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A 4muF capacitor is charged to a potential 10V using a battery. The battery is then removed and then this capacitor is connected parallel to an unchanged capacitor of capacity 6muF. What is the common potential? |
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Answer» Solution :DATA SUPPLIED, `C_(1)=4muF, V_(0)=10V, C_(2)=6muF` Common potential =V By the law of conservation of charges, `C_(1)V_(0)=C_(1)V+C_(2)V=(C_(1)+C_(2))V` `V=(C_(1)V_(0))/(C_(1)+C_(2))=(4 XX 10^(-6) xx 10)/((4+6) 10^(-6))=4V` |
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