Question

The charge flowing in a wire is plotted in fig 1.24. Sketch the corresponding current. Will attach pictures

Answer 1

I actually have the solution, I just don't understand where it comes from..... here is the solution

Answer 2

Current i = \(\Large \frac{dq}{dt}\).
You just have to find the derivative of the 'q' plot to get the "i" plot.
derivative is same as slope.

Answer 3

so what is the q plot? Maybe this is where I am failing. Should I look at this as an transposed absolute value from zero to 4, and then a different absolute function from 4 to 8? Or should I view it as three lines? One from 0 to 2, one form 2 to 6, and a third from 6 to 8? Or am I just burned out from a long study session? lol

Answer 4

The first picture you posted, problem 1.7, is a plot of the charge q against time t.
From time t = 0 to t = 2, the charge increases from 0 to 50 Coulombs.
Slope = (50 - 0) / (2 - 0) = 25 Coulombs per second or 25 Amperes.

Calculate the slope using the same procedure for t between 2 and 6 seconds and then again for t between 6 and 8 seconds.

Answer 5

Ok so when they are asking me to sketch the corresponding current it doesn't have to be on the same graph.... that is where I was getting confused I think. Thank you for your help.

Answer 6

It will be a separate plot because the units along the y-axis are different.
For charge vs time plot, the units are Coulombs vs time.
For current vs time plot, the units are Amperes vs time.

You are welcome.

Answer 7

@ranga one other thing.... why is it in the second picture that they put a line connecting the graph from 25A to -25A? Why wouldn't they put discontinuities? I thought that for something to be continuous it had to be differentiable? It looks to me like at t=2, and t=6 the graph of Q(t) is not differentiable. Wouldn't that imply that there are discontinuities at t=2 and t=6 on the graph of i(t)?

Answer 8

And I was thinking why couldn't you include t=0, and t=8 on the graph of i(t)?

Answer 9

The plot of q vs t is continuous. It can be drawn without having to lift the pen from the paper (that is a simpler way to look at a continuous function). But q vs t plot is not differentiable at t = 2 and at t = 6. If you look a little to the left of t = 2, at t = \(2^-\), the plot has a positive slope and a little to the right of t = 2, at t = \(2^+\), the plot has a negative slope. Therefore at t = 2, the slope is undefined and it is not differentiable. Therefore, at t = 2 and t = 6, the current vs time plot will have a jump discontinuity.

Answer 10

So we have to consider three sub-intervals in the the current vs time plot. It has to be split into three intervals: (0,2); (2,6); and (6,8).

Answer 11

I agree but for some reason there solution shows it being continuous on the current verse time plot. Look at the second picture I included. And why on their piecewise function (again on the second picture) did they disclude t=0 and t=8?

Answer 12

This is how the plot should look like. Discontinuous at t = 2 and t = 6.

Answer 13

If you want you can put a small open circle at t = 0 and at t = 8 to indicate the end points are not included but that is not necessary.

Answer 14

(0,2) positive slope of +25
(2,6) negative slope of -25
(6,8) positive slope of +25

Answer 15

The reason the end points are not included in the current vs time plot is we do not know the behavior of the first plot slightly to the left of 0 and slightly to the right of 8. What if the plot changes direction at those points? Then the slope will be undefined. So we cannot presume the first plot is differentiable at t = 0 and at t = 8 seconds.