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Two capacitors of capacitance C(1) = 2 m...

Two capacitors of capacitance `C_(1) = 2 muF` and `C_(2) = 8 mu F` are connected in series and the resulting combination is connected across 300 volts. Calculate the charge, potential difference and energy stored in the capacitor separately.

Text Solution

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If C is the equivalent capacitance
then `(1)/(C ) = (1)/ (C_(1)) +(1)/(C_(2))=(1)/(2)+(1)/(8)=(5)/(8)`
`:. C = (8)/(5) = 1.6 muF `
Charge q =CV = `1.6xx10^(-6)xx300=4.8xx10^(-4)C`
Potential across `C_(1)=V_(1)=(q)/(C_(1))=240 ` volt
Potential across `C_(2)=V_(2)=(q)/(C_(2))=60` volt
Energyn stored in `C_(1)=(1)/(2) C_(1)V_(1)^(2)=5.76xx10^(-2)` joule
Energy stored in `C_(2)= (1)/(2)C_(2)V_(2)^(2)=1.44xx10^(-2)` joule
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