Questions
Two capacitors of capacitances 3 F and 6 F are charged to a potential to 12 v each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be
detailed solution
Correct option is B
Charge on Ist condenser,Q1=C1V1=3×12=36μCcharge on 2nd condenser,Q2=C2V2=6×12=72μCCommon potential difference of each condenserV=Q2−Q1C1+C2=72−363+6=369=4 volt .Talk to our academic expert!
Similar Questions
Two identical metal plates are given positive charges Q1 and Q2 (<Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is
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