Question

How to solve these two questions ?​

Answer 1

1) See attached graph

To solve this part of the problem, we have to keep in mind the relationship between current and charge:

$i = \frac{\Delta Q}{\Delta t}$

where

i is the current

Q is the charge

t is the time

The equation then means that the current is the rate of change of charge over time.

Therefore, if we plot a graph of the charge vs time (as it is done here), the current at any time will be equal to the slope of the charge vs time graph.

Here we have:

- Between t = 0 and t = 2 s, the slope is $\frac{50-0}{2-0}=25 C/s$, so the current is 25 A

- Between t = 2 s and t = 6 s, the slope is $\frac{-50-(50)}{6-2}=-25 C/s$, so the current is -25 A

- Between t = 6 s and t = 8 s, the slope is $\frac{0-(-50)}{8-6}=25 C/s$, so the current is 25 A

Plotting on a graph, we find the graph in attachment.

2) $15 \mu C$

The relationship we have written before

$i = \frac{\Delta Q}{\Delta t}$

Can be rewritten as

$\Delta Q = i \Delta t$

This is valid for a constant current: if the current is not constant, then the total charge is simply equal to the area under the current vs time graph.

Therefore, we need to find the area under the graph.

Here we have a trapezium, where the two bases are

A = 1 ms = 0.001 s

B = 2 ms = 0.002 s

And the height is

h = 10 mA = 0.010 A

So, the area is

$Area=\frac{(0.001+0.002)\cdot 0.010}{2}=1.5\cdot 10^{-5} C = 15 \mu C$

So, the charge is $15 \mu C$.