The effective capacitances of two capacitors | vedclass

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(C) Let the capacitances of the two capacitors be $C_1$ and $C_2$.When connected in series,the effective capacitance is given by $\frac{C_1 C_2}{C_1 +

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$12\,\mu F, 4\,\mu F$

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(C) Let the capacitances of the two capacitors be $C_1$ and $C_2$.When connected in series,the effective capacitance is given by $\frac{C_1 C_2}{C_1 + C_2} = 3\,\mu F$. (Equation $1$)When connected in parallel,the effective capacitance is given by $C_1 + C_2 = 16\,\mu F$. (Equation $2$)Substituting Equation $2$ into Equation $1$:$C_1 C_2 = 3 	imes 16 = 48$.We have the sum $C_1 + C_2 = 16$ and the product $C_1 C_2 = 48$. These are the roots of the quadratic equation $x^2 - 16x + 48 = 0$.Solving the quadratic equation:$(x - 12)(x - 4) = 0$.Thus,$x = 12$ or $x = 4$.Therefore,the capacitances are $12\,\mu F$ and $4\,\mu F$.

The effective capacitances of two capacitors are $3\,\mu F$ and $16\,\mu F$,when they are connected in series and parallel respectively. The capacitances of the two capacitors are:

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