TOPIC 2: ELECTRICITY
In this section we will learn about the fundamental properties of electricity - what is it? What is actually moving in the wires?
In 1897, J.J. Thompson discovered the electron. You may have met electrons already in your chemistry course, and they are covered in more detail in topic 4. He found that electrons were very small particles with a very small negative charge.
This discovery led to an understanding of electricity - it is caused by a flow of free charged electrons in metals (and many other solid conductors like graphite). The more electrons flow past any point in a second, the higher the current of electricity.
Figure 1. Current in a metallic conductor
The measurement for charge is named after Charles-Augustin de Coulomb. One coulomb (C) is an enormous charge compared to the electron - it would need about 8 billion billion electrons to make 1 coulomb of charge. For historical reasons, the letter given to the measurement of charge is 'Q'.
Current in a circuit is defined as the rate of flow of charge - this means the charge flowing past a point per unit of time.
If 2 coulombs of charge flows through a resistor each second, then the current is 2 amps.
As a formula we can write:
| current = | charge |
| time |
We can rearrange this to give the following:
charge = current x time
Q = I t
[coulombs] = [amps] x [seconds]
Remember that for electrical charge to flow through a closed circuit, the circuit must include a source of potential difference.
Here are a few practice questions to gain confidence with this formula:
Example:
What charge flows through a 6V, 2A motor left on for 30 seconds?
Answer:
Although the voltage is given, this is not needed to solve this question!
We know that charge = current x time = 2A x 30 seconds
Therefore charge flowing is = 2 x 30 = 60 coulombs
Questions:
1. A 0.3 amp household lamp is left on for 5 minutes. Calculate the charge flowing through the lamp in this time.
In 5 minutes, the time in seconds is 5 x 60 = 300 s,
We know that Q = I x t
Therefore Q = 0.3 x 300
Q = 90 coulombs
2. A 30 mA L.E.D. is left on for some time. During this interval, 6 C of charge flows through it. How long was the L.E.D. on for?
30 mA = 0.03 A, or if you are using standard form, 30 x 10-3 A (= 3 x 10-2 A).
If Q = I x t, then:
| t = | Q |
| I |
| t = | 6 |
| 3 x 10-2 |
a) If E = Q x V, then:
| Q = | E |
| V |
| Q = | 180 |
| 1.5 |
Q = 120 C
b) If I = 20 mA (0.02 A) and Q = 120 C,
then using Q = I x t gives:
| t = | Q |
| I |
| t = | 120 |
| 0.02 |
t = 6000 s (100 minutes, or 1 hr 40 mins)
Now test your understanding using these quick, 10 minute questions on this topic from Grade Gorilla:
