Electric Energy and Heat - The law of Joule (2nd part) :

  The law of Joule :

We now know the relationship between voltage and current to electrical power and the energy consumed to produce heat.

However, we know that this heat production is due to the resistance encountered by the electrical charges of the current during their movement through a conductor. The energy consumed therefore depends directly on the current and resistance it encounters.

There is indeed a relationship that links these two magnitudes to energy. To better understand their action vis-à-vis energy, we must first refer to the energy consumed every second, in other words to the electric power.

This relation, stated by the English physicist (yes !) James Prescott JOULE (1818 - 1889), is called Joule's law. As we have already seen, this physicist also gave his name to the unit of measure of energy as well as to the thermal effect of the current called Joule effect.

The law of Joule can be stated as follows :

The electrical power consumed by a resistor to produce heat is obtained by multiplying the value of the resistance by the square of the current flowing through it :

P = R x I²

If the resistance and the current are respectively measured in Ohm and Ampere, the power is expressed in watts.

The law of Joule can be deduced from that already seen : P = V x I. Indeed, Ohm's law, the voltage V across a resistor is given by V = R x I. If we replace V by this product, we get :

P = R x I x I = R x I²

We can make two observations from this formula.

The first is that the power consumed increases in the same proportions as the resistance, for example, if the resistance doubles, the power doubles as well.

If it is not the same when the current increases, indeed, if for example the double current, the power is multiplied by 4.

We can therefore say that :

The electric power P increases as a function of the square of the current.

If in the formula P = V x I we replace not V by its value according to the law of Ohm but I, let I = V / R we find a new formula of the power :

P = V x I = V x V / R = V² / R

So :        P = V² / R

In the case where R doubles, the power P decreases by half whereas if V doubles, the power is multiplied by 4.

We can deduce that :

The electric power P increases according to the square of the voltage.

We are now able to calculate the power and thus the electrical energy consumed to produce heat by knowing two of the three electrical quantities that its the voltage, the current and the resistance on which the operation of any electrical circuit depends.

But how much heat does one get by consuming a determined energy ?

The answer was brought by JOULE following the numerous experiments that he realized.

First of all, to measure a quantity of heat, it must be given a unit of measure.

To do this calculation, we exploit the fact that when a body receives heat, its temperature increases.

Since this increase in heat is measurable with a thermometer, it is possible to deduce the amount of heat received by the body.

Thus was defined the unit of measure of quantity of heat called Calorie (symbol Cal). It was agreed that :

Calorie is the amount of heat needed to raise the temperature of one gram of water by one degree Celsius (for example from 20°C to 21°C).

For the quantities of heat generally encountered in practice, we use the kilocalorie (kcal symbol), this multiple of the calorie is defined by referring not more to a gram of water but to a kilogram of water. The kilocalorie is a thousand times larger than the calorie. We also use the thermie (symbol th) which is worth 1 million calories (a megacalorie).

Joule quantified his experiments precisely, using a known resistance conductor traversed by a known current and this for a given time. He determined that for every joule of energy consumed he obtained 0.238 cal. This amount of heat is called the thermal equivalent of energy. Conversely for a heat quantity of 1 calorie, it takes 4,185 Joules.

RESISTORS AND POWER

Let us now see how the new electrical quantities of power and energy are applied to an element such as a resistance.

Let us take again the circuit used during the analysis of the series links and represented figure 2.

 

Since the battery supplies a voltage of 9 V while the lamp only requires 6 V, we have had to connect the lamp in series with a resistor that causes a voltage drop of 3 V. This determined resistance produces heat by consuming power electrical energy. This energy is consumed unnecessarily since the role of the circuit is not to produce heat but to produce light through the lamp and not by the redness of the resistance.

The energy consumed by the resistance every second, that is to say the electrical power must be considered as dissipated power since it is not used in one way or another. For this reason, the resistors are called : dissipating elements of the power.

Let us now try to calculate the power dissipated by the resistance R (PR) and that dissipated by the lamp L (PL). We assume that the current flowing in the circuit of Figure 2 has an intensity of 0.05 A (50 mA). This current corresponds to the current absorbed by the lamp.

Applying the formula P = V x I, we obtain the values of PR and PL.

PL = VL x I = 6 x 0,05 = 0,3 W = 300 mW

The same values can be obtained, applying the other relations that we know either :

P = R x I²   and    P = V² / R

In these cases, the value of the resistance R and that of the L filament must first be determined by applying Ohm's law.

R = VR / I = 3 / 0,05 = 60 Ohms

RL = VL / I = 6 / 0,05 = 120 Ohms

Which gives for the two powers PR and PL :

PR = R x I² = 60 x (0,05 x 0,05) = 60 X 0,0025 = 0,15 W = 150 mW

and      PL = VL² / R = (6 x 6) / 120 = 36 / 120 = 0,3 W = 300 mW

The resistor R must therefore be able to dissipate a power of at least 150 mW. A resistor is an electronic component characterized not only by its ohmmic value but also by its maximum power that can dissipate without risk of destruction.