Class 10 Physics



Electricity

Heating Effect of Electric Current

When electric current is supplied to a purely resistive conductor, the energy of electric current is dissipated entirely in the form of heat and as a result, resistor gets heated. The heating of resistor because of dissipation of electrical energy is commonly known as Heating Effect of Electric Current. Some examples are as follows:

When electric energy is supplied to an electric bulb, the filament gets heated because of which it gives light. The heating of electric bulb happens because of heating effect of electric current.

When an electric iron is connected to an electric circuit, the element of electric iron gets heated because of dissipation of electric energy, which heats the electric iron. The element of electric iron is a purely resistive conductor. This happens because of heating effect of electric current.


Cause of heating effect of electric current: Electric current generates heat to overcome the resistance offered by the conductor through which it passes. Higher the resistance, the electric current will generate higher amount of heat. Thus, generation of heat by electric current while passing through a conductor is an inevitable consequence. This heating effect is used in many appliances, such as electric iron, electric heater, electric geyser, etc.

Joule’s Law of Heating:

Let; an electric current I is flowing through a resistor having resistance equal to R.

The potential difference through the resistor is equal to V.

The charge Q flows through the circuit for the time t.

Thus, work done in moving of charge Q of potential difference `V = VQ`

Since, this charge Q flows through the circuit for time t

Therefor; power input (P) to the circuit can be given by following equation:

`P=VxxQ/t`--------(1)

We know, electric current `I = Q/t`

Substituting `Q/t = I` in equation (i), we get;

`P = VI` ..........(ii)

Since the electric energy is supplied for time t, thus after multiplying both sides of equation (ii) by time t, we get

`P xx t = VI xx t = VIt` .....(iii)

Thus, for steady current I, the heat produced (H) in time t is equal to VIt

Or, `H = VIt` .........(iv)

We know; according to Ohm's law; V = IR

By substituting this value of V in equation (iv), we get;

`H = IR xx It`

Or, `H = I^2Rt` ........(v)

The expression (v) is known as Joule’s Law of Heating, which states that heat produced in a resistor is directly proportional to the square of current given to the resistor, directly proportional to the resistance for a given current and directly proportional to the time for which the current is flowing through the resistor.

Example 1: If an electric heater consumes electricity at the rate of 500W and the potential difference between the two terminals of electric circuit is 250V, calculate the electric current and resistance through the circuit.

Solution: Given, power input (P) = 500 W
Potential difference (V) = 250 V
Electric current (I) =?
Resistance (R) through the circuit =?

We know that power `(P) = VI`

Or, `500 W = 250 V xx I`

Or, `I = 500 W ÷ 250 V = 2 A`

We know, resistance `R = V/I`

Or, `R = 250 V ÷ 2 A = 125 Ω`


Example 2: An electric geyser consumes electricity at the rate of 1000W. If the potential difference through the electric circuit is 250 V, find the resistance offered by geyser and electric current through the circuit.

Solution: Given, power input (P) = 1000 W

Potential difference (V) = 250 V

Electric current (I) =?

Resistance (R) through the circuit =?

We know that power `(P) = VI`

Or, `1000 W = 250 V xx I`

Or, `I = 1000 V ÷ 250 V = 4 A`

We know, resistance `R = V/I`

Or, `R = 250 V ÷ 4 A = 62.5 Ω`

Example 3: An electric heater having resistance equal to 5Ω is connected to electric source. If it produces 180 J of heat in one second, find the potential difference across the electric heater.

Solution: Given, Resistance (R) = 5 Ω, Heat (H) produced per second by heater = 1800 J, time ‘t’ = 1 s

Potential difference (V) =?

To calculate the potential difference, we need to calculate electric current (I) first.

We know that `H = I^2Rt`

Or, `180 J = I^2 xx 5 Ω xx 1 s`

Or, `I^2 = 180 ÷ 5 = 36`

Or, `I = 6 A`

Now, potential difference `V = IR`

Or, `V = 6 A xx 5 Ω = 30 V`