"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?"

1.04x s

Explanation:

->The maximum current through the resistor is

= V/R = V/= V/R× = V/R

Voltage 'V'=50V

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

Therefore, =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

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 / 1.43

t= 1.04x s

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