CBSE Class 10 Physics Electricity Notes Set 04

Heating Effect of Current: Joule's Law

When a current flows through a conductor it heats it. The heat produced is H = I²Rt, where H is in joule, I in ampere, R in ohm and t in second. The heat is proportional to the square of the current, to the resistance and to the time. To express heat in calories divide the joules by 4.2. Example: 3 A through 20 ohm for 5 minutes gives 9 × 20 × 300 = 54 000 J. If the current is doubled and the resistance halved with the same time, H becomes (2I)² (R/2) t = 2 I²Rt, so the heat doubles.

Applications of the Heating Effect

Electric Power and Energy

Electric power is the rate of using electrical energy, P = VI, in watts (1 W = 1 J/s). 1 kW = 1000 W and 1 H.P. = 746 W. Energy = power × time. The commercial unit is the kilowatt hour (kWh, a unit): 1 kWh = 3.6 × 10⁶ J. Example: eight 100 W bulbs for 40 h use 32 kWh and a 500 W heater for 20 h uses 10 kWh; at Rs. 1.50 per unit the cost is 42 × 1.50 = Rs. 63.

Chemical Effect of Current: Electrolysis

A current passing through an electrolyte splits it. The electrode through which the current enters is the anode (+) and the one through which it leaves is the cathode (-). Hydrogen and metals are deposited at the cathode and non-metals at the anode.

Faraday's first law: the mass m liberated is proportional to the current and the time, m = zIt, where z is the electrochemical equivalent (the mass liberated by one coulomb, in kg/C). For example z is 1.0444 × 10^-8 kg/C for hydrogen, 8.29 × 10^-8 for oxygen and 32.95 × 10^-8 for copper.

Second law: for the same quantity of electricity, the masses liberated are proportional to the chemical equivalents (atomic mass divided by valency), so z / E is the same for all substances. For hydrogen z / E = 1.0444 × 10^-8 / (1.008 × 10^-3), so 96 500 coulomb (one faraday) liberates one gram equivalent of any substance. Example: copper deposited by 0.75 A in 25 min is 32.95 × 10^-8 × 0.75 × 1500 = 3.71 × 10^-4 kg.