A4.0 μF capacitor is connected in series with a 2.0 MΩ resistor, and this combination is connected across an ideal 44.0 V DC battery. Initially the charge on the capacitor is zero. What is the current in the circuit when the capacitor has reached 80 % of its maximum voltage?

  i = 17.6 10⁻⁶ A

Explanation:

The charge of a capacitor in a series circuit is

           i =                     (1)

           R i = V₀  e^{ - \frac{t}{RC} }

            = e^{ - \frac{t}{RC} }

they tell us that the voltage is 80% bone

            = 0.80

we substitute

           0.80 = e^{ - \frac{t}{RC} }

           ln 0.80 =

           t = - RC ln 0.80

in the problem they give the value of R = 2.0 10⁶ Ω and C = 4.0 10⁻⁶ F

          t = - 2 10⁶  4 10⁻⁶ ln 0.8

          t = 1,785 s

we substitute in equation 1

         i =

         i = 22 10⁻⁶

         i = 17.6 10⁻⁶ A

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west