UPDATE: there was a schematic missing. Here is the schematic.

Vin= 16V
r1 =15 Kohms
r2=15 kohms

edit: is it 8 volts?

It is.

 

UPDATE: there was a schematic missing. Here is the schematic.

Vin= 16V
r1 =15 Kohms
r2=15 kohms

edit: is it 8 volts?

It sounds like you might still be guessing more than analyzing. If you are just blindly throwing some equation at it that is labeled "voltage divider equation" then stop. You should be able to derive that formula and if you can't, then you need to take the time to learn how because it is an indicator of very weak analysis skill that will come back to haunt you if you don't.

Given the circuit, what is the total current flowing in it?

Given that current, what is the voltage across the resistor that Vout is taken across?

 

No need to find Current

Simply, R2/R r1 * vin = vout

 

No need to find Current

Simply, R2/R r1 * vin = vout

What if "R r1"?

My point is that I'm not convinced that you understand the formula that you are using. Do you understand where that formula comes from? Can you derive it using fundamental circuit analysis techniques? Or are you just pulling it out of a list of formulas and saying, "Gee, this one looks like maybe it might give me the answer I'm probably looking for?"

 

What if "R r1"?

My point is that I'm not convinced that you understand the formula that you are using. Do you understand where that formula comes from? Can you derive it using fundamental circuit analysis techniques? Or are you just pulling it out of a list of formulas and saying, "Gee, this one looks like maybe it might give me the answer I'm probably looking for?"

Typo , R2/R2+R1 * Vin = Vout

and yes to "Gee, this one looks like maybe it might give me the answer I'm probably looking for?". I am really stressed at this point because I am failing this subject.

 

I am really stressed at this point because I am failing this subject.

You've demonstrated that fact here as well.

In the self assessment of your studies, do you know why your failing? Remember, when you do that self-assessment, it's impossible to lie to yourself, as you already know the effort you placed on your studies.

Do you want to go back to the "other" question you guessed or should we just call it good and let you stress yourself out more?

 

You've demonstrated that fact here as well.

In the self assessment of your studies, do you know why your failing? Remember, when you do that self-assessment, it's impossible to lie to yourself, as you already know the effort you placed on your studies.

Do you want to go back to the "other" question you guessed or should we just call it good and let you stress yourself out more?

I don't know.. I just want to get it done so I can do my other homework.

 

I just want to get it done so I can do my other homework.

Ok. We won't go back.

 

I know what you mean by:

R2/R2+R1 * Vin = Vout

but what you wrote is not what you mean. What you wrote is:

(R2/R2)+(R1 * Vin) = Vout

Remember, multiplication and division take precedence over addition and subtraction.

What you meant was

[R2/(R2+R1)] * Vin = Vout

Although the square brackets truly are redundant, they serve to make the intent that much more explicit.

You need to take more time to be more rigorous in your math. This is for a number of reasons. First, if you are sloppy you can't communicate effectively with others, such as the grader, and won't get the amount of partial credit you might otherwise deserve. Second, being sloppy leads to you truly not seeing that what you wrote and what you meant are different. But if you enter your formula into a program or a spreadsheet or a calculator, it will do what you told it to do, not what you meant for it to do. However, you will have a very hard time detecting this because when you look at the formula, what you see is what you wanted it to say, not what it actually says. We all have that problem because, as humans, we all have a pretty well refined ability to look past what was actually written and see what was intended. Since computers and calculators lack that ability, it is up to us to develop the ability to force ourselves to see things as they are actually written, which is a learned skill.

The reason you are failing is because you have not been learning the material, but rather just trying to memorize a bunch of equations and when to use which one. You can't go very far that way because the overwhelming majority of problems you will face are far too complex to lend themselves to a set of formulas that you pick from -- for even pretty simple circuits your list of formulas would end up being the size of a dictionary very quickly. What you need to do is learn the fundamental analysis techniques well enough so that you can apply them to arbitrarily complex circuits with confidence.

A good exercise to help you catch up to where you should be is to start with the fundamental concepts of Kirchhoff's Voltage Law, Kirchhoff's Current Law, and Ohm's Law and from there derive all of the other relationships involving resistive DC circuits that are in your list of formulas. Once you've derived a relationship, you can then use that relationship in future derivations. That means that you can't use the formula for how to combine resistors in series until after you derive it. Once you start working with capacitors and inductors, you add the constitutive relations for those devices (just as Ohm's Law is the constitutive relation for a resistor), namely Q=CV and V=L(di/dt), to your list of fundamental concepts and proceed as before. What would be even better would be for you to go back to your physics intro E&M text and be sure you are comfortable with the concept of voltage and current in terms of even more fundamental concepts based on the forces between charges and the physical work done in moving one charge around in the presence of other charges.

 

I don't know.. I just want to get it done so I can do my other homework.

You are digging yourself an awfully deep hole. The first step in getting yourself out of it is to stop digging. But as long as you continue with this kind of attitude, all you are doing is digging more and more using bigger and bigger shovels. At some point, the sides of this massive hole are going to come tumbling in on top of you.

Oh, wait, you're failing this course -- which means that the sides are already coming down; yet all you want to do is dig some more. That is most definitely a choice that is yours to make.

 

Yea, I didn't write it the way I put on my calculator.

In this class the test are online, so I have no idea what I got right and what I got wrong. Secondly, I have 50% chance of failing. I need to get 100% on the next quiz in order to pass (this is the last quiz) I have finals in 2-3 weeks. I posted another thread http://forum.allaboutcircuits.com/threads/capacitors-charging-equation-help.103591/#post-783923 which the test is going to be on the exact same question. However I agree with you with all the things you have said.

 

Hello,

From the information given so far it looks like you measure the voltage across the lower resistor, and that means the internal resistance of the meter goes in parallel to the lower resistor before you can calculate the voltage at the junction of the two resistors. That's the voltage you will measure on the meter.
HOWEVER, the actual voltage at the junction is not that voltage, it is the voltage WITHOUT the meter actually connected, so you then have to figure out what the voltage would be if the meter was not connected to the circuit, but do this while using the voltage measured while the meter is actually connected.
This is basically called compensating for the meter's internal resistance.

 

You just need to know how the meter, as a load, affects the reading the meter shows.

If you were using a Simpson 260, it's a 20,000 ohms per volt meter. You would need to know the full scale you are using to figure out the input resistance of the meter. You don't have that problem with the DMMs which are typically 1 Megaohm or 10 Megaohm. I would not be surprised to find a 100 Megaohm input impedance out there.

Meters can load the circuit giving you false readings. If the technician isn't aware of that, they could be troubleshooting a circuit that has not failed.

Here is a question for the OP ...

Using the diagram below. What will the DMM read when the linear potentiometer is at 50%?

 

You just need to know how the meter, as a load, affects the reading the meter shows.

If you were using a Simpson 260, it's a 20,000 ohms per volt meter. You would need to know the full scale you are using to figure out the input resistance of the meter. You don't have that problem with the DMMs which are typically 1 Megaohm or 10 Megaohm. I would not be surprised to find a 100 Megaohm input impedance out there.

Meters can load the circuit giving you false readings. If the technician isn't aware of that, they could be troubleshooting a circuit that has not failed.

Here is a question for the OP ...

Using the diagram below. What will the DMM read when the linear potentiometer is at 50%?

What is the voltage?, What is Fluke 77?

 

What is the voltage?, What is Fluke 77?

Voltage is 10V and fluke77 is voltmeter.

 

You just need to know how the meter, as a load, affects the reading the meter shows.

If you were using a Simpson 260, it's a 20,000 ohms per volt meter. You would need to know the full scale you are using to figure out the input resistance of the meter. You don't have that problem with the DMMs which are typically 1 Megaohm or 10 Megaohm. I would not be surprised to find a 100 Megaohm input impedance out there.

Meters can load the circuit giving you false readings. If the technician isn't aware of that, they could be troubleshooting a circuit that has not failed.

Here is a question for the OP ...

Using the diagram below. What will the DMM read when the linear potentiometer is at 50%?

Hello,

Yes you are right Joe, that's the most important point for this question i think too now.

But what else is interesting is how this question evolves into practical use. In the real world we dont always know the two (or one) resistors values of the circuit being measured, or if it is linear. If we assume it is linear for now though it's still interesting because then we dont have any resistor value to put in parallel with the meter.
For example, using the meter of this thread with 70k resistance, we measure a voltage on an unknown circuit and it reads 10.2 volts. The question then is, what is the real voltage measured? If we see 10.2 on the meter we know that is lower than the true reading, so then the question becomes what is the true reading.
I believe in this case it would take two measurements to be able to calculate the true voltage without knowing the internal impedance of the circuit. This would be interesting to look at too.

 

The DMM will read 5V, the current through the whole circuit is 10mA.

 

The DMM will read 5V, the current through the whole circuit is 10mA.

If that is your "guess", then it is incorrect. You failed to consider the input impedance of the Fluke 77 and it's loading of the circuit.

In real life, you could be "restoring" an old piece of equipment where the service manual has voltages were taken with a Simpson 260 and all you have is your trusty Fluke 77. How would you determine if a voltage you've read at a test point is correct?

This is basic series-parallel circuits. Your struggle with the basics is going to cause you to repeat that class.

I always admire man's credulity when they appear a few days before the deadline and expect someone to work miracles with their gaps of understanding. I sometimes wonder what they have been doing all semester ... because I'm struggling to believe they've been diligently studying. It's not only happening here and other forums, but I see in elsewhere.

on edit ... added the below graph created in excel The simpson 260 was on the 10V scale and the specification is plus or minus 2 percent at full scale. The meter sensitivity is 20kΩ/Volt.