# Voltage Across a Resistor

#### lenoplix

Joined Jun 11, 2020
16
Hi guys,

So I have a dc motor that I am using as a generator, at a 500 RPM if I connect a voltmeter across the terminals I get almost 5 Volts, however the moment I put a resistor across the motor terminals that is rated at 0.5W and 5ohm and try to measure the voltage across the resistor, I get almost 1.5V and lower values... Could this be due to the power rating of the resistor? Will a higher power such as a 10w resistor work better?

Joined Feb 20, 2016
3,595
The power rating of the resistor is not the problem.
The voltage reading is dropping because the generator is loaded. Just the same as a motor can run faster with no load, but will slow down as the load is increased. That is one way to look at it anyway.
If you measure 1.5V and it is a 5 ohm resistor, the current is Voltage/Resistance = 1.5/5 = 0.3Amps.
The power dissipated in the resistor is Voltage x Current = 1.5 x 0.3 = 0.45Watts.
A larger wattage resistor will still dissipate the same, but it is capable of higher power. It will not cause more power to be dissipated.

#### lenoplix

Joined Jun 11, 2020
16
Shouldnt the voltage across the resistor be somewhere near 5V? given that the generator is supplying 5v when not loaded?

#### lenoplix

Joined Jun 11, 2020
16
Shouldnt the voltage across the resistor be somewhere near 5V? given that the generator is supplying 5v when not loaded?
assuming the rotationial speed of the motor shaft is maintained constant

Joined Feb 20, 2016
3,595
You get losses when supplying current. There is internal resistance that is in series with the load, and so the voltage is divided between this internal resistance and the load.
Other losses too.

#### lenoplix

Joined Jun 11, 2020
16
Ofcourse there are other losses too, but i get good readings when I use a higher ohm resistor such as a 220ohm the voltage would be closer to 5v across the resistor.
N.B the resistor is in parallel with the motor.

#### jpanhalt

Joined Jan 18, 2008
10,087
Just use a larger resistor* and the voltage will increase.

*Edit: meaning higher resistance

Last edited:

Joined Feb 20, 2016
3,595
N.B the resistor is in parallel with the motor.
But the current is flowing through the generator then to the load. They are in series. One is the source and the other is the load.

#### lenoplix

Joined Jun 11, 2020
16
Just to be sure, cause electronics is not my thing and I have to write some reports
Just use a larger resistor and the voltage will increase.
To make sure the resistor is suitable I have to check the rated power and calculate if the resistor is suitable using P=V^2/R no?

Joined Feb 20, 2016
3,595
Yes, you have the formula right. It is always a good idea to have the resistor power rating higher that the actual dissipation. So if you are worried about it, use a 1W resistor.
What are you actually trying to run with this setup?

#### jpanhalt

Joined Jan 18, 2008
10,087
Just to be sure, cause electronics is not my thing and I have to write some reports

To make sure the resistor is suitable I have to check the rated power and calculate if the resistor is suitable using P=V^2/R no?
In theory, yes. But you have already determined that generator puts out roughly 0.45 W. Resistor wattage ratings are actually a heat rating. A 0.25 W resistor will get quite warm if you leave it connected and it's dissipating 0.45 W. But for the time needed to measure your output under load, it is probably OK. From a practical and cost standpoint, 0.25 W and 0.5 W resistors are common and cheap.

Moreover, as you increase the resistance current should decrease and heat will likely decrease.

#### MrChips

Joined Oct 2, 2009
21,664
You have to write a report for a school project.

Hence show in your report the output voltage with no resistor connected and with different resistors connected, 200, 100, 50, 22, 10, 5-ohm (all ½W).
Calculate the power delivered to the load for each case.

Discuss your observations. This would make for a nice report.

#### MrAl

Joined Jun 17, 2014
7,757
Hi guys,

So I have a dc motor that I am using as a generator, at a 500 RPM if I connect a voltmeter across the terminals I get almost 5 Volts, however the moment I put a resistor across the motor terminals that is rated at 0.5W and 5ohm and try to measure the voltage across the resistor, I get almost 1.5V and lower values... Could this be due to the power rating of the resistor? Will a higher power such as a 10w resistor work better?

Hello there,

The voltage drops as the load resistance goes down that is because of the internal resistance as others have pointed out.

You measure:
Vout=5v open circuit (almost)
Iout=1.5/5=0.3 amps

The internal resistance is then about:
Rin=(5-1.5)/0.3=35/3 Ohms
which is approximately 11.666667 Ohms.

The output power is (R1 is internal resistance R2 is load resistance and Vcc=5v):
Pout=(Vcc^2*R2)/(R2+R1)^2

and from this we can calculate the max power output occurs when:
R2=R1

and at that point the output power is:
P=(Vcc^2*R2)/(R2+R1)^2

and so the maximum power output is:
0.535714 watts

That is the maximum power output at the speed you are using and that can only be had with a load resistance of about 11.67 Ohms. Any other load resistance and the output power drops because either the output voltage goes down or the output current goes down and that means the product of current times voltage (power) goes down.

#### Taff The Soundman

Joined Jul 16, 2020
3
Simply put, the higher the current that you try to draw from the generator. the lower the voltage that will be available. there is a fixed maximum power that the generator will supply.

#### crutschow

Joined Mar 14, 2008
25,264
To summarize, the generator has an internal resistance that is in series with the external load, which form a simple voltage divider.
The output voltage is thus determined by the voltage division between the internal and external resistances.

Do you know the formula to determine the voltage for a two resistor voltage divider?

#### ci139

Joined Jul 11, 2016
1,675
So I have a dc motor that I am using as a generator
such setup gives you in most cases some 5 to 20 % efficiency

#### Tonyr1084

Joined Sep 24, 2015
4,910
The motor is a power source. ALL unregulated power sources will have its highest voltage output with no load. As the load goes up the voltage will drop. A 5Ω resistor is a bigger load to the power supply than a 220Ω resistor. That's why you're seeing higher voltages with higher ohm resistors.

Keep in mind, the motor is designed to run on a specific voltage. Hence, it is a load to some other power source. Or was at one time. The brushes and connections within the motor were designed to handle the amperage needs of the motor. So if yours is a 5 volt motor that drew 450 mA (0.45 A) then the internal leads are designed to handle 1.3 to 1.5 times that current. Putting a 5Ω resistor in series with a motor that is being forced to spin will drop the voltage considerably.

#### WBahn

Joined Mar 31, 2012
26,064
Hi guys,

So I have a dc motor that I am using as a generator, at a 500 RPM if I connect a voltmeter across the terminals I get almost 5 Volts, however the moment I put a resistor across the motor terminals that is rated at 0.5W and 5ohm and try to measure the voltage across the resistor, I get almost 1.5V and lower values... Could this be due to the power rating of the resistor? Will a higher power such as a 10w resistor work better?

What is it that you are trying to accomplish? Is this an experiment for a school lab assignment? Are you actually trying to use this as a generator to power something? What is the goal?

What would you expect the voltage to be across a zero ohm resistor? Would you expect it to still be 5 V? Of course not, since that would imply infinite current in the resistor. So it shouldn't be surprising that as you go from open circuit to short circuit that the voltage across the load will drop?

What is driving the motor shaft? Is it being regulated to maintain the 500 RPM, or is it allowing the motor speed to decrease as it is loaded down?

Use a variety of different resistors and plot the voltage, the current, and the power for each of them (you only need to measure the voltage in each case). What conclusions can you draw from the plots?