Is My Pull Up Resistor Calculation Correct?

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The discussion revolves around the calculation of resistance using a pull-up resistor and a current source. The user initially calculates a current of 0.0006A, expecting a voltage drop close to 5V, but instead measures 9.83V. This discrepancy leads to a calculated resistance of approximately 16.4k ohms, which the user doubts. A response highlights that the current source may not be precise, noting that real-world current sources require a minimum voltage drop to function correctly. The conversation emphasizes the importance of understanding the operational characteristics of current sources in circuit calculations.
ee1215
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http://i.imgur.com/TbYIHnb.png?1

So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

V=IR
I = 5V/8k = .0006A would result in a voltage drop of close to 5V

Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

But when I read my voltage reader I get 9.83V. Is this correct?

Then my resistance would be 9.83/.0006 ~ 16.4k ?

These calculations do not seem correct.
 
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ee1215 said:
http://i.imgur.com/TbYIHnb.png?1

So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

V=IR
I = 5V/8k = .0006A would result in a voltage drop of close to 5V

Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

But when I read my voltage reader I get 9.83V. Is this correct?

Then my resistance would be 9.83/.0006 ~ 16.4k ?

These calculations do not seem correct.

Sounds like your current source is not very precise. Keep in mind that real-world current sources need to have some voltage drop across them in order to operate correctly. Typically you need a couple of volts across a current source in order for it to say "in compliance".
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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