Electricity Past Paper PDF

Applying Ohm’s law [Eq. (11.5)], we get H = I2 Rt (11.21) This is known as Joule’s law of heating. The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resi...

Applying Ohm’s law [Eq. (11.5)], we get H = I2 Rt (11.21) This is known as Joule’s law of heating. The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor. In practical situations, when an electric appliance is connected to a known voltage source, Eq. (11.21) is used after calculating the current through it, using the Figure 11.13 relation I = V/R. A steady current in a purely resistive electric circuit Example 11.10 An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case? Solution From Eq. (11.19), we know that the power input is P=VI Thus the current I = P/V (a) When heating is at the maximum rate, I = 840 W/220 V = 3.82 A; and the resistance of the electric iron is R = V/I = 220 V/3.82 A = 57.60 Ω. (b) When heating is at the minimum rate, I = 360 W/220 V = 1.64 A; and the resistance of the electric iron is R = V/I = 220 V/1.64 A = 134.15 Ω. Example 11.11 100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor. Solution H = 100 J, R = 4 Ω, t = 1 s, V = ? From Eq. (11.21) we have the current through the resistor as I = √(H/Rt) = √[100 J/(4 Ω × 1 s)] = 5A Thus the potential difference across the resistor, V [from Eq. (11.5)] is V = IR = 5A×4Ω = 20 V. Electricity 189 2024-25