"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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14