2. A 3.6 mH inductor with negligible resistance has a 1.0 A current through it. The current starts to increase at t = 0 s, creating a constant 5.0 mV voltage across the inductor. How much charge passes through the inductor between t = 0 s and t = 5.0 s?

To solve this problem it will be necessary to apply the concepts related to the electric potential in terms of the variation of the current and inductance. From this definition, we will start to find the load, which is dependent on the current as a function of time.

Here,

L = Inductance

Rate of change of current

If we take the equation and put the variation of the current as a function of time, in terms of the voltage in terms of the inductance we would have

The current as a function of time will be then,

The charge is the integral of the current in each variation of the time, then

Equation the terms we will have,

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

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg