The discussion revolves around understanding why a small current is necessary when using an ohmmeter to measure resistance in a wire. Participants explore the principles behind ohmmeters and the implications of current on resistance measurements.
Several participants have provided insights into the operation of ohmmeters, suggesting that understanding the device's function is crucial for answering the original question. There is an ongoing exploration of the consequences of using larger currents in resistance measurements.
Some participants note that the wire being measured acts as a variable resistor, and there is discussion about the impact of current on the wire's properties, such as heating and deformation.
All I get is the resistance = voltage/current. That's it. Everything was from BBC Bitesize, nothing else came up.kuruman said:
That's all you need. An ohmmeter measures voltage V across, for a known current I through the unknown resistor and displays the ratio. You provide the resistor. What, do you think, provides the V and the I once you have have hooked up the resistor and turned the selection switch to "Ohms"? Can you draw a simple circuit diagram with your external resistor also drawn in?IDK10 said:
Its not that sort of circuit, its an ohmmeter, with two wires coming out and connecting to a piece of wire. No exterior battery or resistor. Well the wire acts as a variable resistor, as the crocodile clips connecting the wire to the ohmmeter are moved in increments of 10cm, but no actual resistors were used.kuruman said:
1. Voltage/Currentkuruman said:
OK, tbh, its actually a multimeter, but on the ohmmeter setting.kuruman said:
Yes, indeed. I indicated as much in post#4. They are all multimeters, but we call them voltmeters when we use them to measure voltage, ammeters when we use them to measure current and ohmmeters when we use them to measure resistance. Do you think you can answer my questions now?IDK10 said:
What I found was 'Most ohmmeters work by passing a (usually small) current through the sample and measuring the voltage drop - thus finding the resistance by Ohm's law V=IR. Any internal resistance of the ohmmeter should be well known and included in the calculation (usually automatically).', so is it to give a large enough resistance so that, when the internal resistance is taken into account, the final resistance is significant.kuruman said:
Good. Can you now draw a simple basic circuit of the innards of the ohmmeter and find an expression for the measured resistance? Include the internal resistance and ask yourself (and answer) the question, "What if the current is not small and the internal resistance is not that well known or even not known at all?"IDK10 said:
I don't understand what this means. Draw the circuit to focus on what you wish to say.IDK10 said:
Finding a diagram is going to be hard. I have no idea of what the inside of an ohmmeter looks like, so I can't find a diagram of the inside of one. We also don't need to know about the inside of one.kuruman said:
I can try, keep in mind that the resistor, isn't an actual resistor, but a metal wire, which acts like a variable resistor.kuruman said:
I think you can safely assume that the wire has a fixed length which means a fixed non-zero resistance for answering this question. If the resistance varied while you are trying to measure it, you would be wasting your time.IDK10 said:
The resistance would vary while measuring it, we record the resistance of different lengths of wire.kuruman said:I think you can safely assume that the wire has a fixed length which means a fixed non-zero resistance for answering this question. If the resistance varied while you are trying to measure it, you would be wasting your time.