question_answer

In the following diagram, the charge and potential difference across \[8\,\mu F\,\] capacitance will be respectively                     

A) \[320\,\mu \,C,\,\,40\,\,V\]          

B) \[420\,\mu \,C,\,\,50\,\,V\]

C) \[214\,\mu \,C,\,\,27\,\,V\]          

D) \[360\,\mu \,C,\,\,45\,\,V\]

Correct Answer: C

Solution :

Given circuit can be redrawn as follows capacitors, \[9\,\,\mu F,\,\,9\mu F\,\,and\,\,7\mu F\,\] are short circuited. So they are deleted. \[{{V}_{1}}+{{V}_{2}}=40\,\,V\] And \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{36}{18}\,\,=\,\,2\] \[{{V}_{1}}\,\,=\frac{80}{3}\,\,V\] \[{{V}_{2}}\,\,=\frac{40}{3}\,\,V\] Charge on \[8\,\,\mu F\] capacitor \[8\times \frac{80}{3}=213.3\,\mu F\approx \,\,214\,\mu F\]