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.

Subscribe on YouTube

Get free tutorials, exam tips & updates.

Subscribe

Join WhatsApp Channel

Stay updated with instant alerts.

Follow

Free SSC Mock Tests

Boost your preparation with free practice tests.

Start Now

Govt Job Alerts

Get free daily updates on government jobs.

Get Alerts

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