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A 10-μF capacitor in an LC circuit made entirely of superconducting materials ( R = 0 Ω ) is charged to 100 μC. Then a superconducting switch is closed. At t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. At a later time, Vab = +10V . At that time, Vdc is

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kunsoil

Answer:

Vdc=10V

Explanation:

in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,

if the loop remains closed for  a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

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