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
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(d) ${C_1}+{C_2}+{C_3} = 12$....$(i)$
${C_1}.{C_2}.{C_3} = 48$....$(ii)$
$C_1 + C_2 = 6$ ....$(iii)$
From equation $(i)$ and $(iii)$
$C_3 = 6$ ....$(iv)$
From equation $(ii)$ and $(iv)$ $C_1C_2 = 8$
Also ${({C_1} - {C_2})^2} = {({C_1} + {C_2})^2} - 4{C_1}{C_2}$
${({C_1} - {C_2})^2} = {(6)^2} - 4 \times 8 = 4$
$==>$ $ C_1 -C_2 = 2$.....$(v)$
On solving $(iii)$ and $(v)$ $C_1 = 4,\,\, C_2 = 2 $
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