question_answer

In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state  

A) \[{{V}_{AB}}={{V}_{BC}}=100V\]    

B) \[{{V}_{AB}}=75,\,{{V}_{BC}}=25V\]

C) \[{{V}_{AB}}=25,\,{{V}_{BC}}=75V\]

D)             \[{{V}_{AB}}=\,{{V}_{BC}}=50V\]

Correct Answer: C

Solution :

\[{{C}_{eq}}=\frac{(3+3)\times (1+1)}{(3+3)+(1+1)}+1=\left( \frac{6\times 2}{6+2} \right)+1=\frac{5}{2}\mu F\] \[\therefore \,\,\,Q=C\times V=\frac{5}{2}\times 100=250\mu C\] Charge in \[6\mu F\]branch \[=VC=\left( \frac{6\times 2}{6+2} \right)100=150\mu C\] \[{{V}_{AB}}=\frac{150}{6}=25V\] and \[{{V}_{BC}}=100-{{V}_{AB}}=75V\]