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

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.