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 $

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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 $



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