Whenever an electric current passes through a material that has some resistance (i.e., except a superconductor), it creates heat. This resistive heating is the result of “frictions”, created by microscopic phenomena such as retardation forces and collisions involving charge carriers (usual electrons); informal terminology, heat is the work done by charge carriers in order to travel to a lower potential.
This heat generation can be provided by design, as in any heating device (for example, a toaster, electric heater, or electric blanket).
Such an apparatus consists essentially of a conductor whose resistance is chosen so as to produce the desired amount of resistive heating. In other cases, resistive heating may be undesirable. Power lines are a classic example. On the one hand, their purpose is to transmit energy, not to dissipate it; the energy converted into heat along the way is indeed lost (hence the term resistive losses). Also, resistive heating of transmission and distribution lines is undesirable because it causes thermal expansion of conductors, causing them to sag. In extreme cases such as failures, resistive heating can literally melt wires.
Calculation of resistive heating
There are two simple formulas for calculating the amount of heat dissipated in a resistor (that is, an object with a certain resistance). This heat is measured in terms of power, which is energy per unit of time. So, we calculate the rate at which energy is converted into heat inside a conductor. The first formula is:
P = I x V
where P is the power, I the current through the resistance, and V the voltage drop across the resistance.
Power is measured in units of watts (W), which corresponds to amps x volts. So, a current of one ampere flowing through a resistor across a voltage drop of one volt produces one watt of heat. Units of watts can also be expressed in joules per second. To conceptualize the magnitude of a watt, it helps to consider the heat created by a 100-watt bulb or a 1000 watt freestanding heater.
The relationship P = I x V makes sense if we recall that voltage is a measure of energy per unit charge, while current is charge flow. So the product of current and voltage tells us how many electrons are “going through,” multiplied by the amount of energy each electron loses as heat, resulting in an overall rate of heat production. We can write this as:
and see that, with the cancellation of the load, the units of current multiplied by the units of voltage effectively give us units of power.
The second formula for calculating resistive heating is as follows:
P = I 2 x R
where P is power, I is current and R is resistance. This equation could be deduced from the first one by replacing I. R for V (according to Ohm’s law). This second formula is more frequently used in practice to calculate resistive heating, while the first formula has other more general applications.
As one might deduce from the equation, the units of watts also correspond to amps 2 x ohms (A 2 x Ω). So a current of one ampere flowing through a wire with an ohm resistance would heat that wire at a rate of one watt. Because the current is squared in the equation, two amps passing through the same wire would heat it at a rate of 4 watts, and so on.
A toaster oven consumes a current of 6 A at a voltage of 120 V. It dissipates 720 W in the form of heat. We can see it in two ways:
First, using P = I x V, 120 V x 6 A = 720 W. We could also use the resistor, which is 20 V (20 V x 6 A = 120 V), and write P = I 2 x R / (6A) 2 x 20 Ω = 720 W.
It is important to clearly distinguish how power depends on resistance, current, and voltage, as they are all interdependent. Obviously, the dissipated power will increase with increasing voltage and current. From the formula P = I 2 x R, we can also expect the power to increase with the increasing resistance, assuming that the current remains constant. However, it may be incorrect to assume that we can vary the resistance without varying the current.
In particular, in many situations, it is the voltage that remains (approximately) constant. For example, the voltage at the customer’s wall outlet ideally remains at 120V, regardless of how much power is consumed. Resistance is determined by the physical properties of the device: its intrinsic design and, if applicable, a power setting (such as “high” or “low”). Depending on the standard voltage, the resistance, therefore, determines the amount of current “consumed” by the device according to Ohm’s law: a higher resistance means a lower current, and vice versa. In fact, resistance and current are inversely proportional: if one double, the other is halved.
So what is the effect of resistance on energy consumption?
The key here is that resistive heating is dependent on the square of the current, which means that power is more sensitive to changes in current than resistance. Therefore, at constant voltage, the effect of a change in current is greater than the effect of a corresponding change in resistance. For example, a decrease in resistance (which in itself would tend to decrease resistive heating) results in an increase in current, which accordingly increases resistive heating. Thus, at constant voltage, the decreasing resistance has the net effect of increasing power consumption. A device that consumes more power has lower internal resistance.
For an intuitive example, consider the extreme case of a short circuit, caused by effectively zero resistance (usually unintentional). Suppose a thick metal bar is placed between the terminals of a car battery. A very large current would flow, the metal would become very hot, and the battery would be discharged very quickly. If a similar experiment were done on a wall outlet by sticking a fork in it, for example, the high current would be interrupted by the circuit breaker before the fork or wires melted (DO NOT try this!).
The other extreme case is simply an open circuit, where the two terminals are separated and where the air resistance between them is infinite: here the current and the consumption are obviously zero.
Consider two incandescent bulbs, with 240V and 480V resistors. How much power do they each draw when plugged into a 120V outlet?
We must first calculate the current through each bulb, using Ohm’s law: by substituting V ¼ 120 V and R1 ¼ 240 V in V = I x R, we get I 1 = 0.5 A. For R 2 = 480 V, we have I 2 = 0.25 A.
Now we can use these values for I and R in the power formula, P = I 2 x R, which gives P 1 = (0.5 A) 2 x 240 V = 60 W and P 2 = (0.25 A) 2 x 480 V = 30 W.
We see that at constant voltage, the bulb with twice the resistance draws half the power.
There are, however, other situations where current rather than voltage is constant. Transmission and distribution lines are important cases. Here, the reasoning suggested above does apply, and the resistive heating is directly proportional to the resistance. The important difference between power lines and appliances is that for power lines, the current is not affected by the resistance of the line itself; rather, it is determined by the load or power consumption at the end of the line (because the resistance of the line itself is very small and insignificant compared to that of devices at the end, so any reasonable change in the resistance of the line will have a negligible effect on the overall resistance, and therefore the current flowing through it).
However, the voltage drops along the line (i.e., the voltage difference between its ends, not to be confused with the line voltage with respect to earth) is not constrained and vary in line current and resistance. So Ohm’s law is still valid, but now it is I who is fixed and V and R which vary. Apply the formula P = I 2 x R for resistive heating with the current kept constant, we see that doubling the resistance of the line will double the resistive losses.
In practice, it is desirable to minimize resistive losses on power transmission and distribution lines, these conductors are chosen with the minimum practically and economically achievable resistance.