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When two capacitors are connected in series and connected across 4kV line, the energy stored in the system is 8 J. the same capacitors, if connected in parallel across the same line, the energy stored in 36 J. find the individual capacitances. |
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Answer» SOLUTION :`E=(1)/(2)CV^(2)` finding `C_(1)+C_(2)=4.5xx10^(-6)F` finding `C_(1)+C_(2)=4.5xx10^(-12)` Final answer `C_(1)3muF` and `C_(2)=1.5muF` Detailed `E_(S)=(1)/(2)C_(S)V^(2)=(1)/(2)[(C_(1)C_(2))/(C_(1)+C_(2))]V^(2)` . . (1) `E_(p)=(1)/(2)C_(p)V^(2)=(1)/(2)[C_(1)+C_(2)]V^(2)`. . (2) `8=(1)/(2)[(C_(1)C_(2))/(C_(1)+C_(2))]xx(4xx10^(3))^(2)`. . (4) `36=(1)/(2)[C_(1)+C_(2)]xx(4xx10^(3))`. . (5) on SOLVING equation (4) and (5) we get `C_(1)=3muF,C_(2)=1.5muF`. |
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