(C) Given,resultant capacity,$C_{eq} = 2 \mu F$. From the figure,all five capacitors are connected in series. The equivalent capacity of capacitors in

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(C) Given,resultant capacity,$C_{eq} = 2 \mu F$. From the figure,all five capacitors are connected in series. The equivalent capacity of capacitors in series is given by: $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}$ $\frac{1}{C_{eq}} = \frac{5}{C}$ Substituting the value of $C_{eq}$: $\frac{1}{2 \mu F} = \frac{5}{C}$ $C = 5 \times 2 \mu F = 10 \mu F$ Therefore,the capacity of each capacitor is $10 \mu F$.

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

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Five capacitors each of capacity $C$ are connected as shown in the figure. If their resultant capacity is $2 \mu F$,then the capacity of each capacitor is: (in $\mu F$)

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A
$2.5$
B
$2$
C
$10$
D
$5$

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Electrostatic Potential and Capacitance

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Equivalent Capacitance of Capacitor connected in Series and Parallel

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Five capacitors each of capacity $C$ are connected as shown in the figure. If their resultant capacity is $2 \mu F$,then the capacity of each capacitor is: (in $\mu F$)

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