# Role of a voltmeter in an initially open circuit

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?

JT

edit: Sorry if this is in the wrong subforum (I suspect it belongs in the HW subforum)

Low-Q
Gold Member
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?

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 in to 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