Role of a voltmeter in an initially open circuit

AI Thread Summary
In an open circuit swung through a magnetic field, the induced electromotive force (emf) can lead to varying voltage readings, potentially resulting in either spikes or a flat graph depending on the conditions. Connecting a voltmeter across the open ends of the circuit effectively closes it, allowing for current to flow, albeit minimally due to the voltmeter's high input impedance. This high impedance means the voltmeter typically does not significantly affect the circuit under normal conditions. The degree of impact from the voltmeter depends on its specifications and the characteristics of the circuit being measured. Understanding these principles is crucial for accurate voltage measurement in open circuit scenarios.
justing943
Messages
2
Reaction score
0
Hi!

Could someone briefly explain the following scenario to me please :)

If an open circuit (with the 2 ends named X and Y) hung on a string is swung through an into page magnetic field, what would the emf be? How would that change if you are asked about the voltage 'measured' between X and Y?

The answers that are debatable are between a graph where there are two spikes, one positive one negative, and a completely flat voltage graph.

If an open circuit is connected to a voltmeter (one end to each open end), how would that affect the circuit?

Thanks in advance :)
JT

edit: Sorry if this is in the wrong subforum (I suspect it belongs in the HW subforum)
 
Physics news on Phys.org
justing943 said:
Hi!

Could someone briefly explain the following scenario to me please :)

If an open circuit (with the 2 ends named X and Y) hung on a string is swung through an into page magnetic field, what would the emf be? How would that change if you are asked about the voltage 'measured' between X and Y?

The answers that are debatable are between a graph where there are two spikes, one positive one negative, and a completely flat voltage graph.

If an open circuit is connected to a voltmeter (one end to each open end), how would that affect the circuit?

Thanks in advance :)
JT

edit: Sorry if this is in the wrong subforum (I suspect it belongs in the HW subforum)
Hooking up a volt meter will close the circuit. A volt meter is actually measuring a current through a known impedance, but the scale is tuned into read voltage. However, the current is very low due to the high input impedance of the volt meter. So under "normal" conditions the current that flows through the volt meter is therfor not taken into account. Modern volt meters is controlled by an electronic circuit that has a few million Ohms input impedance.

How much the volt meter affect the circuit will therefor depend on the voltmeter properties and the circuit that is measured.

Vidar
 
Thanks very much for the reply :). I do believe that solves my problem :D
 
I was using the Smith chart to determine the input impedance of a transmission line that has a reflection from the load. One can do this if one knows the characteristic impedance Zo, the degree of mismatch of the load ZL and the length of the transmission line in wavelengths. However, my question is: Consider the input impedance of a wave which appears back at the source after reflection from the load and has traveled for some fraction of a wavelength. The impedance of this wave as it...
Back
Top