Electrostatics Question 647
Question: The equivalent capacitance of three capacitors of Capacitance $ C _{1}:C _{2} $ and $ C _{3} $ are connected in parallel is 12 units and product$ C _{1}.C _{2}.C _{3}=48 $ . When the capacitors $ C _{1} $ and $ C _{2} $ are connected in parallel, the equivalent capacitance is 6 units. Then the capacitance are
Options:
A) 2, 3, 7
B) 1.5, 2.5, 8
C) 1, 5, 6
D) 4, 2, 6
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Answer:
Correct Answer: D
Solution:
[d] $ C _{1}+C _{2}+C _{3}=12 $ … (i) $ C _{1}C _{2}C _{3}=48 $ … (ii) $ C _{1}+C _{2}=6 $ … (iii)
From equations (i) and (iii) $ C _{3}=6 $ … (iv)
From equations (ii) and (iv) $ C _{1}C _{2}=8 $ Also $ {{(C _{1}-C _{2})}^{2}}={{(C _{1}+C _{2})}^{2}}-4C _{1}C _{2} $
$ {{(C _{1}-C _{2})}^{2}}={{(6)}^{2}}-4\times 8=4 $
$ \Rightarrow C _{1}-C _{2}=2 $ … (v)
On solving (iii) and (v) $ C _{1}=4 $ , $ C _{2}=2 $
