# What does a voltmeter read?

1. Oct 30, 2009

### JustinLevy

When working out circuits problems, can we consider a voltmeter to read the difference in the scalar potential between two points
$$Voltmeter = V(b) - V(a),$$
or does it tell us a "path potential" / work to move a test charge along a path
$$Voltmeter = -\int_a^b \mathbf{E} \cdot d\mathbf{l}$$
?

Last edited: Oct 30, 2009
2. Oct 30, 2009

### ideasrule

What's the difference?

3. Oct 30, 2009

### JustinLevy

In general, the potentials are defined as such:
$$\mathbf{E} = - \nabla V - \frac{\partial}{\partial t} \mathbf{A}$$
$$\mathbf{B} = \nabla \times \mathbf{A}$$
That doesn't define them uniquely, but at least up to a gauge transformation.

In cases where an eletric field is being generated by a changing magnetic field (such as in an inductor), then those two suggestions in the openning post will differ. Therefore this is relevant for asking what a voltmeter would read across an inductor.

4. Oct 30, 2009

### JustinLevy

I just realized that the answer cannot be V(b)-V(a), because then the measurements would not be gauge invariant. So the answer must be the E.dl one.

Is that correct?

5. Oct 30, 2009

### JustinLevy

That simple post has got me so confused now.

Consider this:

Let's say that this loop has inductance L, no resistance, and that initially there is no current in the loop. Then at time t=0, a current source is used to increase the current at a steady rate (I = a t).

Current goes up the green wire, around the blue loop, and then up the red wire. Additionally, consider a voltmeter connected with the ground on the red point, and the "positive" terminal on the green point.

What does the volt meter read?

Well, when the current starts increasing, a magnetic field will be generated in the positive direction. It appears that this increase in flux will cause an eletric field around the loop against the current, which would mean the integral of - E.dl from the red to the green point is negative.

This seems to argue the voltmeter would read Voltage = - L dI/dt.

This is clearly wrong, as we should get:
Voltage = L dI/dt

What the heck is going on here?