Class 10 Physics



Electricity

Heating Effect of Electric Current-Practical Application

Practical Application of Heating Effect of Electric Current & Electric Power

For exploiting the heating effect of electric current, the element of appliances must have high melting point to retain more heat. The heating effect of electric current is used in the following applications:

Electric Bulb: In an electric bulb, the filament of bulb gives light because of heating effect of electricity. The filament of bulb is generally made of tungsten metal; having melting point equal to 3380°C.


Electric iron: The element of electric iron is made of alloys having high melting point. Electric heater and geyser work on the same mechanism.

Electric fuse: Electric fuse is used to protect the electric appliances from high voltage; if any. Electric fuse is made of metal or alloy of metals, such as aluminium, copper, iron, lead, etc. In the case of flow of higher voltage than specified, fuse wire melts and protects the electric appliances.

Fuse of 1A, 2A, 3A, 5A, 10A, etc. are used for domestic purpose.

Suppose, if an electric heater consumes 1000W at 220V.

Then electric current in circuit `I = P/V`

Or, `I = 1000 W − 220 V = 4.5 A`

Thus, in this case a fuse of 5A should be used to protect the electric heater in the case of flow of higher voltage.

Electric Power:

SI unit of electric power is watt (W).

`1W = 1 text(volt) xx 1 text(ampere) = 1V xx 1A`

1 kilo watt or 1kW = 1000 W

Consumption of electricity (electric energy) is generally measured in kilo watt.

Unit of electric energy is kilo watt hour (kWh)

`1 kWh = 1000 text(watt) xx 1 text(hour) = 1000 W xx 3600 s`

Or, `1kWh = 3.6 xx 10^6 text(watt second) = 3.6 xx 10^6  J`

Example 1: If the potential difference is 220V and the power of bulb is 110W, what is the electric current flowing in the circuit?

Solution: Given, Potential difference, V = 220V, Power of bulb , P = 110 W

Electric current (I) =?

We know that `P = VI`

Or, `110 W = 220 V xx I`

Or, `I = 110 W ÷ 220 V = 0.5 A`

Example 2: If the power of an electric heater is 1000W and electricity of 240 V is flowing through it, find the electric current in the electric heater.

Solution: Given, Power (P) = 1000W, Potential difference (V) = 240V, Electric current (I) =?

We know that `P = VI`

Or, `1000 W = 240 V xx I`

Or, `I = 1000 W ÷ 240 V = 4.16 A`


Example 3: What is the electric current through an electric geyser of 500W, if the potential difference across the electric circuit is 250V?

Solution: Given, P = 500W, V= 250 V, therefore, Electric current (I) =?

We know that `P = VI`

Or, `500 W = 250 V xx I`

Or, `I = 500 W ÷ 250 V = 2 A`

Example 4: If the electric current is 10A and potential different between two terminals is 240V, find the power of the electric appliance.

Solution: Given, Electric current (I) = 10A, Potential difference (V) = 240V, Power (P) =?

Since, `P = V xx I`

Therefore, `P = 240V xx 10A = 2400 W`

Example 5: Find the power of electric iron, if electric current through the circuit is 5A and potential difference is 220V.

Solution: Given, Electric current (I) = 5 A, Potential difference (V) = 220 V, Power(P) =?

We know that, `P = V xx I`

Or, `P = 220 V xx 5 A = 1100 W`