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.your right.i'll give up
and now i'll carry on
so battery voltage /total restistance x r1 = voltage
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
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.I'm just looking for the answer.
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.nobody has been able to give me that.
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.its a difficult problem,fair enough,its got us all puzzled
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.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
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.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
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.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
Yes, but that will do you no good until you understand how to get that result correctly.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?
|Thread starter||Similar threads||Forum||Replies||Date|
|A||Basic Inverter Voltage regulation||Power Electronics||26|
|Basic distribution problem for a ripple voltage||Homework Help||3|
|S||Voltage and Current across a basic DC circuit||Test & Measurement||16|
|M||Basic: MOSFET negative gate voltage rating question||Power Electronics||16|
|M||Basic Question on voltage||General Electronics Chat||5|
by Wes Brodsky
by Jake Hertz
by Luke James