Question

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answer 1

Answer:

$v = 250[1 - {e^{-6000t}}]$ mV

Explanation:

The voltage across a capacitor at a time t, is given by:

$v(t) = \frac{1}{C} \int\limits^{t}_{t_0} {i(t)} \, dt + v(t_0)$                 ----------------(i)

Where;

v(t) = voltage at time t

$t_{0}$ = 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$e^{-6000t}$ mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

$v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt + v(0)$    

$v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt + 0$

$v = \frac{1}{2*10^{-6}} \int\limits^{t}_{0} {3e^{-6000t}} \, dt$            

$v = \frac{3}{2*10^{-6}} \int\limits^{t}_{0} {e^{-6000t}} \, dt$             [Solve the integral]

$v = \frac{3}{2*10^{-6}*(-6000)} {e^{-6000t}}|_0^t$

$v = \frac{-3000}{12} {e^{-6000t}}|_0^t$

$v = -250 {e^{-6000t}}|_0^t$

$v = -250 {e^{-6000t}} - [-250 {e^{-6000(0)}]$

$v = -250 {e^{-6000t}} - [-250]$

$v = -250 {e^{-6000t}} + 250$

$v = 250 -250 {e^{-6000t}}$

$v = 250[1 - {e^{-6000t}}]$

Therefore, the voltage across the capacitor is $v = 250[1 - {e^{-6000t}}]$ mV