charge vs current

[logistic_guy]

The charge entering a certain element is shown in the figure. Find the current at:

\(\displaystyle \bold{(a)} \ t = 1 \ \text{ms}\)
\(\displaystyle \bold{(b)} \ t = 6 \ \text{ms}\)
\(\displaystyle \bold{(c)} \ t = 10 \ \text{ms}\)

 

[khansaheb]

The charge entering a certain element is shown in the figure. Find the current at:

\(\displaystyle \bold{(a)} \ t = 1 \ \text{ms}\)
\(\displaystyle \bold{(b)} \ t = 6 \ \text{ms}\)
\(\displaystyle \bold{(c)} \ t = 10 \ \text{ms}\)

View attachment 39237

show us your effort/s to solve this problem.

 

[logistic_guy]

\(\displaystyle \bold{(a)}\)

The current is defined as:

\(\displaystyle I = \frac{dq}{dt}\)

So the slope of the charge at any moment of time is the current.

From the figure, we see that the charge has the same slope between \(\displaystyle t = 0 \ \text{ms}\) and \(\displaystyle t = 2 \ \text{ms}\).

Then, the current at \(\displaystyle t = 1 \ \text{ms}\) is:

\(\displaystyle I = \frac{30 - 0}{2 - 0} = \frac{30}{2} = 15 \ \frac{\text{mC}}{\text{ms}} = 15 \ \text{A}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

As we said before, \(\displaystyle I = \frac{dq}{dt}\).

Since the figure shows a constant value of \(\displaystyle q\) between \(\displaystyle t = 2 \ \text{ms}\) and \(\displaystyle t = 8 \ \text{ms}\),

\(\displaystyle I = 0\) at \(\displaystyle t = 6 \ \text{ms}\)

 

[logistic_guy]

\(\displaystyle \bold{(c)}\)

We can see in the figure that the slope of the charge \(\displaystyle q\) is the same between \(\displaystyle t = 8 \ \text{ms}\) and \(\displaystyle t = 12 \ \text{ms}\).

Therefore, the current (the slope) at \(\displaystyle t = 10 \ \text{ms}\) is:

\(\displaystyle I = \frac{30 - 0}{8 - 12} = \frac{30}{-4} = -7.5 \ \text{A}\)