Question

"An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has an internal resistance of 10.0 Ω and the open-circuit voltage across its terminals is 50.0 V. The leads have no appreciable resistance. At time t = 0, the switch is suddenly closed." "When does the maximum current occur?"

Answer 1

Answer:

1.04x$10^{-3}$ s

Explanation:

->The maximum current through the resistor is

$I_{max}$ = V/R = V/$Re^{-t/RC}$= V/R×$e^{0}$ = V/R

Voltage 'V'=50V

Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω

Therefore, $I_{max}$=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 $10^{-3}$

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 $10^{-3}$/ 1.43

t= 1.04x$10^{-3}$ s