Answer:
a) 0.048A
b) 0.18µC
c) 1.85A
Explanation:
The discharged current of the capacitor as a function of time is given by:
[tex]i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\[/tex]
[tex]\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s[/tex]
a)
[tex]i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A[/tex]
b)
[tex]q=q_o*e^{-\frac{t}{\tau}}[/tex]
[tex]q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC[/tex]
c) the maximum current occurs when t=0
[tex]i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A[/tex]
