(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

Seven capacitors,each of capacitance $2\,\mu F$,are to be connected to obtain a total equivalent capacitance of $10/11\,\mu F$. Which of the following combinations is possible?

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Two capacitors each having a capacitance $2 \times 10^{-6} \ F$ and a breakdown voltage $5000 \ V$,are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

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The effective capacitance between $A$ and $B$ in the given figure is:

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Four capacitors are connected as shown in the figure. The equivalent capacitance between the points $P$ and $Q$ is ....... $\mu F$.

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Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

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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:

Similar Questions

Seven capacitors,each of capacitance $2\,\mu F$,are to be connected to obtain a total equivalent capacitance of $10/11\,\mu F$. Which of the following combinations is possible?

Two capacitors each having a capacitance $2 \times 10^{-6} \ F$ and a breakdown voltage $5000 \ V$,are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

The effective capacitance between $A$ and $B$ in the given figure is:

Four capacitors are connected as shown in the figure. The equivalent capacitance between the points $P$ and $Q$ is ....... $\mu F$.

Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

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Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$