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Suppose the initial charge on the capacitor in Exercise 7 is 6mC. What is the total energy at later time ? |
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Answer» Solution :Given, charge on the capacitor `Q=6mC = 6XX10^(-3)C` `C=30mu F=30xx10^(-6)F` Energy stored in the circuit `E =(Q^(2))/(2C)=((6xx10^(-3))^(2))/(2xx30xx10^(-2))=(36)/(60)=0.6J` After some time, the energy is shared between C and L, but the total energy remains CONSTANT. So, we assume that there is no loss of energy. |
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