3
\$\begingroup\$

I am having difficulty reconciling resistance which provides resistance to current flow and resistance which is used to generate something useful, such as heat or a magnetic field which rotates a motor, etc.

Work requires power and, per P = V*I, power requires current. A resistor limits the amount of amps being drawn, so it would seem the resistor, in the case of an electric load, impedes its own ability to perform work? Intuitively, a load that has the least amount of resistance, drawing more amps, should produce more work. But, useful work is dependent on resistance, so proportional to it. Hence my confusion.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Power lost in an undersized wire going from Point A to Point B is not doing useful work. \$\endgroup\$
    – Hot Licks
    Apr 26, 2020 at 21:16
  • \$\begingroup\$ @HotLicks It can keep the feet of cold birds warm in the winter. \$\endgroup\$
    – pipe
    Apr 26, 2020 at 22:26
  • \$\begingroup\$ @pipe - Maybe 0.01 degree warmer. \$\endgroup\$
    – Hot Licks
    Apr 26, 2020 at 22:36
  • \$\begingroup\$ How much work do you want your lightbulb to do? Do you want the entire power output of the National Grid to light up your living room? \$\endgroup\$
    – user253751
    Apr 27, 2020 at 0:29

3 Answers 3

5
\$\begingroup\$

The electrical power put into a load, V*I, is transferred from the source of the voltage to the load. What happens to that energy put into the load depends, of course, on what that load is.

If the load is a simple resistor, then that energy just goes into producing heat. You are correct that for a pure voltage source with no internal resistance of its own connected to a simple resistor as a load, the higher the resistance, the less energy is dissipated in the resistor as heat (\$P=V^2/R\$). In this scenario, lowering the resistance just keeps increasing the power dissipated in the load.

The situation is more subtle when you consider that all real voltage sources have their own internal resistance. In that case, you find that you get the most energy into the load resistor if the resistance of the load equals the internal resistance of the voltage source (look up Maximum Power Transfer Theorem for more info). What happens is that the power is shared between the internal resistance of the source and the load resistance. If you lower the resistance of the load well below that of the source, then most of the energy from the source voltage just goes into the source resistance instead of the load resistance, and so lowering the load resistance no longer keeps increasing power dissipated in the load resistor.

\$\endgroup\$
4
  • \$\begingroup\$ I think my question is a more basic one. Resistance in the case of a light bulb is what allows it to dissipate heat and light. So, resistance is allowing for the useful work to occur. Yet, resistance decreases the amount of current, which limits "power", or the ability to perform useful work. I am having troubles reconciling the seemingly contradictory parts. \$\endgroup\$
    – Crosshair
    Apr 26, 2020 at 13:34
  • \$\begingroup\$ @Crosshair, read again. This answer really does contain all you need to know to reconcile those facts. It all hinges on the fact that your voltage source is never ideal...The source output always drops if you try to draw too much current from it. You might want to research the maximum power transfer theorem for more detail. \$\endgroup\$
    – The Photon
    Apr 26, 2020 at 13:41
  • \$\begingroup\$ @Crosshair Perhaps it would help to think of what you are holding fixed when you increase resistance. If you consider current fixed, then increasing resistance increases power. If you consider holding voltage fixed across the load, then increasing resistance also has the effect of lowering the current. Since power is proportional to the square of current, this effect wins out over the increase in R. \$\endgroup\$
    – rpm2718
    Apr 26, 2020 at 13:52
  • \$\begingroup\$ In the case of utility power, it is not far wrong to consider it to be a perfect voltage source. Yes, a 100W bulb (in the days of incandescents) had a high enough resistance to not draw more power. You could also buy 200W bulbs if you wanted more light. They had half the resistance so dissipated twice the power. The resistance is chosen to be the right value, not to maximize the power dissipated by the resistor. Hair dryers tend to run 1500-1700 Watts in the US because that is the maximum one can draw on a standard 15A circuit. It is limited not by source resistance but by the breaker. \$\endgroup\$ Apr 27, 2020 at 2:22
4
\$\begingroup\$

Work requires power and, per P = V*I, power requires current.

This is true.

A resistor limits the amount of amps being drawn, so it would seem the resistor, in the case of an electric load, impedes its own ability to perform work?

Consider the casual understanding of "a short circuit": power-supplying wires of opposite polarity touch, there's a bright flash, stuff is melted. This is what happens when you don't have enough resistance: too much current flows, things heat up, it's generally uncontrolled and dangerous. In a practical load, we want a controlled amount of power.

Not every load can be described as a resistor, but in those that can, the resistance is chosen to set the amount of power drawn (given the known supplied voltage). \$P = VI\$ and \$V = IR\$, so we can put those equations together to get

$$R = \frac{V^2}{P}$$

which tells us how to choose the load resistance to choose the power we want.

(However, not every load is a resistor. For example, a DC motor more-or-less maintains a speed determined by the voltage only, while the current varies with the mechanical load: it will draw as much power as necessary to reach and maintain that speed, limited mainly by the series resistance of the windings — which is just wasting heat, and not relevant to the useful load. Thus, motors are designed and selected for a maximum power needed in the application, while the actual electrical power varies as the mechanical load does. Computing devices (that use CMOS logic) follow a similar principle: moment-to-moment, they draw as much power as needed to perform the work, plus some inefficiency.)

Intuitively, a load that has the least amount of resistance, drawing more amps, should produce more work.

If your power source is an ideal (theoretical) voltage source connected directly to your load resistance, then this is true (up to the point where the load resistance is zero and the mathematical solution is undefined).

However, there are several practical considerations:

  • As I mentioned above, you don't want to use unlimited power. You want to use enough that the rest of your machine can survive it, and that it does the job controllably.

  • All wires have resistance, including the wires from your power source to your load. If the load resistance is less than the resistance of the wires, then more than half the power is being dissipated in the wiring — even if they can take the heat, it's not efficient to operate in that condition. And, in practical electrical service, current limits are set by circuit breakers to well under that point.

  • In the case of small power sources like batteries or solar panels such that using all of their capacity is a reasonable idea rather than an explosive one, there are other considerations why you don't want to just minimize the load resistance.

    If your power source is a battery, then there's an effective resistance ("internal resistance") in the construction of the battery (which is partly due to the actual internal conductors and partly due to the maximum rate of the electrochemical reactions). Just like the resistance in the wires, this makes drawing more power inefficient. The maximum power transfer theorem tells us that the maximum possible power drawn is when the load resistance is equal to the source resistance (both internal and in wiring). However, this is also a condition of exactly 50% efficiency (half the power is lost in heating the battery and wiring), whereas much higher efficiency can be obtained if the load resistance is higher than the source resistance.

    For this reason, we have to use bigger batteries for higher-power devices, even if run time were not an issue — we want a battery whose internal resistance is low enough to supply the needed power (and without overheating), and even bigger than that will be more efficient (thus requiring less power in recharging, in the long run).

No matter what the power source, using thicker (lower resistance) power wires will reduce power wasted in heating the wires, and increase the maximum power practically available. To link this back to your original wondering: we do want to have the least amount of resistance, everywhere but the load. The load needs to have a controlled amount of resistance to control how much power is delivered.

\$\endgroup\$
2
  • \$\begingroup\$ This was very helpful, thank you! Why is an inductive load presented on a schematic diagram as a resistor when it isn't a resistor load? Is it because we need to consider on the schematic a given current, and, with a voltage source, current is controlled by resistance, so we treat the inductive load as a resistor? Presumably, there is an actual resistor being used to control the amount of current powering the magnet in a electric motor, but we just treat the entire system (actual resistor + inductive load) as one resistor on the schematic? \$\endgroup\$
    – Crosshair
    Apr 26, 2020 at 16:18
  • \$\begingroup\$ @Crosshair Those are two different questions, really, and could be answered at length in themselves. I recommend you post them separately. Though, to correct an assumption — using a variable resistor to control a motor is an obsolete wasteful way to do it. And in the case I was describing there is no controller — the motor just maintains a single speed. That's all you need for many appliances and power tools. The motor is designed for that speed at that voltage; you choose a different motor instead of adding a resistor. \$\endgroup\$
    – Kevin Reid
    Apr 26, 2020 at 16:28
1
\$\begingroup\$

If you assume the source of power is a reasonably good "voltage source", then the voltage remains reasonably constant as you draw more power from it. Most power sources especially the power grid, act like this. (Some, like solar cells, don't; we can ignore them for now.)

You quote P = V*I, which is correct - power needs both voltage and current.

You also say "Intuitively, a load that has the least amount of resistance, drawing more amps, should produce more work" and indeed this intuition is true.

Let's express the bold section mathematically : I = V/R, or V = I*R - this is Ohm's Law. At a constant voltage, reducing the load resistance increases the current - exactly as per your intuition.

Now put these together : P = V * I and V = I * R, to get P = I * I * R or P = I^2*R.

You can see that as you say, power is proportional to resistance. But it's also proportional to the square of current, so as you reduce resistance, current increases, and power increases faster thanks to the square.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.