Solution :
The equivalent capacity of\[10\,\,\mu F\]and\[10\mu F\]capacitors connected in series \[{{C}_{1}}=\frac{10\times 10}{10+10}=5\mu F\] Now equivalent capacitance between\[B\]and\[C\]which are connected in parallel \[{{C}_{2}}=5+10=15\mu F\] Total capacitance of the circuit between\[A\]and\[C\]when\[{{C}_{2}}\] and\[5\mu F\]are connected in series. \[{{C}_{3}}=\frac{5\times 15}{5+15}=\frac{15}{4}\mu F\] \[\therefore \] \[Q={{C}_{3}}V=\frac{15}{4}\times 2000=7500\mu C\] Potential difference across\[A\]and\[B\] \[{{V}_{AB}}=\frac{7500}{5}\] \[i.e.,\] \[{{V}_{A}}-{{V}_{B}}=1500\,\,V\] \[\therefore \]\[2000-{{V}_{B}}=1500\]or\[{{V}_{B}}=500\,\,V\]
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