However, in circuits containing more than one resistor, we must be careful in how we apply Ohm's Law. In the three-resistor example circuit below, we know that we have 9 volts between points 1 and 4, which is the amount of electromotive force trying to push electrons through the series combination of R1, R2, and R3. However, we cannot take the value of 9 volts and divide it by 3k, 10k or 5k Ω to try to find a current value, because we don't know how much voltage is across any one of those resistors, individually.

This page says that we can calculate the voltage drop across multiple resistors in series with each other, by using a total current calculation figure, but the total current calculation figure seems like you just pick and arbitrary number, so I'm not sure I understand.


If you wanted to offer a different example, that applies to a circuit I am currently working on, I am trying to calculate the voltage drop across different resistors on a 5vdc regulated circuit. Does the math work out to (with a 20k total resistance) you'd lose .250vdc for each 1k of resistance?



http://www.allaboutcircuits.com/vol_1/chpt_5/2.html

Now we have all the necessary information to calculate circuit current, because we have the voltage between points 1 and 4 (9 volts) and the resistance between points 1 and 4 (18 kΩ):

Knowing that current is equal through all components of a series circuit (and we just determined the current through the battery), we can go back to our original circuit schematic and note the current through each component:

Now that we know the amount of current through each resistor, we can use Ohm's Law to determine the voltage drop across each one (applying Ohm's Law in its proper context):

Notice the voltage drops across each resistor, and how the sum of the voltage drops (1.5 + 5 + 2.5) is equal to the battery (supply) voltage: 9 volts. This is the third principle of series circuits: that the supply voltage is equal to the sum of the individual voltage drops.

The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a series circuit, and because free electrons flow through conductors like marbles in a tube, the rate of flow (marble speed) at any point in the circuit (tube) at any specific point in time must be equal.
From the way that the 9 volt battery is arranged, we can tell that the electrons in this circuit will flow in a counter-clockwise direction, from point 4 to 3 to 2 to 1 and back to 4. However, we have one source of voltage and three resistances. How do we use Ohm's Law here?
An important caveat to Ohm's Law is that all quantities (voltage, current, resistance, and power) must relate to each other in terms of the same two points in a circuit. For instance, with a single-battery, single-resistor circuit, we could easily calculate any quantity because they all applied to the same two points in the circuit:

Since points 1 and 2 are connected together with wire of negligible resistance, as are points 3 and 4, we can say that point 1 is electrically common to point 2, and that point 3 is electrically common to point 4. Since we know we have 9 volts of electromotive force between points 1 and 4 (directly across the battery), and since point 2 is common to point 1 and point 3 common to point 4, we must also have 9 volts between points 2 and 3 (directly across the resistor). Therefore, we can apply Ohm's Law (I = E/R) to the current through the resistor, because we know the voltage (E) across the resistor and the resistance (R) of that resistor. All terms (E, I, R) apply to the same two points in the circuit, to that same resistor, so we can use the Ohm's Law formula with no reservation.

 

You've answered your own question within your post, so I'm not sure if you still have a question or you've simply stated a learnig path.

 

Ok, I guess I simply stated a learning path, haha

Scenario - I have a 5v regulator and need to output 1.75v and 1.5v . I've been using potentiometers, but they are getting wiped somehow and not offering any resistance.

So If I wanted to use less than (20) 1k resistors in series, I could use a 12k resistor to drop to 2 vdc and then to step it down to 1.75vdc and 1.5vdc, using (2) 1k resistors respectively?

How do I know what wattage to use, for this?

You've answered your own question within your post, so I'm not sure if you still have a question or you've simply stated a learning path.

 

say because .25 volts is your smallest increment, then the 5v supply / .25 = 20. I suspect thats how you came out with the 20 1k resistors.

So starting from common (or ground as some would say), working back towards the supply.

1.5 volts require 1.5/.25 = 6 1Kohm. Because your next level is .25 volts higher, we'll add another 1K in series to obtain that voltage. Your next step is the 5v supply, which is a difference of (5-1.75)/.25 = 13 1kohm. So now add them up and you get your 20 1K resistors.

Calculating this example, the resistors in series is a total of 20Kohm. At the supply of 5v that will deliver 5/20K = .25mA, so a total of .25mA*1K = .25v drop across each resistor.

Wattage = VA, so your 13Kohm resistor drops 3.25v at .25mA, your dispatted wattage being .8 milliwatt.

So now you have your refernce voltages, but they are only that. Once you add any load, you will need to recalculate based on the effects that the load contribute to the circuit.

 

Hi,

Thanks for the reply, so am I understanding you correctly, per my diagram?

(Keeping the multimeters negative probe on the negative side of the battery and moving the positive probe to the different locations in between the resistors that I have marked?)

say because .25 volts is your smallest increment, then the 5v supply / .25 = 20. I suspect thats how you came out with the 20 1k resistors.

So starting from common (or ground as some would say), working back towards the supply.

1.5 volts require 1.5/.25 = 6 1Kohm. Because your next level is .25 volts higher, we'll add another 1K in series to obtain that voltage. Your next step is the 5v supply, which is a difference of (5-1.75)/.25 = 13 1kohm. So now add them up and you get your 20 1K resistors.

Calculating this example, the resistors in series is a total of 20Kohm. At the supply of 5v that will deliver 5/20K = .25mA, so a total of .25mA*1K = .25v drop across each resistor.

Wattage = VA, so your 13Kohm resistor drops 3.25v at .25mA, your dispatted wattage being .8 milliwatt.

So now you have your refernce voltages, but they are only that. Once you add any load, you will need to recalculate based on the effects that the load contribute to the circuit.

 

I guess I did something wrong, because when I tried to do this in practical application, it does not work out to even close to the right voltages?

 

Hello,

What kind of multimeter did you use?
The total current is only 0.25 mA.
Any load will lead to a misreading of the voltages.

Greetings,
Bertus

 

I used an auto ranging digital multimeter, why, should I use something different?

Hello,

What kind of multimeter did you use?
The total current is only 0.25 mA.
Any load will lead to a misreading of the voltages.

Greetings,
Bertus

 

Hello,

Do you know the input resistance of your meter?
This resistance is parallel to the resistors when you measure the voltage.

Greetings,
Bertus

 

Most decent DVMs have a 10MΩ input impedance. VOMs are considerably lower, around 100KΩ is typical (or less).

 

Question - how close is "not even close"? What is the tolerance of those resistors? How close to 5 VDC is your source - how many decimal places does your meter give? What are all the actual voltages measured?

 

Which leads me to another thought. If you are touching the meter leads your body's resistance could be pretty low, throwing off the readings. People in general are pretty variable as far as their internal resistance goes.

 

Can't I calibrate my DMM and be ok as far as resistance from the leads and wire?

My voltage values do not have to be exact, but I believe my hurdle is understanding different ways I could have 1.5v and 1.75v and 2.0v in my arrangement of resistors?

Do I have to have (20) 1k resistors, or could I have (1) 10k resistor and (10) 1k resistors? And in the second scenario, does the large resistor go closer to the positive voltage or ground side of the resistor arrangement?

 

The DVM input resistance is fixed, defined by design. Most cases this is 10MΩ, which is pretty invisable for resistance measurements. Since we don't know the meter you're using we can only speculate, but it is an educated guess.

I haven't been following the rest of this thread too closely, so I'll let someone else pick up the slack.

I will say this, figure total resisance, then resisance to the common point, and you can calculate your voltage pretty simply.

 

I'm using an inexpensive radio shack multimeter

I just went out and bought (20) 1k resistors, and I'm able to get close to the voltage resolution I need, but I'm sure this is an inefficient way of doing this, and I could use larger value resistors in place of some of the resitors (to replace the areas that I don't need to any voltage value from) to get the same resolution ... but I don't know how to calculate for this?

 

Ok,

This seems to easy, but maybe this is correct

Take the dc voltage and divide it by the number of total resistance and you have your resolution.

Then find the window of that voltage that you want and you can use large resistor values until you get down do the voltage window that you need small increments in?

(E/R) * % of total voltage that window resolution is

??

 

There are several problems in previous posts to choose from. Which one are you referring to? Or are you trying another example?

 

Let's go back to your:

Scenario - I have a 5v regulator and need to output 1.75v and 1.5v . I've been using potentiometers, but they are getting wiped somehow and not offering any resistance.

First, let's decide on a range of current that you need in this scenario. Let's just say you simply needed the voltages present, and you weren't going to power anything. Somewhere around 1mA should be enough.

So the question now is, how much resistance is needed to limit current to 1mA with a 5v source?
R = E/I, or Resistance(Ohms) = Voltage/Current(Amperes)
Since E=5v, and I=1mA = 0.001 Amperes...
R = 5,000 Ohms.

So to get 1.5v, we know that R = E/I, or Resistance(Ohms) = Voltage / Current(Amperes).
Since E=1.5v and I = 1mA, R= 1.5/1mA = 1.5/0.001 = 1,500 Ohms (1.5k)

Now you need 1.75v - but that's just 0.25v higher than the 1.5v you already have.
So to go 0.25 higher than the 1.5v to get the 1.75v, you'll use R=E/I again.
R = 0.25v/1mA = 250 Ohms.

We're up to 1.75v, and have to drop the rest of the 5v. 5v-1.75v=3.25v.
R = 3.25v/1mA = 3250 Ohms.

Another way to look at it would be since 5v/5,000 Ohms = 1mA, and 1mV/1 Ohm = 1mA, for each Ohm resistance there is 1mV dropped in this particular circuit that has a total resistance of 5,000 Ohms.

 

I'm trying to see if I understand the calculations correctly by presenting a different example.

I was able to raise my resolution by raising my resistance to 40k, and then the 5v signal was decreasing by .125 for each 1k of resistance. Since I wanted to key off of the voltage values of 2v, 1.75v and 1.5v, I used (2) 10k resistors close to the voltage regulator, then (10) 1k resistors in series following that and then (1) more 10k resistor after that to finalize the 40k of total resistance. This gave me 2vdc @ 24k of resistance, 1.75vdc at 26k of resistance and 1.5vdc at 28k of resistance.

If I had a 24k resistor and a 12k resistor, I'd only have to use (6) total resistors, but unfortunately I don't have those values, so I had to use more resistors in series.

Am I understanding what I did correctly?

There are several problems in previous posts to choose from. Which one are you referring to? Or are you trying another example?

 

You had a schematic for the previous example. Schematics are the language of electronics, they have no ambiguity. So lets start from this example:

You tell us the values of Vcc and R1-6. We'll show you how to solve for them.