Effects of Electric Current: Heating, Power and Electrolysis
This domain tests the heating and chemical effects of current in the Kerala SSLC physics paper: Joule's law numericals, fuse and filament design, and the working of an electrolyte.
Heating effect
When charge flows through a resistor, electrical energy is converted into heat. Joule's law gives H = I²Rt: the heat is proportional to the square of the current, the resistance and the time. Using Ohm's law this becomes H = VIt = V²t/R, so a device rated 60 W, 230 V has R = V²/P ≈ 882 Ω and draws I = P/V ≈ 0.26 A.
A fuse is a thin wire of a tin–lead alloy, chosen for its low melting point, connected in the live wire so that it melts when the current exceeds its rating. The filament of a bulb is tungsten, chosen for its very high melting point (about 3400 °C) and high resistivity.
Chemical effect
In an electrolyte, current is carried by ions: cations move to the cathode and anions to the anode. This is used in electroplating and in refining metals; the mass deposited depends on the current and the time of passage.
Worked example
An electric iron rated 1000 W, 230 V is used 2 h a day for 30 days. Energy = 1 kW × 2 h × 30 = 60 kWh = 60 units; at ₹6.50 per unit the cost is ₹390. The resistance is R = V²/P = 230²/1000 ≈ 52.9 Ω and the current 4.35 A, so a 5 A fuse is correct - a 3 A fuse would blow in normal use.
Traps that cost marks
- Substituting time in minutes into H = I²Rt instead of seconds.
- Writing 1 kWh = 1000 J instead of 3.6 × 10⁶ J.
- Claiming a fuse has a high melting point.
- Confusing power (watt) with energy (kWh).