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 7. 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 **negatively 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**

In atoms, protons are positive charges, and electrons are negative. 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 x t**

[coulombs] = [amps] x [seconds]

Here are a few practice questions to gain confidence with this formula:

**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} |

Before electrons were discovered, scientists were drawing diagrams with current flowing from positive to negative around a circuit. The discovery of electrons confused things, because it meant the charge carriers in metals were negative, and must flow from negative to positive around the circuit! This problem was side-stepped completely - we still draw the conventional current direction as positive to negative, and just ignore the flow of electrons in the opposite direction. Stick with your current arrows as positive to negative!

Electricity is a source of energy for us all. When we connect devices to batteries or a mains supply, the electric current does electrical work. But what is the connection between the energy transferred and the charges flowing around the circuit?

As you might guess, the more charges that flow, the more energy is transferred. However this energy transfer also depends on the 'kick' given to each charge by the battery. This kick is the voltage.

Therefore the energy supplied depends on both the charge flowing and the voltage.

The formula linking energy, charge and voltage is quite straightforward:

**energy transferred = charge x voltage**

**E = Q x V**

[joules] = [coulombs] x [volts]

if we rearrange this formula to put voltage as the subject, we get:

V = | E |

Q |

The formula above explains what voltage is in more technical language. Voltage is the energy transferred to each unit of charge. This leads to a few key definitions, that are connected to this formula:

**Voltage is the energy transferred per unit charge passed.**- In terms of units, the
**volt is one joule per coulomb**.

You will need to learn these two definitions for the concept of voltage.

**Questions**

3. A kettle needs 20 kJ to boil the water inside. If the kettle voltage is 240 V, calculate the total charge flowing through the kettle.

20 kJ is equal to 20 000 J.

We know E = Q x V,

So :

Q = | E |

V |

Q = | 20 000 |

240 |

Q = 83.333 coulombs, or **83.3 C** to 3 sig figs.

4. A small cell is used to run a camping lamp. The cell stores 180 J of energy, and is rated as 1.5 V.

- a) What is the total charge that can be delivered by the cell before it runs out? (Assume when running, the cell voltage remains constant and the cell is 100% efficient).
- b) The lamp needs 20 mA to operate. Using this information and your answer from part (a), calculate the time for which the lamp can remain lit.

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:**