Electrostatics Question 634
Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 $ \mu C $ . The value of $ C _{1} $ is [Kerala PMT 2005]
Options:
A) $ 2.5\mu F $
B) $ 15\mu F $
C) $ 1.5\mu F $
D) $ 0.1\mu F $
Show Answer
Answer:
Correct Answer: D
Solution:
The capacitance across A and B $ =\frac{C _{1}}{2}+C _{1}+C _{1}=\frac{3}{2}C _{1} $
As Q = CV, $ 1.5\mu C=\frac{5}{2}C _{1}\times 6 $
therefore $ C _{1}=\frac{1.5}{15}\times {{10}^{-6}} $
$ =0.1\times {{10}^{-6}}F=0.1\mu F. $
BETA
