basic voltage query

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WBahn

Joined Mar 31, 2012
26,398
your right.i'll give up

and now i'll carry on

so battery voltage /total restistance x r1 = voltage

12/300,000 =0.00004

0.00004 x r1(100,000 ohm)= 4v

I understand v=ir I understand that,if you look at my replies ive even been able to manipulate that formula.i just haven't spelled it out
No, you don't understand v=ir. Even if the total resistance was 300 kΩ (which, again, it is not), application of Ohm's Law would yield the total current in the system and NOT the current through the meter resistance.

By pure happenstance, you ended up with an answer that is equal to the correct answer, but you did not arrive at a correct answer. What I mean by this is that you might as well have rolled a pair six-headed dice and concluded that the answer was 4 V because that was the total number of spots on the top faces of the dice after they stopped. That approach would be equally valid to the method you used.
 

WBahn

Joined Mar 31, 2012
26,398
I'm just looking for the answer.
That's a big part of the problem. You are looking for someone to give you the answer instead of seeking to understand how to approach the problem and solve it.

nobody has been able to give me that.
You are correct that no one has give you the answer. It is not because no one has been able to, it's that people here recognize that doing so would not help you. Consider that you have been shown how to work problems like this in class and you have example problems that have been worked out in your book. You have access to hundreds, if not thousands, of explanations and worked examples on the Internet, including the E-book available on this very site. Despite all of that, you are at a complete loss how to work this basic problem. So what is the likelihood that having one more problem solution just given to you will make some lightbulb come on above your head? Pretty close to zero. The only way you will learn this material is to struggle with it. We have been trying to guide you along a path that will get you there, but you have been stubbornly refusing to take the many suggestions that have been given to you and follow them. Instead, you keep throwing out effectively random rubbish hoping that someone will throw up there hands and just give you the solution.
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
I totally rolled the dice,sure the formula was a wild guess,a complete gamble,dice roll and hit the jackpot

your brilliant at everything you do,im absolutely rubbish at everything

your a supreme being,im nothing...is that what you want to hear
 

WBahn

Joined Mar 31, 2012
26,398
its a difficult problem,fair enough,its got us all puzzled
This is a cheap ploy that gets tried on a regular basis -- when no one will just give you the answer, assert that it is because no one else knows it either hoping that someone will feel the need to prove that they DO know the answer by posting it. Hint: We see right through that childish trick and all it will accomplish is driving away the people that have been trying to help you.

Trust me, no one that has been trying to help you is puzzled and it is NOT a difficult problem at all for any of us. But we are not taking a course in electronics for the first time, either. It's supposed to be challenging for you. But you need to approach the problem systematically and using techniques that you can defend.
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
ive come onto an electronics forum for answering homework style questions because I'm an expert,because I know the answers,

it seems to me you don't know the answer.im 2 days here,and no better off.

don't help me
 

WBahn

Joined Mar 31, 2012
26,398
I totally rolled the dice,sure the formula was a wild guess,a complete gamble,dice roll and hit the jackpot

your brilliant at everything you do,im absolutely rubbish at everything

your a supreme being,im nothing...is that what you want to hear
It has nothing to do with brilliance or supremacy or anything of the kind. It has to do with expecting you to put forth the effort to systematically work through the problem instead of just throwing nearly random values and nearly random equations and using the results in nearly random ways.

Consider an analogy. You go to the doctor with a set of symptoms and your doctor looks them up in one of his references. It tells him what the condition is and says that the treatment can be found on page 3964. But he looks on page 3694 and writes down the drug and dosage information, but in doing so misspells the drug name and writes the prescription for 10 g instead of 10 mg. However, as it just happens to turn out, the misspelled drug name and the incorrect dosage turn out to be the correct drug and dosage that were shown on the correct page (the one that the doctor never looked at). So, by coincidence, you got the correct treatment for your condition. You now find out that this is what happened. Are you going to go back to that doctor? Do you think that doctor should be practicing medicine?
 

WBahn

Joined Mar 31, 2012
26,398
ive come onto an electronics forum for answering homework style questions because I'm an expert,because I know the answers,

it seems to me you don't know the answer.im 2 days here,and no better off.

don't help me
We obviously CAN'T help you because you have a worthless attitude. Get that straightened out and maybe there will be some possibility. Until then, there isn't much anyone can do for you because you are being your own worst enemy.
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
ok I understand.i wouldn't go to that doctor.

I Agree With Everything You Are Saying.

You Are Right

I am Not Right.

you suggested using the voltmeter as a 100k resistor,that doesn't seem sensible
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
you are right,i am my own worst enemy

everything you say is right.

now back to the circuit.the voltage is 4v with voltmeter connected?
 

JoeJester

Joined Apr 26, 2005
4,390
WBahn is in the top 5 who visit the homework forum to help. In fact, in my opine, he is at the top of the list.

You just wanted answers and not work with us, we had to drag info from you contantly.

In the homework forum, we prefer you to demonstrate the requisite knowledge when answering the problem or members questions.

Now, change the sensitivity of your meter to that of a Simpson 260. Using the 25V scale, what would the voltage drop be on either resistor.

Draw your circuit. Show the math, and show the percent of error.
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
ive drawn the circuit.your trying to drag info out because I don't know

wbahn is great I'm sure,he could be the best on here,ive no problem with that.

he's detailed ok,thats no problem

I'm googling the voltmeter query as we speak.

if I cant drive and I have a bmw,it wont make a difference if you ask me to drive a toyota
 

WBahn

Joined Mar 31, 2012
26,398
ok I understand.i wouldn't go to that doctor.

I Agree With Everything You Are Saying.

You Are Right

I am Not Right.

you suggested using the voltmeter as a 100k resistor,that doesn't seem sensible
It's not a matter of using the voltmeter as a resistor. It's a matter of modeling the effect that the voltmeter has on the circuit as being the same effect that would be seen if you used a resistor in it's place. As far as the circuit goes, both are indistinguishable.

How many older analog voltmeters works is you have a "meter movement" that deflected a needle on the display in response to the current flowing in it. These were linear so that if you doubled the current you got twice the needle deflection. The meter movement was designed to give a full-scale deflection at a particular current, say 1 mA just for illustration. The meter movement's coil also had a certain amount of resistance, let's say 1 kΩ for our hypothetical movement. That means that 1 V applied to the movement would produce a full scale deflection while a 0.5V application would deflect it half way. Write some numbers on a piece a paper behind the needle and you now have a voltmeter on the 1 V range. But what if you want to measure 100 V? If you apply that to the meter movement, you will have a destroyed device. But what if you place a 99 kΩ resistor in series with the meter? Now when you apply your 100 V you will get 1 mA flowing through the meter and it will deflect full scale. Viola! You now have a meter that has a 100 V range.

We can make the meter read any full-scale reading we want (above 1 V) by simply adding enough resistance in series with the meter to make it so that a total of 1 mA of current flows through the meter when the full-scale voltage is applied. In our meter's case, we need 1 kΩ of resistance for every volt of the full-scale reading, so our sensitivity is 1 kΩ/V.

If we build this into a meter with a switch for different ranges, then what that switch does is switch in different resistors for the different ranges. On the 10 V scale, it would switch in a 9 kΩ resistor such that the total resistance seen between the voltmeter's probe tips is 10 kΩ.

When we connect the voltmeter to a circuit, the circuit behaves as if we have just connected a 10 kΩ resistor between the points where the probe tips are located for the simple reason that this is precisely what we have actually done.
 

WBahn

Joined Mar 31, 2012
26,398
you are right,i am my own worst enemy

everything you say is right.

now back to the circuit.the voltage is 4v with voltmeter connected?
Yes, but that will do you no good until you understand how to get that result correctly.

And you can drop the attitude any time.
 
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