A system of 2 capacitors of capacitance 2 mu | vedclass

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(C) The effective capacitance $C$ of two capacitors connected in series is given by $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the giv

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$8 \mu C$ and $24 \mu J$

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(C) The effective capacitance $C$ of two capacitors connected in series is given by $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the given values: $\frac{1}{C} = \frac{1}{2 \mu F} + \frac{1}{4 \mu F} = \frac{2+1}{4 \mu F} = \frac{3}{4 \mu F}$.Thus,$C = \frac{4}{3} \mu F = \frac{4}{3} 	imes 10^{-6} 	ext{ F}$.The electric charge $Q$ stored in the series combination is $Q = C V$.$Q = (\frac{4}{3} 	imes 10^{-6} 	ext{ F}) 	imes 6 	ext{ V} = 8 	imes 10^{-6} 	ext{ C} = 8 \mu C$.The energy $U$ stored in the system is $U = \frac{1}{2} C V^2$.$U = \frac{1}{2} 	imes (\frac{4}{3} 	imes 10^{-6} 	ext{ F}) 	imes (6 	ext{ V})^2 = \frac{1}{2} 	imes \frac{4}{3} 	imes 10^{-6} 	imes 36 = 24 	imes 10^{-6} 	ext{ J} = 24 \mu J$.

$A$ system of $2$ capacitors of capacitance $2 \mu F$ and $4 \mu F$ is connected in series across a potential difference of $6 \text{ V}$. The electric charge and energy stored in the system are

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