MP Board · Class 10 · Science · ElectricityWhat is Joule's law of heating? Derive the expression for heat produced in a resistor. An electric iron consumes energy at a rate of $840\, \text{W}$ when heating is at the maximum rate and $360\, \text{W}$ when the heating is at the minimum. The voltage is $220\, \text{V}$. Calculate the current and the resistance in each case.

Joule's Law of Heating\nJoule's law of heating states that the heat produced in a resistor is:

1. Directly proportional to the square of current ($I^2$) flowing through the resistor.
2. Directly proportional to the resistance ($R$) for a given current.
3. Directly proportional to the time ($t$) for which the current flows through the resistor.

Derivation of Formula\nLet a current $I$ flow through a resistor of resistance $R$ for time $t$, under a potential difference $V$.

• The work done $W$ in moving a charge $Q$ through potential difference $V$ is given by: $$W = V \times Q$$
• Since $Q = I \times t$, we can write: $$W = V \times I \times t$$
• According to Ohm's law, $V = IR$. Substituting this in the above equation: $$W = (IR) \times I \times t = I^2Rt$$\nAssuming all electrical work is converted into heat energy ($H$), we get: $$H = I^2Rt$$\nThis is Joule's law of heating.

Numerical Problem

Given Data:

• Voltage ($V$) = $220, \text{V}$

Case 1: Maximum heating rate ($P_1 = 840, \text{W}$)

1. Current ($I_1$): $$P_1 = V \times I_1 \implies I_1 = \frac{P_1}{V}$$ $$I_1 = \frac{840, \text{W}}{220, \text{V}} = 3.82, \text{A}$$
2. Resistance ($R_1$): $$R_1 = \frac{V}{I_1} = \frac{220, \text{V}}{3.82, \text{A}} = 57.59, \Omega$$

Case 2: Minimum heating rate ($P_2 = 360, \text{W}$)

1. Current ($I_2$): $$P_2 = V \times I_2 \implies I_2 = \frac{P_2}{V}$$ $$I_2 = \frac{360, \text{W}}{220, \text{V}} = 1.64, \text{A}$$
2. Resistance ($R_2$): $$R_2 = \frac{V}{I_2} = \frac{220, \text{V}}{1.64, \text{A}} = 134.15, \Omega$$