Examville.com - Heating Effect Of Electric Current

Heating Effect of Electric Current 1. Electric Power P = I2R = V2/R watts Electrical energy consumed, W = I2 x R x t = (V2/R) x t Joules The formula applies only to devices (heater, electric kettles, electric bulb etc.) and resistors, where whole of electrical energy consumed is converted into heat. 2. Electric power = P = V x I watts Electrical energy consumed, W = V / t joules. These two formulas apply to all kind of loads. 3. Power of electric bulb, P = V2 / R or R α 1/P for constant V. So, for the same voltage rating, the bulb with smaller power will have more resistance. 4. 1KWh = 1KW x 1hr. = 1000W x 3600s = 36 x 105 joules or Ws 1Kcal = 4180J and 1KWh = 860Kcal. 5. In a house, all the electrical appliances are joined in parallel. When electrical appliances are joined in parallel, the working of each appliance is independent of each other. A 40W electric bulb means that it converts 40J of electrical energy into heat in one second. The brightness of an electric bulb is directly proportional to the rate of production of heat by the bulb. 6.When a number of electric bulbs of power P1, P2,P3,P4, …… rated at the same voltage V are connected in series across voltage V, then the total power PT is PT = (1/ P1) + (1/P2 )+ (1/P3 )+ (1/P4) + ……….. 7. When a number of electric bulbs of power P1, P2,P3,P4, …… rated at the same voltage V are connected in parallel across voltage V, then the total power PT is PT = P1 + P2 + P3 + P4 +………..

8. When two bulbs of different wattage but of the same rated voltage v are connected in series across voltage 2V, then bulb of smaller rating (wattage) will fuse. 9. In series combination of electric bulbs, the brightness of 40W bulb is more than that of 60W bulb. When the bulbs are connected in series, the rate of heats generated in them is directly proportional to their resistance. (This is because current in them is the same.) 10. The electrical energy supplied to the heating appliance forms the input energy. The heat we get from the device is the output energy. Efficiency = η = (Output energy) / (Input energy) 11. Two coils of resistance R1 and R2 take time T1 and T2 respectively to produce a given amount of heat H, when connected to the same supply. (a). If the coils are connected in series across the same supply, then time T taken to produce the same amount oh heat H is T = T1 + T2. (b). If the coils are connected in parallel across the same supply then time T taken to produce the same amount of heat H is (1 / T) = (1/T1) + (1/T2) Or, T = (T1T2) / (T1 + T2) 12. If x equal resistors are connected in series across a given supply and dissipate power P, these resistors if connected in parallel across the same supply, will dissipate power = x2P. 13. If H is the heat lost per second per unit surface area of a fuse wire, then H = (I2ρ) / (2π2r3), or I2 α r3 So, I α r3/2 Fusing current I is independent of the length of the wire.

14. When two resistors R1 and R2 are connected in parallel across a supply and power dissipated in them are P1 and P2 respectively, then P1R1 = P2R2 15. Electrical appliances in a house are connected in parallel because different appliances have the same voltage rating but different current rating. 16. The hot resistance of an electric bulb is more than 10 times its cold resistance. ____________________________________