This is probably a pretty stupid question...

[mort]

... but I'm gonna ask it anyway.

If I have the voltages before and after a resistor, can I calculate how many ohms the resistor is actually providing in a live circuit?

Before voltage is 440v. After the resistor it is 430v.

Is this enough information to be able to get the active resistance taking place?

[diagrammatiks]

no you need the current draw as well.

[mort]

That's what I was just beginning to figure out with a little googling. Just simple Ohm's Law right? Or do I need a more complex formula, given that I have 2 voltages??

All of the excersizes I am finding don't quite capture what I'm trying to learn here...

[Jazz P Bass]

Well, if you pull up one end of the resistor & insert an ammeter in the circuit, you will know the current.
Voltage divided by Current equals Resistance.
I do believe that you must have two of three to compute Ohm's Law.

[Zipslack]

Well, if you pull up one end of the resistor & insert an ammeter in the circuit, you will know the current.
Voltage divided by Current equals Resistance.
I do believe that you must have two of three to compute Ohm's Law.

Of course, if you're going to lift one leg, might as well just measure it. Or better yet, use the color code and don't lift at all.

[diagrammatiks]

you need to know at least 2 of three things

the voltage drop, the current, the resistance.

[g1]

Is this enough information to be able to get the active resistance taking place?

A resistor is a passive device. It's resistance should not change (unless defective). The resistance is the same whether the voltage is AC or DC. There is no "active resistance". Unless you are talking about inductors or capacitors, then that is called impedance, not resistance.
Reading the colour code or measuring it with one end lifted is all you need to do to determine the resistance.

[mort]

The resistor in question is labelled as a 1K 5W resistor. It's the first resistor on the filter section of this circuit, where the voltage is labelled as +420v.

http://www.300guitars.com/wp-content...164_layout.gif


There's a 10v drop across that resistor. It's probably fine. I just wanted to run through the excersize of determining its acual resistance for learning purposes. I guess the most practical way is to lift and read.

I'm not at my house at the moment to see, but I don't know if I can read amp draw with the probes of my meter. It has a clamp on type device for reading amps, which would not work in this situation.

http://www.fieldpiece.com/PDF/Manual...n-SC5X-web.pdf

edit: looks like my meter will only read ADC up to 4000 microamps, so I need another meter for reading DC current.

It's funny, you never know what you will learn. I start out looking for resistance and figure out I need a new meter

[Enzo]

Just measure the part by lifting one end from the circuit to isolate it. Then the exercise becomes calculating the current through it. I can;t recall ever needing to measure the current through a power resistor to calculate its resistance. Its resistance will be whatever it is regardless of current.

As was mentioned above, Ohm's Law is very powerful and very useful, but you always need to know two of the three parameters to get the third.


A plumbing equivalent would be measuring the water pressure in a pipe, and the flow rate to calculate the diameter of the pipe. That instead of the more reasonable measuring the pipe diameter and pressure to calculate what the flow rate would be.