Electrostatics Question 560
In the following circuit, the resultant capacitance between $ A $ and $ B $ is 1mF. Then value of $ C $ is [IIT 1977]
Options:
A) $ \frac{32}{11}\mu F $
B) $ \frac{11}{32}\mu F $
C) $ \frac{23}{32}\mu F $
D) $ \frac{32}{23}\mu F $
Show Answer
Answer:
Correct Answer: D
Solution:
12 mF and 6mF are in series and again are in parallel with 4mF.
Therefore, resultant of these three will be $ =\frac{12\times 6}{12+6}+4=4+4=8\mu F $
This equivalent system is in series with 1 µF. Its equivalent capacitance $ =\frac{8\times 1}{8+1}=\frac{8}{9}\mu F $ ….(i)
Equivalent of 8mF, 2mF and 2mF $ =\frac{4\times 8}{4+8}=\frac{32}{12}=\frac{8}{3}\mu F $ …..(ii) (i) and (ii) are in parallel and are in series with C $ \frac{8}{9}+\frac{8}{3}=\frac{32}{9} $ and $ C _{eq}=1=\frac{\frac{32}{9}\times C}{\frac{32}{9}+C} $
therefore $ C=\frac{32}{23}\mu F $
BETA
