‘A’ and ‘B’ are two condensers of capacities 2 μF and 4μF They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :
24μC from A to B
8μ C from A to B
8μC from B to A
24μC from B to A
A
24μC from B to A
B
24μC from A to B
C
8μ C from A to B
D
8μC from B to A
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Solution
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QA=CAVA =2×10−6×12 QA=24×10−6
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