Additional Physics Topic 11  Current, Charge, and Power
Power of an Appliance
The power of an appliance is how much energy it transforms every second. For example an electric drill converts 500 joules every second into kinetic energy.
The units for power are watts (W) or kilowatts (kW) where:
1 W = 1 J/s
1 kW = 1000 W
The equation for electrical power is:
power (watt, W) = current (ampere, A) × potential difference (volt, V)
In Physics Code:
P = IV
In triangle form:
Worked Example What is the current taken by a 500 W drill from a 230 V AC supply? 
Answer I = P/V I = 500 W ÷ 230 V = 2.17 A 
A lorry starter motor runs off a 24 V DC battery. Under load it takes a current of 500 A. What power does it develop as it turns the lorry's engine? 
This starter motor can start a petrol engine between 9 and 26 litres (pretty big).
Its power is 7.76 kW. What current does it take from a 12 V battery to turn the engine? 
In fact the starter motor in Question 2 takes a current of 1650 A. What is the power taken from the battery? What happens to the power? What is the efficiency of the motor? 
You can see from the result of the last calculation that the starter motor will soon get hot if the engine does not start. Components that take heavy currents need thick and heavy wires to take the current to them. They also need to be very heavily built. This one has a mass of 29 kg.
Heating elements are resistors. In this case electrical energy is converted straight to heat.
The heating effect from a current goes up as the square of the current. That means that if you double the current, the heating effect (power) goes up four times. This can lead to a component overheating as seen in this picture.
It does not take a genius to see that this can be dangerous. High currents due to faults can easily set a house on fire.
Choosing the Right Fuse for an Appliance
Domestic appliances are connected to the mains by means of a fused plug, which we saw last topic. Remember that a fuse is a weak link that melts when the appliance takes too big a current. The BS1362 fuses are available in the following sizes;
1 A
2 A
3 A  common
5 A  common
7 A
10 A  common
13 A  very common
Plugs are mostly supplied with a 13 A fuse, and most people don't bother replacing the fuse with one of the correct value. This can be dangerous for the following reason. Suppose we have an appliance that takes 250 W from the mains. Its current would be 1.1 A (how?). Suppose it developed a fault that made it take 750 W. Its current would now be 3.3 A. There would be 500 W being used to heat the appliance up.
What do you think would happen to the appliance? 
If there were a 13 A fuse, would it blow? 
If there were a 3 A fuse, would it blow? 
The result of using a 13 A fuse to protect a 1.1 A appliance could be disastrous.
There are three steps in working out the fuse value for an appliance:
Calculate the current from the power. Current = power ÷ 230 V
Look at the fuse values in the BS1362 series.
Choose the next highest value above your current.
Worked example What value fuse would you choose for a 500 W appliance? 
Answer

If your current worked out to be 5 A, you wouldn't choose the 5 A fuse, as it would keep blowing. You would choose the 7 A fuse.
A fan heater is rated at 2000 W. Which is the best fuse to use in its plug? 
Charge and Current
In the last topic we saw that in a current of 1 amp, 6 × 10^{18} electrons flowed every second. We called a packet of 6 × 10^{18} electrons a coulomb. Therefore a current of 1 amp is a flow of charge of 1 coulomb every second.
1 A = 1 C/s
Therefore we can write an equation:
charge (coulomb, C) = current (ampere, A) × time (second, s)
In Physics code:
Q = It
In triangle form:
Worked Example A current of 4.5 A flows for 20 s. How much charge has flowed? 
Answer Q = It Q = 4.5 A × 20 s = 90 C 
A current of 2.5 A is flowing. How long does it take for 20 C to flow? 
It takes 12 s for 108 C to flow. What is the current? 
Energy and Charge (HT Only)
Why do we want to know the charge? If we know the charge and the voltage, we can work out the energy transformed.
Energy (J) = power (W) × time (s)
Power (W) = voltage (V) × current (A)
So it doesn't need a genius to see that:
Energy (J) = voltage (V) × current (A) × time (s)
When we charge up batteries, we want to know the charge that they hold.
When charge passes through a load (e.g. a resistor) it gives off energy that can be used to do a useful job of work (e.g. heat up some water). If 1 coulomb of charge goes through a resistor which has a p.d. of 1 volt across it, it gives off 1 joule of energy.
From this we see that that:
Energy (J) = voltage (V) × current (A) × time (s)
Now
Charge = current × time
So it doesn't take a genius to see that:
Energy = voltage × charge.
We can write this as an equation:
energy transformed (joule, J) = potential difference (volt, V) × charge (coulomb, C)
In Physics Code
E = VQ
In triangle form:
Worked Example How much energy does 150 C charge carry from a 12 volt battery? 
E = VQ E = 12 V × 150 C = 1800 J 
20000 J of energy is transformed by a 230 V light bulb. How much charge is this? 
Question 11 
Complete the interactive gapfill exercise 
Question 12 
Have a go at the interactive crossword that gets you to think about mains electricity. 
Summary
