When potential difference is applied across the two ends of a conductor, an electric field is setup across its ends. Under the effect of the electric field the electrons get accelerated and move in a direction opposite to the direction of the electric field.

As they collide against the ions or atoms in the conductor, the electrons transfer their kinetic energy to the atoms.Thus, the average kinetic energy of random vibration of the atoms increases and consequently the temperature of the conductor rises. **It is termed as Heating effect of the electric currents.**

Suppose that a steady current (I) is flowing though a resistance (R). The potential drop (V) across the ends of the resistance satisfies, according to ohm’s law ,

V = IR

Suppose that this steady current I flows through the resistance for a time t . Then, the total charge that crosses through the resistance q = It

The energy will be consumed from the source of emf and it will appear as heat energy (W) across the resistance.

W = Vq = V (It) = V It . . .(i)

W = IR × It = I^{2}Rt . . . . . (ii)

Therefore, rate of dissipation of electrical energy across the resistance, i.e. electric power is given by

$\large P = \frac{W}{t} $

Using equation (i) we have

P = VI = I^{2}R

The SI unit of electric power is watt

1 watt = 1 volt × 1 ampere

The energy dissipated in the resistance R is

$\large H = W = I^2 R t \; Joule = \frac{I^2 R t}{4.2} calorie$

This relation was first recognized and verified experimentally by Joule and so is called Joule Heating Effect.

**Fuse :**

(i) Fuse is a safely device used in the household power supply to break the circuit as soon as the current exceeds a previously determined safe limit due to the over loading or short circuit. It consists of a wire, whose melting point is slightly above the temperature reached when the maximum safe current is set up through is, Hence, when the current exceeds the safe limit, the wire melts and the circuit breaks. The fuse is said to have blown off.

(ii) Safe current and radius of the fuse wire. The maximum current that can pass through the fuse wire without melting it is called safe current.

The fuse wire is made of an alloy of lead and tin.

Example : If two bulbs of 25 W and 100 W rated at 400 V are connected in series across a 440 V supply, will both the bulb fuse ? If not which one ?

Solution: As resistance of the bulb $ \displaystyle R_B = \frac{V_s^2}{P}$

$ \displaystyle \frac{R_{25}}{R_{100}} = \frac{100}{25} = 4 $

The bulbs are connected in series

$ \displaystyle V_{25} = \frac{R_{25}}{R_{100} + R_{25}}\times V $

$ \displaystyle V_{25} = \frac{4}{5} \times 440 = 352 volt $

$ \displaystyle V_{100} = \frac{R_{100}}{R_{100} + R_{25}}\times V $

$ \displaystyle V_{100} = \frac{1}{5} \times 440 = 88 volt $

Thus neither bulb will fuse.

Example :Two bulbs rated at 60 W & 100 W (220 V) are connected (a) in series (b) in parallel. Find the heat dissipated in the bulbs and also the total heat dissipated in the circuit.

Solution: $ \displaystyle P_1 = \frac{V^2}{R_1} = 60 W$

$ \displaystyle P_2 = \frac{V^2}{R_2} = 100 W$

When the bulbs are connected in parallel, a p.d. of 220 V appears across each bulb.

The power dissipated is : 60 W and 100 W and the total power dissipated is 160 W.

When the bulbs are connected in series, the same current flows through them.

$ \displaystyle R_1 = \frac{V^2}{P_1} $ and $ \displaystyle R_2 = \frac{V^2}{P_2} $ gives the resistances .

The current in the bulbs is

$ \displaystyle i = \frac{V}{R_1 + R_2}$

and the power dissipated is $ \displaystyle i^2 R_1 \; , i^2 R_2 \; and , i^2 (R_1 + R_2) $

$ \displaystyle \frac{V^2 R_1}{(R_1 + R_2)^2} \; , \frac{V^2 R_2}{(R_1 + R_2)^2} \; and , \frac{V^2 }{(R_1 + R_2)} $

$ \displaystyle \frac{1/P_1}{(\frac{1}{P_1}+ \frac{1}{P_2})^2} $ , $ \displaystyle \frac{1/P_1}{(\frac{1}{P_2}+ \frac{1}{P_2})^2} $ and $ \displaystyle \frac{1}{(\frac{1}{P_1}+ \frac{1}{P_2})} $

or , 23.5 W, 14.1W, 37.5 W