A capacitor of capacitance C = 6 μF is connected in series with a resistor R = 5 MΩ to a battery of emf ξ = 15 V. What is the value of the current in the circuit when the capacitor has reached 20% of its maximum charge?
Maximum Charge: Q = C x V = 90 μC
20% of 90 μC : 20%Q = 18 μC
Q(t) = ξC(1-1/e^(t/RC))
18 μC = 90 μC (1-1/e^(t/30))
t = 7 ms
I = dQ/dt = ξ/(R x e^(t/RC))
I = 15/(5000 x e^(7/30)) = 2.37x10^-6, or 2.4 μA