Answer:
[tex]4.71\times 10^{-3}A[/tex]
Explanation:
[tex]Q_{max}[/tex] = Maximum charge stored by capacitor = 6 μC = 6 x 10⁻⁶ C
[tex]t[/tex] = time taken for charge on the capacitor to become zero = 2 ms = 2 x 10⁻³ s
Time period is given as
[tex]T = 4t[/tex]
[tex]T = 4(2\times 10^{-3})[/tex]
[tex]T = 8\times 10^{-3} s[/tex]
Angular frequency is given as
[tex]w = \frac{2\pi }{T}[/tex]
[tex]w = \frac{2(3.14) }{8\times 10^{-3}}[/tex]
[tex]w[/tex] =785 rad/s
Charge at any time is given as
[tex]Q(t) = Q_{max}Coswt[/tex]
Taking derivative both side relative to "t"
[tex]\frac{\mathrm{d}Q(t) }{\mathrm{d} t} = \frac{\mathrm{d}(Q_{max}Coswt) }{\mathrm{d} t}[/tex]
[tex]i(t)= -Q_{max} w Sinwt[/tex]
Amplitude of the current is given as
[tex]i_{max}= Q_{max} w[/tex]
[tex]i_{max}= (6\times 10^{-6}) (785)[/tex]
[tex]i_{max}= 4.71\times 10^{-3}A[/tex]
