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Two condensers of capacity 0.3 \mu \mathrm{F} and 0.6 \mu \mathrm{F} respectively are connected in series. The combination is connected across a source of potential difference of 6 volt. The ratio of energy stored by the first to that by the second condenser will be------

 

Option: 1

9


Option: 2

2


Option: 3

6


Option: 4

4


Answers (1)

best_answer

\mathrm{Given, C_{1}=0.3 \mu \mathrm{f}\: and \: C_{2}=0.6 \mu \mathrm{F}}

since two capacitors are connected in series,
\mathrm{q_{1}=q_{2}=q}

Case I Energy in first capacitor is \mathrm{E_{1}=\frac{q^{2}}{2 C_{1}}}

Case II  Energy in second capacitor is \mathrm{E_{2}=\frac{2^{2}}{2 c_{2}}}

\mathrm{\therefore \quad \frac{E_{1}}{E_{2}}=\frac{q^{2} / 2 c_{1}}{q^{2} / 2 c_{2}} }
\mathrm{\frac{E_{1}}{E_{2}}=\frac{q^{2}}{2 c_{1}} \times \frac{2 c_{2}}{q^{2}} }
\mathrm{\frac{E_{1}}{E_{2}}=\frac{c_{2}}{c_{1}}=\frac{0.6 \mu \mathrm{F}}{0.3 \mu \mathrm{F}} }
\mathrm{\frac{E_{1}}{E_{2}}=\frac{6}{3}=2 }

\mathrm{E_{1}: E_{2}=2: 1}  Ans.

Posted by

rishi.raj

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