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Net capacitance of three identical capacitors placed in series combination is \alpha \mu F. The three capacitors then are rearranged and placed in parallel. Find the ratio of energy stored between the two configurations if they are both connected to the same source of battery.

Option: 1

1: 9


Option: 2

1: 4


Option: 3

9: 1


Option: 4

4: 1


Answers (1)

best_answer

For series combination, total capacitance is given as C_s=\frac{C}{n} .

For parallel, it is C_p=\mathrm{nC}

\frac{C_p}{C_s}=\frac{n C}{\frac{C}{n}}=C_p=n^2 C_s

For same voltage, energy stored in capacitor is: 

U=\frac{1}{2} C V^2=U \propto C

C_s=\alpha \mu F, n=3

According to problem, 

\begin{gathered} C=n C_s=3 \alpha \mu F \\ C_p=n C=3 \times 3 \alpha \mu F=9 \alpha \mu F \end{gathered}

For same voltage: 

\begin{aligned} & \frac{U_s}{U_p}=\frac{C_s}{C_p}=\frac{1}{n^2}=\frac{(1 \alpha \mu F)^2}{(3 \alpha \mu F)^2} \\ & \frac{U_s}{U_p}=\frac{1}{9} \text { or } U_s: U_p=1: 9 \end{aligned}

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Deependra Verma

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