(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{

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(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the given values $C_1 = 4\,\mu F$ and $C_2 = 6\,\mu F$:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\,\mu F$.The charge $Q$ on each capacitor in a series combination is the same and is given by $Q = C_{eq} \times V$.Given $V = 500\;V$,we have:$Q = 2.4\,\mu F \times 500\;V = 1200\;\mu C$.

Class 12 Physics Electric Potential and Capacitance Equivalent Capacitance of Capacitor connected in Series and Parallel

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Two capacitors of capacitance $4\,\mu F$ and $6\,\mu F$ are connected in series. $A$ potential difference of $500\;V$ is applied to the outer plates of the two-capacitor system. The charge on each capacitor is numerically:

A
$6000\;C$
B
$1200\;C$
C
$1200\;\mu C$
D
$6000\;\mu C$

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Electric Potential and Capacitance

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Equivalent Capacitance of Capacitor connected in Series and Parallel

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Two capacitors of capacitance $4\,\mu F$ and $6\,\mu F$ are connected in series. $A$ potential difference of $500\;V$ is applied to the outer plates of the two-capacitor system. The charge on each capacitor is numerically:

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