Question

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?

Answer 1

i = 17.6 10⁻⁶ A

Step-by-step explanation:

The charge of a capacitor in a series circuit is

i =
`(v_o)/(R)`
`e^{ - (t)/(RC) }` (1)

R i = V₀ e^{ - \frac{t}{RC} }

`(V)/(V_o)` = e^{ - \frac{t}{RC} }

they tell us that the voltage is 80% bone

`(V)/(V_o)` = 0.80

we substitute

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

ln 0.80 =
`- (t)/(RC)`

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 =
`(44.0)/(2 \ 10^6)`
`e^{- (1.785)/(2 \ 4 ) }`

i = 22 10⁻⁶
`e^(-0.2231)`

i = 17.6 10⁻⁶ A