2.5 know and use the relationship between power, current and voltage and apply the relationship to the selection of appropriate fuses

2. The power of an electrical appliance can be calculated from the current that flows through it and the potential difference (voltage) across it.
   You can work out power using this equation:
   P = I × V
   P is the power (in watts, W)
   I is the current (in amperes/amps, A)
   V is the potential differences (in volts, V)
   
   
   Example
   • What is the power of a 5 A 1.5 V lamp?
   • Power = 5 × 1.5 = 7.5 W
   
   

   Working out the best fuse to use

   
   
   The equation P = I × V can be rearranged to find the current if the power and potential difference (voltage)are known:
   I = P ÷ V
   
   
   Example
   What current flows through a 1.15 kW electric fire at a potential difference of 230 V? (Remember that 1.15 kW is 1,150 W)
   Current = 1150 ÷ 230 = 5A
   
   
   Fuses come in standard ratings of 3 A, 5 A or 13 A.
   The best fuse to use in this example would be the 13A fuse. The 3A and 5A fuses would blow even when the fire was working normally.