Electrostatics Question 592
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 [KCET 1999]
Options:
A) 2, 3, 7
B) 1.5, 2.5, 8
C) 1, 5, 6
D) 4, 2, 6
Show Answer
Answer:
Correct Answer: D
Solution:
$ {C _{\text{1}}}+{C _{\text{2}}}+{C _{\text{3}}}=\text{12} $ ….(i)
$ {C _{\text{1}}}{C _{\text{2}}}{C _{\text{3}}}=\text{ 48} $ ….(ii)
$ {C _{\text{1}}}+{C _{\text{2}}}=\text{ 6} $ ….(iii)
From equation (i) and (iii) $ {C _{\text{3}}}=\text{ 6} $ ….(iv)
From equation (ii) and (iv) $ {C _{\text{1}}}{C _{\text{2}}}=\text{ 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 $
therefore $ {C _{\text{1}}}{C _{\text{2}}}=\text{ 2} $ …..(v)
On solving (iii) and (v) $ {C _{\text{1}}}=\text{ 4},~{C _{\text{2}}}=\text{ 2} $
