Electrostatics Question 274
Question: Four metallic plates each with a surface area of one side A are placed at a distance d from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $ a $ and $ b $ is
Options:
A) $ V _{AB}=V _{BC}=100V $
B) $ V _{AB}=75V,\ V _{BC}=25\ V $
C) $ V _{AB}=25V,\ V _{BC}=75\ V $
D) $ V _{AB}=V _{BC}=50\ V $
Show Answer
Answer:
Correct Answer: C
Solution:
$ C _{eq}=\frac{(3+3)\times (1+1)}{(3+3)+(1+1)}+1=( \frac{6\times 2}{6+2} )+1 $
$ =\frac{5}{2}\mu F $ $ Q=C\times V=\frac{5}{2}\times 100=250\mu C $
Charge in $ 6\mu F $ branch $ =VC=( \frac{6\times 2}{6+2} )100 $
$ =150\mu C $
$ V _{AB}=\frac{150}{6}=25V $ and $ V _{BC}=100-V _{AB}=75V $
