An RC circuit has a time constant of 2.10 s. At t = 0, the capacitor is fully charged. At what time will the energy stored in the capacitor be half of its initial value?
Given that:
U=q^2/2C
q(t)=V*C(1-e^-t/RC)

Question

An RC circuit has a time constant of 2.10 s. At t = 0, the capacitor is fully charged. At what time will the energy stored in the capacitor be half of its initial value?
Given that: 
U=q^2/2C 
q(t)=V*C(1-e^-t/RC)

anonymous · Answer

how is the capacitor fully charged at t=0, can you somehow verify that?

anonymous · Answer

actually i don't know and i kind of didnt understand what he meant by saying the capacitor is fully charged at t=0  :S