(A) Let the two capacitors have capacitances $C_1$ and $C_2$.The equivalent capacitance in parallel is $C_p = C_1 + C_2$.The equivalent capacitance in

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(A) Let the two capacitors have capacitances $C_1$ and $C_2$.The equivalent capacitance in parallel is $C_p = C_1 + C_2$.The equivalent capacitance in series is $C_s = \frac{C_1 C_2}{C_1 + C_2}$.Given that $C_p = 4 C_s$,we substitute the expressions:$C_1 + C_2 = 4 \left( \frac{C_1 C_2}{C_1 + C_2} \right)$$(C_1 + C_2)^2 = 4 C_1 C_2$$C_1^2 + C_2^2 + 2 C_1 C_2 = 4 C_1 C_2$$C_1^2 + C_2^2 - 2 C_1 C_2 = 0$$(C_1 - C_2)^2 = 0$Therefore,$C_1 = C_2$.

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

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The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that:

A
Their capacities are equal
B
Their capacities are $1\,\mu F$ and $2\,\mu F$
C
Their capacities are $0.5\,\mu F$ and $1\,\mu F$
D
Their capacities are infinite

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Class 12

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Electric Potential and Capacitance

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Equivalent Capacitance of Capacitor connected in Series and Parallel

The equivalent capacitance between points $A$ and $B$ as shown in the figure is:

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If the initial charge on all the capacitors was zero,the work done by the battery in the circuit shown is ........... $mJ$.

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Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of point $B$ is $V_2$. The potential at point $D$ will be

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$A$ potential difference of $300\, V$ is applied to a combination of $2.0\,\mu F$ and $8.0\,\mu F$ capacitors connected in series. The charge on the $2.0\,\mu F$ capacitor is

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The resultant capacitance of the given circuit is

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The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that:

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The equivalent capacitance between points $A$ and $B$ as shown in the figure is:

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