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

`v = 250[1 - {e^(-6000t)}]` mV

Step-by-step explanation:

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

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

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

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

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

`v = (3)/(2*10^(-6)*(-6000)) {e^(-6000t)}|_0^t`

`v = (-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