The equivalent capacitance of three capacitor | vedclass

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(D) Given that $C_1 + C_2 + C_3 = 12$ ....$(i)$$C_1 \cdot C_2 \cdot C_3 = 48$ ....$(ii)$$C_1 + C_2 = 6$ ....$(iii)$From equations $(i)$ and $(iii)$,we

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$4, 2, 6$

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(D) Given that $C_1 + C_2 + C_3 = 12$ ....$(i)$$C_1 \cdot C_2 \cdot C_3 = 48$ ....$(ii)$$C_1 + C_2 = 6$ ....$(iii)$From equations $(i)$ and $(iii)$,we get $C_3 = 12 - 6 = 6$ units.Substituting $C_3 = 6$ into equation $(ii)$,we get $C_1 \cdot C_2 \cdot 6 = 48$,which implies $C_1 \cdot C_2 = 8$.We know that $(C_1 - C_2)^2 = (C_1 + C_2)^2 - 4C_1C_2$.Substituting the values,$(C_1 - C_2)^2 = (6)^2 - 4(8) = 36 - 32 = 4$.Thus,$C_1 - C_2 = 2$ ....$(iv)$.Solving equations $(iii)$ and $(iv)$ by adding them: $2C_1 = 8 \implies C_1 = 4$.Substituting $C_1 = 4$ into $(iii)$,we get $4 + C_2 = 6 \implies C_2 = 2$.Therefore,the capacitances are $4, 2, 6$.

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