Heating effect of Current : Explained

When a volt­age source, such as a bat­tery, cre­ates cur­rent for a cir­cuit, the major­i­ty of the ener­gy is con­sumed to keep the cur­rent flow­ing. A por­tion of the ener­gy, how­ev­er, is wast­ed in the form of heat. The full ener­gy in a cir­cuit com­pris­ing sim­ply a bat­tery and resis­tance is dis­si­pat­ed as heat. This is the cur­ren­t’s heat­ing impact.

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Heat­ing effect of Current

The heat­ing effect of an elec­tric cur­rent depends on three factors:

  • The resis­tance, R of the con­duc­tor. A high­er resis­tance pro­duces more heat.
  • The time, t for which cur­rent flows. The longer the time the larg­er the amount of heat produced
  • The amount of cur­rent, I. the high­er the cur­rent the larg­er the amount of heat generated.

Pow­er = Work Done / Time TakenorP = W / t

Since, Poten­tial Dif­fer­ence = Work Done / ChargeorV = W / Q

P = VQ / tor P = VI, where I (cur­rent) = Charge / Time Taken

Ener­gy sup­plied by pow­er source to a cir­cuit in time t is, H = P x t = VIt

Or from Ohm’s law, H = I2Rt

This is known as Joule’s law of heat­ing. It states that heat pro­duces in a cir­cuit is direct­ly pro­por­tion­al to the square of cur­rent flow­ing, resis­tance for cur­rent and time for which cur­rent flows.

Dis­ad­van­tages of heat­ing effect

  • Loss of ener­gy in the unwant­ed heat.
  • Wear and tear of components.

There are four effects of elec­tric­i­ty :
The ‘effects of elec­tric­i­ty’ are:

  • mag­net­ic effect.
  • heat­ing effect.
  • chem­i­cal effect.
  • elec­tric shocks.

Exam­ple

  1. An elec­tri­cal bulb is labeled 100W, 240V. Calculate:

a)The cur­rent through the fil­a­ment when the bulb works nor­mal­ly b)The resis­tance of the fil­a­ment used in the bulb.

Solu­tion

  1. I = P/V = 100/240 = 0.4167A
  2. R = P/I2 = 100/ 0.41672 = 576.04Ω or R = V2/P =2402/100 = 576Ω
  3. Find the ener­gy dis­si­pat­ed in 5 min­utes by an elec­tric bulb with a fil­a­ment of resis­tance of 500Ω con­nect­ed to a 240V sup­ply. { ans. 34,560J}

Solu­tion

E = Pt = V2/R *t = (2402 *5*60)/500 = 34,560J.

  1. A 2.5 kW immer­sion heater is used to heat water. Calculate:
  1. The oper­at­ing volt­age of the heater if its resis­tance is 24Ω
  2. The elec­tri­cal ener­gy con­vert­ed to heat ener­gy in 2 hours.

{ans. 244.9488V, 1.8*107J}

Solu­tion

  1. P=VI=I2R

I = (2500/24)1/2 =10.2062A

V=IR= 10.2062 * 24 = 244.9488V

  1. E = VIt = Pt = 2500*2*60*60 = 1.8 * 107J

OR E= VIt = 244.9488 * 10.2062 * 2 * 60 * 60 = 1.8 * 107JAn elec­tric bulb is labeled 100W, 240V. Cal­cu­late:
The cur­rent through the fil­a­ment
The resis­tance of the fil­a­ment used in the bulb.

Solu­tionP = VI I = P/V = 100/240 =0.4167A
From Ohm’s law, V =IR R=V/I =240/0.4167 = 575.95Ω

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