basic voltage query

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WBahn

Joined Mar 31, 2012
29,978
10,000 x 10=100,000 ohms
10,000 x 10 is equal to 100,000. It is a pure number multiplied by a pure number and therefore produces a pure number. You can't just go tacking units onto an answer because you want, hope, and pray that the answer is supposed somehow magically grow those particular units at the end.

You have a sensitivity of 10,000 Ω/V.

What does that mean? What is the definition of the sensitivity of a voltmeter? Why won't you look it up?

http://encyclopedia2.thefreedictionary.com/voltmeter+sensitivity

The sensitivity of a voltmeter is, by definition, the total resistance of the meter divided by the full scale reading of the meter. If you ponder it for just a moment, this is the same thing as the amount of current that must flow through the meter in order to produce a full scale reading.

What you are needing in order to do your calculation is the total resistance of the meter. Well, if you know the ratio of the total resistance to the full-scale reading and you know what the full scale reading is, you can get the total resistance.

To get your 100,000 Ω, multiply the sensitivity by the full-scale voltage of that range:

(10,000 Ω/V)(10 V) = (10,000 Ω·V/V) = 10,000 Ω

Do you see how nicely the units work out?

and 1000 / 100 = 10%
Where did this come from?

Where did the 1000 come from?

Where did the 100 come from?

Why are you dividing them?

1000/100 = 10. Not 10%, which is 0.10. 1000/10 is 1000%.
 

WBahn

Joined Mar 31, 2012
29,978
Missed it by a lot. 1000 / 10 = 100 = 10000% but that is not the answer to the question.

By inspection, what is the voltage across R2. Then, what is the voltage you read on the meter when you place the meter leads across R2.
He has 1000 / 100, but your point is still completely valid and relevant.
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
ok your backing up what your saying,im starting to understand this.i can only humbly say,thanks for your wisdom,knowledge,and possibly years of expertise in this area
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
what about the resistors,i mean don't they need to be added?your answer doesn't seem to include this.i mean they where given in the question.
we seem to have abandoned the 12v
 

JoeJester

Joined Apr 26, 2005
4,390
I haven't abandoned anything.

Without the meter connected, what is the voltage across R2?

With the meter connected, what is the voltage across R2?

Answer those two questions.
 

WBahn

Joined Mar 31, 2012
29,978
meter is connected in series, 12v
How do you figure that?

Sketch the circuit with and without the meter.

Model the meter as a 100,000 Ω resistor (that was the whole point of determining it's resistance) connected across one of the circuit resistors. Then analyze the resulting circuit to find the voltage that would appear across the meter resistance. That's the voltage that the meter will read. What is that voltage?
 

WBahn

Joined Mar 31, 2012
29,978
the resistance of the meter and r1 is
12 / 200 = 0.06 x 10 = 0.6volts with meter is 6-0.6 =5.4 v
You say "the resistance of the meter and r1 is" and then follow it with a bunch of stuff that doesn't end up with a resistance.

Are you aware of what an equals sign means?

It means that whatever is on the left is equal to whatever is on the right. You seem oblivious to that fact.

12/200 is NOT equal to 0.06 x 10 !!!!

And 0.06 x 10 is just a number, it is NOT a voltage !!!! So it cannot be equal to 0.6volts because that is a voltage and not a number!!!!

And, even if I use my crystal ball to infer what you meant to say, it doesn't make any sense.

This is a simple circuit analysis problem. Such problems are best started with a diagram of the circuit. But you seem absolutely opposed to providing a sketch of the circuit you are working with.
 

Thread Starter

Albert Symes

Joined Oct 22, 2015
41
the question,it doesn't provide a circuit sketch.and in fact it asks you to draw the circuit
your absolutely right,they aren't equal,this is why I'm asking the question and your answering it
 
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WBahn

Joined Mar 31, 2012
29,978
This is like pulling teeth!

Now do what I recommended back in Post #32.

Model the voltmeter as a 100 kΩ resistor and determine the voltage across it.
 
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