Answer:
i = 17.6 10⁻⁶ A
Explanation:
The charge of a capacitor in a series circuit is
i = [tex]\frac{v_o}{R}[/tex] [tex]e^{ - \frac{t}{RC} }[/tex] (1)
R i = V₀ e^{ - \frac{t}{RC} }
[tex]\frac{V}{V_o}[/tex] = e^{ - \frac{t}{RC} }
they tell us that the voltage is 80% bone
[tex]\frac{V}{V_o}[/tex] = 0.80
we substitute
0.80 = e^{ - \frac{t}{RC} }
ln 0.80 = [tex]- \frac{t}{RC}[/tex]
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 = [tex]\frac{44.0}{2 \ 10^6}[/tex] [tex]e^{- \frac{1.785}{2 \ 4 } }[/tex]
i = 22 10⁻⁶ [tex]e^{-0.2231}[/tex]
i = 17.6 10⁻⁶ A
