Answer:
[tex]v = 250[1 - {e^{-6000t}}][/tex] mV
Explanation:
The voltage across a capacitor at a time t, is given by:
[tex]v(t) = \frac{1}{C} \int\limits^{t}_{t_0} {i(t)} \, dt + v(t_0)[/tex] ----------------(i)
Where;
v(t) = voltage at time t
[tex]t_{0}[/tex] = initial time
C = capacitance of the capacitor
i(t) = current through the capacitor at time t
v(t₀) = voltage at initial time.
From the question:
C = 2μF = 2 x 10⁻⁶F
i(t) = 3[tex]e^{-6000t}[/tex] mA
t₀ = 0
v(t₀ = 0) = 0
Substitute these values into equation (i) as follows;
[tex]v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt + v(0)[/tex]
[tex]v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt + 0[/tex]
[tex]v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt[/tex]
[tex]v = \frac{3}{2*10^{-6}} \int\limits^{t}_{0} {e^{-6000t}} \, dt[/tex] [Solve the integral]
[tex]v = \frac{3}{2*10^{-6}*(-6000)} {e^{-6000t}}|_0^t[/tex]
[tex]v = \frac{-3000}{12} {e^{-6000t}}|_0^t[/tex]
[tex]v = -250 {e^{-6000t}}|_0^t[/tex]
[tex]v = -250 {e^{-6000t}} - [-250 {e^{-6000(0)}][/tex]
[tex]v = -250 {e^{-6000t}} - [-250][/tex]
[tex]v = -250 {e^{-6000t}} + 250[/tex]
[tex]v = 250 -250 {e^{-6000t}}[/tex]
[tex]v = 250[1 - {e^{-6000t}}][/tex]
Therefore, the voltage across the capacitor is [tex]v = 250[1 - {e^{-6000t}}][/tex] mV
