From the relation P = ^V²/_R R= ^V²/_R

Substituting the values V and P in above equation, we get
R = ^120×120/₆₀ =240 Ω

If this bulb is connected to the 12V a battery, the consumption by thebulb is given by

P = ^V²/_R = ^12×12/₂₄₀= 3/5 = 0.6 W

Since watt is a small unit of power, a bigger unit Kilowatt is generallyused to express power consumption.

1 KW = 1000 W = 1000 J/S

You might have seen the current bill that comes to your home everymonth. In that bill, consumption of electricity is marked in units. Whatdoes the unit represent?

The unit of electric power consumption is equal to 1 KWH (one KiloWatt Hour).

1 KWH = (1000 J/S) (60 x 60 S)
= 3600 x 1000 J
= 3.6 x 106 J

• What do you mean by overload?
• Why does it cause damage to electric appliances?
We frequently hear news about overload of current and damages causedby this overload.

Electricity enters our homes through two wires called lines. Theseline wires have low resistance and the potential difference between thewires is usually about 240V. These two line wires run throughout thehousehold circuit, to which we connect various appliances such as fan, TV,refrigerator etc.

All the electric devices of our home are connected at different pointsbetween these two wires. This means all the electric appliances are inparallel connection. Hence, potential drop across each device is 240V. Ifwe know the value of resistance of the electric device, we can calculatethe current passing through it using the equation I = V/R. For example, thecurrent passing through a bulb with resistance 240 Ω is 1 A.

Based on the resistance of each electric device, it draws some currentfrom the supply. Total current drawn from the mains is equal to the sum ofthe currents passing through each device (Junction law).