What is Joule's law of heating? Derive the expression for heat produced in a resistor. An electric iron consumes energy

Question

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

Accepted Answer

Joule's Law of Heating
Joule's law of heating states that the heat produced in a resistor is: Directly proportional to the square of current ($I^2$) flowing through the resistor. Directly proportional to the resistance ($R$) for a given current. Directly proportional to the time ($t$) for which the current flows through the resistor. Derivation of Formula
Let 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: $$W = V 	imes Q$$ Since $Q = I 	imes t$, we can write: $$W = V 	imes I 	imes t$$ According to Ohm's law, $V = IR$. Substituting this in the above equation: $$W = (IR) 	imes I 	imes t = I^2Rt$$
Assuming all electrical work is converted into heat energy ($H$), we get: $$H = I^2Rt$$
This is Joule's law of heating. --- Numerical Problem Given Data: Voltage ($V$) = $220\, 	ext{V}$ Case 1: Maximum heating rate ($P1 = 840\, 	ext{W}$) Current ($I1$): $$P1 = V 	imes I1 \implies I1 = \frac{P1}{V}$$ $$I1 = \frac{840\, 	ext{W}}{220\, 	ext{V}} = 3.82\, 	ext{A}$$ Resistance ($R1$): $$R1 = \frac{V}{I1} = \frac{220\, 	ext{V}}{3.82\, 	ext{A}} = 57.59\, \Omega$$ Case 2: Minimum heating rate ($P2 = 360\, 	ext{W}$) Current ($I2$): $$P2 = V 	imes I2 \implies I2 = \frac{P2}{V}$$ $$I2 = \frac{360\, 	ext{W}}{220\, 	ext{V}} = 1.64\, 	ext{A}$$ Resistance ($R2$): $$R2 = \frac{V}{I2} = \frac{220\, 	ext{V}}{1.64\, 	ext{A}} = 134.15\, \Omega$$

Electricity

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