Answer:
1.04x[tex]10^{-3}[/tex] s
Explanation:
->The maximum current through the resistor is
[tex]I_{max}[/tex] = V/R = V/[tex]Re^{-t/RC}[/tex]= V/R×[tex]e^{0}[/tex] = V/R
Voltage 'V'=50V
Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω
Therefore, [tex]I_{max}[/tex]=50/35=> 1.43 A
->The maximum charge can be determined by
Q = CV
where,
Capacitance of the capacitor 'C' = 30.0µF = 30×10-⁶F
Therefore,
Q=30×10-⁶ x 50=>1.5 x [tex]10^{-3}[/tex]
In order to find that when does the maximum current occur, the time taken given the quantity of charge and the electric current is:
t= Q / I=> 1.5 x [tex]10^{-3}[/tex]/ 1.43
t= 1.04x[tex]10^{-3}[/tex] s
