something i havenot undeestand yet

[7447]

DEAR MR.Nigel Goodwin:
you have wrote that: for a 200V range the ratio between the bottom and top resistor is 1/199 .
could you please exepress me why was writen:10,000,000/200 = 50,0007
and thanks again for help
:wink:
7474

 

[mstechca]

DEAR MR.Nigel Goodwin:

maybe you should use the PM system if this question is for him.

could you please exepress me why was writen: 10,000,000/200 = 50,0007
and thanks again for help

10,000,000/200 = 50,0007?

How about 10,000,000/200 = 500,000?

Nigel, maybe you know what he is talking about since he specifically asked you.

 

[Nigel Goodwin]

DEAR MR.Nigel Goodwin:
you have wrote that: for a 200V range the ratio between the bottom and top resistor is 1/199 .
could you please exepress me why was writen:10,000,000/200 = 50,0007
and thanks again for help
:wink:
7474

It would really have been better to continue this in the same thread!.

The maths is very simple - the bottom resistor is one 200th of the total resistor chain, and the top one is 199 200th's of the total chain. As the standard input resistance for a digital meter is 10Mohm you simply divide that by 200 to get the 'one' part - you can either multiply that by 199 to get the top part, or simply subtract it from 10Mohm.

If you want a different input resistance, simply substitute the value you need and divide that by 200.

 

[Roff]

I thought he was asking why there was a 7 after 50,000 in Nigel's original post (there wasn't). Maybe that was just a typo on his part - although it's difficult to imagine how you could fat-finger that one.

 

[Roff]

DEAR MR.Nigel Goodwin:

maybe you should use the PM system if this question is for him.

could you please exepress me why was writen: 10,000,000/200 = 50,0007
and thanks again for help

10,000,000/200 = 50,0007?

How about 10,000,000/200 = 500,000?

Nigel, maybe you know what he is talking about since he specifically asked you.

Errrmm.... 10,000,000/200=50,000.

 

[mstechca]

I must have divided wrong.

But let me simplify nigel's explanation about voltage dividers.

To obtain a voltage from another voltage, you take two resistors. Connect them in series with the battery. The point where two resistors meet is the output (the voltage you want).

The resistor from the output to +ve will be called R1 and
The resistor from the output to -ve will be called R2 for now.

If you use the equation:

Output voltage = (R1 / (R1 + R2)) * (battery voltage)

you will begin to understand where Nigel obtains his numbers.