Question

A consumer has the following appliances in the house:

• An electric iron rated 1500 W
• A water heater rated 500 W
• An electric cooker rated 2500 W
• Three bulbs each rated 60 W.
  
  The house is fitted with a 12 A fuse. Determine:
  1. whether the consumer can connect all the appliances to the 240 V power supply at the same time; 
  2. the resistance of the heating element used in the electric cooker. 

Answer 1

1. Total power = 1500 + 2500 + 500 +( 60x3)
   = 4680 W
   Total current required = ⁴⁶⁸⁰/₂₄₀= 19.5A
                                         
   Hence fuse blows and disconnects the current when it exceeds 10A ie, all appliances can't be connected at the same time/ max current for all appliances connected is higher than the fuse rating.
2. I =  ^P/_V   = ²⁵⁰⁰/₂₄₀
   R= 240 ÷ ( ²⁵⁰⁰/₂₄₀)
   = 240 x 240
         2500
   = 23.04Ω