Heating Effects of Electric Current

The heating effect of electric current is a fundamental concept in physics where electrical energy is converted into heat energy as current flows through a conductor. This effect is widely used in everyday appliances like electric heaters, toasters, and electric irons. Understanding this principle helps explain energy consumption, resistance, and the working of various electrical devices.Understanding the heating effect of current is crucial for analyzing power consumption, electrical safety, and the design of electrical circuits. It also plays a vital role in understanding the concepts of Joule’s law of heating, resistance, and energy efficiency in electrical systems.

1.0Electrical Power

Energy liberated per second in a device is called its power. The electrical power P delivered or consumed by an electrical device is given by 

$P=Vi$

where V = Potential difference across the device and 

i = CurrentPower (P) = $\frac{V\cdot dq}{dt}=Vi$

If the current enters the higher potential point of the device then electric power is consumed by it (i.e. acts as load). If the current enters the lower potential point then the device supplies power (i.e. acts as source).

2.0Cause of Heating

The potential difference applied across the two ends of the conductor sets up an electric field. Under the effect of electric field, electrons accelerate and as they move, they collide against the ions and atoms in the conductor, the energy of electrons transferred to the atoms and ions appears as heat.

Joules's Law of Heating

Let voltage 'V'is applied across the conductor of resistance R and a charge dq flows through it in time dt then work done by the battery is

$dw=dqV=idtV$

$\int dw=\int i(iR)dt=\int i^{2}Rdt$

Since Heat energy released from the conductor is equal to work done by the battery unit,

H=Heat released= $W=\int i^{2}Rdt$

If i=constant, $H=i^{2}Rt=\frac{V^2}{R}t=(Vi)t$

Heat produced in a conductor does not depend upon the direction of current.

SI Unit:Joule

$1kWh=3.6\times 10^{6}Joule=1\text{ unit; }1BTU(\text{ British Thermal Unit })=1055J$

Power delivered to the load resistance

$P=\frac{dw}{dt}$

$P=\frac{i^{2}Rdt}{dt}=i^{2}R=\frac{V^2}{R}=Vi$

3.0Fusing of an Electrical Device

Every Electrical device like a bulb,heater,geyser,etc comes with a design/rated voltage $\left(V_0\right)$ and design /rated wattage or power $\left(P_0\right)$.

If the applied voltage V across a device becomes greater than $V_0$, then it will not work or we can say the device will be fused.

The resistance of an electrical device can be found using

$P_0=\frac{V_0^2}{R}$

$R=\frac{V_0^2}{P_0}$

If V is the applied voltage across the device and if  P is the power consumed by it, then

$P=\frac{v^2}{R}\text{ i.e }P=\left(\frac{v^2}{v_0^2}\right)P_0$

Fuse wire

The fuse wire for an electric circuit is chosen keeping in view the value of safe current through the circuit. The fuse wire should have high resistance per unit length and low melting point. However, the melting point of the material of the fuse wire should be above the temperature that will be reached during the passage of the current through the circuit.

A fuse wire is made of alloys of lead (Pb) and tin (Sn). Length of fuse wire is immaterial. The material of the filament of a heater should have high resistivity and high melting point.

4.0Equivalent Power in Combination of Bulbs

1. Series Combination

Let,two bulbs with $\left(V_0\right)$ but $P_{01}>P_{02}$ are connected in series.

Here, $P_{01}>P_{02}\Rightarrow R_1<R_2$

Now, since series $\Rightarrow P_1=I^2{R}_{1}and{P}_2=I^2{R}_2$

$\therefore P_1>P_2$

In series combination of bulbs since 

$Brightness\propto PowerconsumedbyBulb\propto R\propto \frac{1}{P_{\text{Rated }}}$

Bulb of lesser wattage will shine

P= Total Power of Combinations $=\frac{v^2}{R_1+R_2}$

$P=\frac{V^2}{\frac{V_0^2}{P_{01}}+\frac{V_0^2}{P_{0z}}}={\left(\frac{V}{V_0}\right)}^2\left(\frac{P_{01}{P}_{02}}{P_{01}+P_{02}}\right)$

If $V=V_0$ then $\frac{1}{P}=\frac{1}{P_{01}}+\frac{1}{P_{02}}$

2. Parallel Combination

Similarly,let two bulbs with $V_0$ but $P_{01}>P_{02}$ are connected in parallel

Since $P_{01}>P_{02}$

∴ $R_1<R_2$

Now $P_1=\frac{V^2}{R_1};P_2=\frac{V^2}{R_2}$

∴ $P_1<P_2$a

In a parallel combination of bulbs since

$\text{ Brightness }\propto \text{ Power consumed by bulb }\propto I\propto \frac{1}{R}\propto P_{\text{Rated }}$

So a bulb of greater wattage will shine more.

$P_{\text{total }}=\frac{V^2}{R_1}+\frac{V^2}{R_2}={\left(\frac{V}{V_0}\right)}^2\left[P_1+P_2\right]$

If $V=V_0$ then,

$P_{\text{total }}=P_1+P_2$