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A 600 pF capacitor is charged by a 200 V supply. Calculate the electrostatic energystored in it.It is then disconnectedfrom the supplyand is connected in parallel to another uncharged600 pF capacitor . What is the energystored in the combination? |
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Answer» Solution :Given : - `V_1`=2000 V `C_1=C_2`=600 PF When the 600 pF capacitor is CONNECTED to 200 V ,charge on thecapacitor is `Q=CV=600xx10^(-12)xx200=12xx10^(-8)` C Energy stored initially in the capacitor `U=1/2CV^2=1/2(600xx10^(-12))(200)^2` `U=12xx10^(-6)` j When the SECOND uncharged capacitor is connected in parallel to the first ,charge gets distributedamong them. Capacitors are having equal capacitances , hence Charges in each capacitor in the same =`Q_1=Q/2` `Q_1=6xx10^(-8)C` Now , energy stored in each capacitor `U_t=1/2 Q_1^2/C=1/2((6xx10^(-8))^2)/((600xx10^(-12)))=3xx10^(-6) J` `therefore` Total energy of the combination = `6xx10^(-6)`J |
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