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


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)


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) = 3e^(-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

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The phasor diagram of the currents is attached.


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