show all work Suppose R 240 ohm and C 140 muF in a chargin

show all work

Suppose R = 240 ohm and C = 140 muF in a (charging) RC series circuit with a 12 V power source. What is the maximum charge on the capacitor? After 3.2 time constants, the capacitor will reach what percent of its nmaximum charge? max xharge 0.002846_C 0.002668_C 0.001552_C 0.00168_C 0.00186_C 0.002222_C Percent 161.1% 56.69% 95.92% 122.9% 85.85% 157.8%

Solution

14.

The Maximum capacitor charge is given by the equation

Q_max = CV

where C is the given capacitance and V is the given voltage.

Using the data provided we get,

Q_max = 140*10^-6 *12 =0.00168 C.

Thus option d) is the correct choice.

15.

The capacitor discharge equation is given by

Q = Q_max*exp(-t/RC)

where Q_max is the maximum charge which was computed in 14., t is the time and the product of resistance and capacitance (RC) is known as the time constant of the circuit.

Using the data given and computed we get

Q = 0.00168*exp(-3.2) = 0.00006845

Therefore the percentage of the maximum charge that is there on the capacitor is

Q/Q_max *100= 4.07%

None of the choices given in the problem match the answer.