Whenever an electric current flows through a material that has some resistance (i.e., anything but a superconductor), it creates heat. This resistive heating is the result of “friction,” as created by microscopic phenomena such as retarding forces and collisions involving the charge carriers (usually electrons); in formal terminology, the heat corresponds to the work done by the charge carriers in order to travel to a lower potential.
This heat generation may be intended by design, as in any heating appliance (for example, a toaster, an electric space heater, or an electric blanket).
Such an appliance essentially consists 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. For one, their purpose is to transmit energy, not to dissipate it; the energy converted to heat along the way is, in effect, lost (thus the term resistive losses). Furthermore, resistive heating of transmission and distribution lines is undesirable, since it causes thermal expansion of the conductors, making them sag. In extreme cases such as fault conditions, resistive heating can literally melt the wires.
Calculating Resistive Heating
There are two simple formulas for calculating the amount of heat dissipated in a resistor (i.e., any object with some resistance). This heat is measured in terms of power, which corresponds to energy per unit time. Thus, we are calculating a rate at which energy is being converted into heat inside a conductor. The first formula is:
P = I x V
where P is the power, I is the current through the resistor, and V is the voltage drop across the resistor.
Power is measured in units of watts (W), which correspond to amperes x volts. Thus, 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 as joules per second. To conceptualize the magnitude of a watt, it helps to consider the heat created by a 100-watt light bulb, or a 1000-watt space heater.
The relationship P = I x V makes sense if we recall that voltage is a measure of energy per unit charge, while the current is the flow rate of charge. The product of current and voltage therefore tells us how many electrons are “passing through,” multiplied by the amount of energy each electron loses in the form of heat as it goes, giving an overall rate of heat production. We can write this as:
and see that, with the charge canceling out, units of current multiplied by units of voltage indeed give us units of power.
The second formula for calculating resistive heating is:
P = I2 x R
where P is the power, I is the current, and R is the resistance. This equation could be derived from the first one by substituting I . R for V (according to Ohm’s law). This second formula is more frequently used in practice to calculate resistive heating, whereas the first formula has other, more general applications.
As we might infer from the equation, the units of watts also correspond to amperes2 x ohms (A2 x Ω). Thus, a current of one ampere flowing through a wire with one ohm resistance would heat this wire at a rate of one watt. Because the current is squared in the equation, two amperes through the same wire would heat it at a rate of 4 watts, and so on.
A toaster oven draws a current of 6 A at a voltage of 120 V. It dissipates 720 W in the form of heat. We can see this in two ways:
First, using P= I x V, 120V x 6 A = 720 W. Alternatively, we could use the resistance, which is 20 V (20 V x 6 A = 120 V), and write P = I2 x R / (6A)2 x 20 Ω = 720 W.
It is important to distinguish carefully how power depends on resistance, current, and voltage, since these are all interdependent. Obviously, the power dissipated will increase with increasing voltage and with increasing current. From the formula P = I2 x R, we might also expect power to increase with increasing resistance, assuming that the current remains constant. However, it may be incorrect to assume that we can vary resistance without varying the current.
Specifically, in many situations it is the voltage that remains (approximately) constant. For example, the voltage at a customer’s wall outlet ideally remains at 120 V, regardless of how much power is consumed. The resistance is determined by the physical properties of the appliance: its intrinsic design, and, if applicable, a power setting (such as “high” or “low”). Given the standard voltage, then, the resistance determines the amount of current “drawn” by the appliance according to Ohm’s law: higher resistance means lower current, and vice versa. In fact, resistance and current are inversely proportional in this case: if one doubles, the other is halved.
What, then, is the effect of resistance on power consumption?
The key here is that resistive heating depends on the square of the current, meaning that the power is more sensitive to changes in current than resistance. Therefore, at constant voltage, the effect of a change in current outweighs the effect of the corresponding change in resistance. For example, decreasing the resistance (which, in and of itself, would tend to decrease resistive heating) causes the current to increase, which increases resistive heating by a greater factor. Thus, at constant voltage, the net effect of decreasing resistance is to increase power consumption. An appliance that draws more power has a lower internal resistance.
For an intuitive example, consider the extreme case of a short circuit, caused by an effectively zero resistance (usually unintentional). Suppose a thick metal bar were placed across the terminals of a car battery. A very large current would flow, the metal would become very hot, and the battery would be drawn down very rapidly. If a similar experiment were performed on a wall outlet by sticking, say, a fork into it, the high current would hopefully be interrupted by the circuit breaker before either the fork or the wires melted (DO NOT actually try this!).
The other extreme case is simply an open circuit, where the two terminals are separate and the resistance of the air between them is infinite: here the current and the power consumption are obviously zero.
Consider two incandescent light bulbs, with resistances of 240 V and 480 V. How much power do they each draw when connected to a 120 V outlet?
First we must compute the current through each bulb, using Ohm’s law: Substituting V ¼ 120 V and R1 ¼ 240 V into V = I x R, we obtain I1 = 0.5 A. For R2 = 480 V, we get I2 = 0.25 A.
Now we can use these values for I and R in the power formula, P = I2 x R, which yields P1 = (0.5 A)2 x 240 V = 60 W and P2 = (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 other situations, however, where the current rather than the voltage is constant. Transmission and distribution lines are an important case. Here, the reasoning suggested earlier does in fact apply, and resistive heating is directly proportional to resistance. The important difference between power lines and appliances is that for power lines, the current is unaffected by the resistance of the line itself, being determined instead by the load or power consumption at the end of the line (this is because the resistance of the line itself is very small and insignificant compared to that of the appliances at the end, so that any reasonable change in the resistance of the line will have a negligible effect on the overall resistance, and thus the current flowing through it).
However, the voltage drop along the line (i.e., the difference in voltage between its endpoints, not to be confused with the line voltage relative to ground) is unconstrained and varies depending on current and the line’s resistance. Thus, Ohm’s law still holds, but it is now I that is fixed and V and R that vary. Applying the formula P = I2 x R for resistive heating with the current held constant, we see that doubling the resistance of the power line will double resistive losses.
Since in practice it is desirable to minimize resistive losses on power transmission and distribution lines, these conductors are chosen with the minimal resistance that is practically and economically feasible.
Superb article, but unfortunately the problem with the V in place of the Omega is still there – so you have volts and resistance listed with the same unit (four years on from it being pointed out!)
But as long as you are aware of the error as it were, the rest of it is very clear and excellently explained.
I am confused on when to use either of the equations of power
What would the adequate working temperature of an electrical wire leading to a submersed element of 5Kw. (I am not talking about the rated working temperature, but the actual measured temperature). Because if load is adequate, current, resistance, amperage, no shorts and wire are all adequate…what could cause an adequate copper wire to heat (rated 90 Celsius) over the operating temperature of the breaker (rated 40 Celsius)?
I don’t know if someone has been messing with or editing the site but in the worked examples it seems V is used in place of the omega symbol (ohms), except near the end of the first one. That makes it a bit confusing between resistance and voltage. This was the case as at 07/03/2018. Please delete my comment if you manage to correct this.
Electrical heating is explained nicely,however I would like to know the following–
Purpose is heating a block of steel–
assume — uniform cross section eg.round / square etc
— irregular cross sections
Question is – which is preferrable AC or DC supply ?
resistance of the item being heated being fixed what type of control is preferable Current or Voltage control ?
We understood how active power will be lost as heating in transmission lines, the question how the reactive power will be loss? as heat? or another shape?
i am working in dubai as an electrical engineer and i got a job to calculate all the electricity consumption of the industry…i started with a machine autoclave the total connected load as per that installed machine for diffrent loads like heaters motor dryer and air compressers and cooling tanks like 260kw 38kw 8kw 75kw 5kw so i just added them and multiplied by 12 hrs for 12 hrs of process but the result is a large no. of kw its not possible pplease let me know how cn i calculate if not from rated powers on the machines then shud i need to take readings with clamp metrer diffrently and then calculate with pf or else please do helpp…. your reply is highly awaited
You have to check which load is working for how many hours in a day. Whether all the load are fully loaded. Whether all the loads running simultaneously? Check these aspect to calculate power consumption.
A simple but basic question has struck in my mind that given a costant value of voltage if an electric heater has high resistance and hence a low current flowing through it due to high resistance, then how the power consumed by the electric heater is large according to the ohm’s law ???
this question is also haunting me out for the past two days and i still did not get any answer.
Kindly Attn:- Managing Director
This is to inform you that, we are dealing in used Injection Moulding Machinery from 40T to 1000T and Blow Moulding Machine
Injection household like bucket,tubs,stools,containers moulds and Blow moulds refer attachment excel file
Plastics products manufacture investment, we can set the complete project for household plastics products manufacturing
Pls let us know your view , if any doubt or any related issue.
Sir! pls let us know your area plastic industries may require above machinery
save the attachment in your computer for future refer
“WE ASSURE BEST SERVICE AT ALL THE TIME”
If you have household market pls refer attachment which project for HOUSEHOLD plastics manufacturer
For Ampees Enterprises
pls feel free to contact mr patel at 918097321822
Sir,isnt it just a short cicuit dat hapen in electric heaters?plz explain in detail.
In the above example the bulbs resistances were shown as 120 v and 480 v instead of 120 ohms and 480 ohms.
Pl arrange to correct them
Small correction on the bulbs resistance it is 240 ohms and 480 ohms not 120 ohms pl note.
Your explanation if outstanding! But if power drawn increases as resistance decreases, don’t you think it is better to use a metal that has lower resistivity for resistive heating? Such as using a heating element with low resistance in an electrical kettle.
I am thinking as you think at first. But then I realized that when resistance of the heating element is too low, power drawn will be too high and can cause excessive heating on the element. So, the resistance is chosen as low as possible to produce more heat, but with caution to the risk of damage to the element if the power drawn is too much that can’t be withstood by the element. Hope it can help you