# Ohm's law and heat generation

#### Loker

Joined Apr 29, 2015
6
I'm reading a textbook with a section on DC motors that claims the following:

"A seemingly obvious approach to control the speed of a dc motor would be simply to limit the current flow by using a potentiometer, as shown in the circuit to the left in the figure (the figure is a motor and potentiometer in series). According to Ohm’s law, as the resistance of the pot increases, the current decreases, and the motor will slow down. However, using a pot to control the current flow is inefficient. As the pot’s resistance increases, the amount of current energy that must be converted into heat increases."

But according to Ohm's law, shouldn't the power through the resistor (and thus the heat generated) increase as the resistance decreases? What am I missing?

#### ericgibbs

Joined Jan 29, 2010
10,189
hi Loker,
For the resistor W= I^2 * R.
E

#### ericgibbs

Joined Jan 29, 2010
10,189
hi Loker,
Look at this LTS sim of the set up. Rs = 1R to 10R
L1 is the motor.
E

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#### Loker

Joined Apr 29, 2015
6
Hi Eric, thanks for the response. Per Ohm's law, if you have a fixed voltage, the current decreases when the resistance increases, so to suggest the power will increase because the resistance increases is false:

With a fixed voltage (say 9V) across a variable resistor (say 10k - 100k), to find the power dissipated by the resistor at 10k, we first find the current : V=I*R so the current is 900mA. Since P=IV,

The power dissipated by the 10k resistor is 8.1mW.

To find the power dissipated by the resistor at 100k, we first find the current : V=IR so the current is 90mA. Since P=IV,

The power dissipated by the 100k resistor is 810uW.

Since the power dissipated by the larger resistor is substantially smaller, the heat generated is also substantially smaller. This seems to contradict what the textbook claims, and I can't find any logical reconciliation.

#### ericgibbs

Joined Jan 29, 2010
10,189
hi L,
A 100K series resistor is a very high value, what is the motor resistance.?
BTW: using a series resistor for motor speed control is a bad idea, when you consider the effect it will have on the motor Torque.
E

#### Loker

Joined Apr 29, 2015
6
The textbook is explaining why a series variable resistor is a bad choice for motor control, I'm just trying to explain to myself why it seemingly contradicts Ohm's law (I'm sure it doesn't, but I can't prove that mathematically). My example above explains the "apparent" contradiction:

When you increase the resistance of a resistor, the power through it drops. I can't grasp why this should change just because the resistor is in series with a motor.

The apparent contradiction is independent of the size of the resistor.

#### BR-549

Joined Sep 22, 2013
4,938
The power of the circuit did not increase. It shifted......from the motor to the resistor.

#### ericgibbs

Joined Jan 29, 2010
10,189
hi,
'I think' what the writer of that opening paragraph was trying to explain, is the ratio of the power lost in the resistance compared to the usable power of the motor becomes more inefficient as the series resistance increases.
E

#### Loker

Joined Apr 29, 2015
6
Got it, thanks a bunch!

#### kubeek

Joined Sep 20, 2005
5,724
The power dissipated in the resistor starts at 0W for 0ohms, then increases and peaks at the point where the resistor is the same resistance as the motor. After that the dissipated power decreases again as the resistance goes higher.

#### MrChips

Joined Oct 2, 2009
21,325
The power dissipated in the resistor starts at 0W for 0ohms, then increases and peaks at the point where the resistor is the same resistance as the motor. After that the dissipated power decreases again as the resistance goes higher.
That is true for a resistive load. Is it true for a motor?

#### Lo_volt

Joined Apr 3, 2014
97
I'll agree, the textbook writer's explanation is a ambiguous. The writer likely didn't want to spend too much time on the least efficient method of controlling motor speed.

The better way to look at the efficiency is to calculate the power coming from the supply, simply volts x current, and to compare that to the power being consumed by the motor, again volt x current. The difference is the power burned in the resistor. The ratio is the efficiency. Note that this simplifies the problem by ignoring that fact that motors are inductive loads, but it still illustrates efficiency. Putting a resistor in the current path simply turns the energy into heat that is lost. There are more efficient ways to control motor speed.

Loker you have a good grasp of Ohm's law. You appear to have caught the author in a slip-up. Read on, I'm sure the book has more to offer!

#### ebeowulf17

Joined Aug 12, 2014
3,274
Hi Eric, thanks for the response. Per Ohm's law, if you have a fixed voltage, the current decreases when the resistance increases, so to suggest the power will increase because the resistance increases is false:

With a fixed voltage (say 9V) across a variable resistor (say 10k - 100k), to find the power dissipated by the resistor at 10k, we first find the current : V=I*R so the current is 900mA. Since P=IV,

The power dissipated by the 10k resistor is 8.1mW.

To find the power dissipated by the resistor at 100k, we first find the current : V=IR so the current is 90mA. Since P=IV,

The power dissipated by the 100k resistor is 810uW.

Since the power dissipated by the larger resistor is substantially smaller, the heat generated is also substantially smaller. This seems to contradict what the textbook claims, and I can't find any logical reconciliation.
You're neglecting two things in your mathematical description here.

First, the motor itself has resistance, so you must add that in to your total resistance in order to calculate current through the circuit (as @MrChips hinted above, motors can have very complicated non-ohmic behavior, so you can't always think of them simply as a resistor in these calculations. Nevertheless, they'll present some impedance/resistance of some sort.)

The voltage dropped across the resistor is the total voltage minus the voltage dropped across the motor. The balance of these voltage splits depends on the resistances of each item.

What all this means is that power dissipation in the resistor as a function of resistance isn't linear - it's low at both extremes, with a high spot in the middle:

Very high pot resistance: the high resistance limits current such that power dissipation is very low.

Very low pot resistance: although current through the circuit is very high, most of the resistance is coming from the motor, which means most of the voltage is dropped across the motor. Since voltage drop across the resistor is very low, dissipation there is also very low.

Pot resistance equal to motor resistance: in this scenario, the resistor is dropping half the total circuit voltage, and current is only reduced to half of its highest possible value. As such, significant dissipation occurs in the pot.

#### WBahn

Joined Mar 31, 2012
25,910
Hi Eric, thanks for the response. Per Ohm's law, if you have a fixed voltage, the current decreases when the resistance increases, so to suggest the power will increase because the resistance increases is false:

With a fixed voltage (say 9V) across a variable resistor (say 10k - 100k), to find the power dissipated by the resistor at 10k, we first find the current : V=I*R so the current is 900mA. Since P=IV,

The power dissipated by the 10k resistor is 8.1mW.

To find the power dissipated by the resistor at 100k, we first find the current : V=IR so the current is 90mA. Since P=IV,

The power dissipated by the 100k resistor is 810uW.

Since the power dissipated by the larger resistor is substantially smaller, the heat generated is also substantially smaller. This seems to contradict what the textbook claims, and I can't find any logical reconciliation.
You've gone and changed the problem. The problem had the pot in series with a motor, your new problem here has the pot connected across a fixed voltage source directly. Different problem, different result.

The use of a motor is a poor choice for this problem if the intent it to make a point like this about heat dissipation in the pot. A better choice would be perhaps a heating element or even just a "device" that has a fixed resistance.

Put your 100 kΩ pot in series with a fixed 50 kΩ resistor and pot the combination across a fixed voltage source. Then plot the power dissipated in each, as well as the total power dissipation, as you sweep the pot from 0 Ω to 100 kΩ.

Joined Jul 18, 2013
20,877
The other thing that makes it a bad example of motor control, is that with a DC motor, BEMF is generated that opposes the applied voltage and varies with RPM and load.
Max.

#### crutschow

Joined Mar 14, 2008
25,116
That is true for a resistive load. Is it true for a motor?
It's true when the resistive motor impedance, due to its load, equals the resistor value.
That point, of course, depends upon the nature of the motor load.

#### kubeek

Joined Sep 20, 2005
5,724
That is true for a resistive load. Is it true for a motor?
it is true for a motor without a load. And I think it is adequate to explain the reasoning to the the OP. How a loaded motor would behave is not possible to answer without knowing what is the load and therefore how it behaves. If it is a constant power load like a power generator then if the series resitance goeas above some point the motor will suddenly stop and all power will be dissipated by the series resistor. Something like a fan or propeller will slow down a bit more steeply than it would if it were just the power dissipated on a fixed resistor.

Anyway, I was hoping to show OP where and how the original text had ommited the area of operation, where the increase/decrease in resistance does not match the behavior described in the book.

#### wayneh

Joined Sep 9, 2010
16,398
I don't know if it helps the TS grasp the issue but you run into a similar thing when you pair a generator to a load. Both have some impedance. The most power you can drive into the load happens when they both have the same impedance. If the load is bigger than that (less resistance), the voltage from the generator falls off. The current goes up all the way to a dead short, but the power dissipated by the load goes down because of the voltage sag. If the load is smaller (higher resistance) than the generator impedance, the voltage across the load increases but the current drops too much and again the power goes with it.

Point is, there's an optimum load resistance, and the response to a change in load resistance depends on which side of the peak you're on.

#### sparky 1

Joined Nov 3, 2018
256
@Loker
yes, one example is an automobile starter motor that has sloppy old bearings.
Higher battery voltage (doing a jump start from a fresh battery) is only a slight improvement.

You are correct to consider heat. A DC motor can either make heat or make torque.
The voltage meter will reflect a significant drop in voltage that reflects the drag (heat)
The drag is a resistance that results in heat. It can be studied with more elaborate
equation than Ohms law.

The VFD motors (variable frequency drive motors) that are beginning to show up in power tools are really cool, they have torque at low speeds as well. An example might be found in a cordless wood router with VFD controller, very impressive capabilities.

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#### crutschow

Joined Mar 14, 2008
25,116
This adjustment of load impedance to source impedance is commonly done with solar panels and an MPPT battery charger.
The MPPT algorithm constantly adjusts the load so that it matches the solar panel impedance to extract the maximum power from the panel.
This MPPT adjustment has to be constantly done, since the effective panel impedance varies with the sun intensity and angle.