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Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially ? What is the total energy at later time? |
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Answer» Solution :Here, `q_(m) = 6mC = 6 xx 10^(-3) C` `THEREFORE` TOTAL energy stored in the circuit initially = total energy stored at any LATER time `= U = 1/2 .q_(m)^(2)/C = ((6 xx 10^(-3))^(2)/(2 xx 30 xx 10^(-6))) = 0.6 J` |
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