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 $



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