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NCERT Class XII Chapter
Alternating Current
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Question : 8 of 26
Marks: +1, -0
Suppose the initial charge on the capacitor given in question 7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Solution:  
Initial energy on capacitor, UEU_E = q022C\frac{q_0^2}{2C} = (6×10−3)22×30×10−6\frac{(6 \times 10^{-3})^2}{2 \times 30 \times 10^{-6}} = 0.6 J
Any time total energy in the circuit is constant, hence energy later is 0.6 J.
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