Question

show that 50%of energy is lost when a capacitor is connected to a cell

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Answer 1

Answer:

Suppose the capacitance of the capacitor is C and the emf of the battery is V. The charge given to the capacitor is Q = CV. The work done by the battery is W = QV. The battery supplies this energy. The battery supplies this energy. The energy stored in the capacitor is U = (1)/(2)CV^2 = (1)/(2) QV. The remaining energy QV - (1)/(2) QV = (1)/(2) QV is lost as heat. thus, half energy supplied by the battery is lost as heat.

Explanation:

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