# Primary voltage drop in current transformer

Following on from this question, Current transformer energy harvesting from a mid-voltage line, I've never seen any mention of primary voltage drop in a current transformer circuit.

simulate this circuit – Schematic created using CircuitLab

Figure 1. A typical CT application using a 5 A meter to measure the current through a load.

Since the burden resistance of the ammeter is reflected back onto the primary side by the inverse turns ratio squared there has to be a voltage drop on the primary side. This raises a few questions.

1. Where does the voltage drop occur? Does it fade in and out with a peak right inside the CT?
2. Would we see a larger reading on VM2 compared with VM1 (10 mm and 5 mm from the plane of the CT, for example)?
3. What is the relationship with the dimensions of the CT?
4. What is the effect of the angle between the axis of the CT and that of the cable. (CT

Where does the voltage drop occur? Does it fade in and out with a peak right inside the CT?

The maximum volt drop/metre ($$\\frac{dv}{d\ell}\$$) is in the middle of the core and rapidly becomes the normal volt drop/meter for a wire beyond the core.

Would we see a larger reading on VM2 compared with VM1 (10 mm and 5 mm from the plane of the CT, for example)?

Yes we would because of the normal external volt drop of a longer cable (Ω per metre and inductance per metre).

What is the relationship with the dimensions of the CT?

If you are asking at what point the internal higher volt drop per metre becomes regular wire volt drop per metre for a given size of CT, I can't say. I can't imagine that much beyond twice the core width it would be very much influenced by the core. Good question though.

What is the effect of the angle between the axis of the CT and that of the cable.

I don't think there will be much effect given that in practice, the angle probably isn't more than about 45 °. Another good one!

Would we see a larger reading on VM2 compared with VM1 (10 mm and 5 mm from the plane of the CT, for example)?

Unlikely. Each volt-meter together with its leads and corresponding section of the wire passing through the core form a loop. That loop is cut by the varying magnetic field inside the core of the CT. Thus, they form a one loop secondary of the transformer, and will see a voltage almost identical to the voltage drop of the primary. Almost all of the magnetic field is confined to the core, so the area of the loop will have little effect, although theoretically, if the loop for vm2 encloses that for vm1, it might encircle more magnetic lines of force.

Where does the voltage drop occur? Does it fade in and out with a peak right inside the CT?

That is problematic, see further down the question. You could probe a very small section of the wire in the CT core, and depending upon whether the test probes enter the core from either side, or whether they enter from the same side you will get a different answer. If the test probes enter from opposite sides, you will get a voltage approximately equal to the full voltage drop of the CT. In fact you could touch these probes together and still get the full voltage drop, because the probes are forming a 1 turn secondary. However, if the probes enter the CT from the same side, you will get almost 0 voltage. What you do get will be mostly the resistive drop.

The question you ask poses a measurement conundrum. The circuit consisting of either voltmeter, its leads, and the conductor through the CT core is "cut" by the changing flux of the CT core, in effect making a transformer secondary.

From one point of view, the voltage between the contact points of the voltmeter leads is undefined. To explain that point of view, consider Maxwell's(*) third equation, i.e. Faraday's Law.

$$\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}$$

If there is no time-varying magnetic field, this implies that

$$\nabla \times E = 0$$

which means that E is a conservative field. Being a conservative field implies that E is the gradient of a scalar function, (which we call V)

$$E = \nabla V$$

V, being a scalar function implies that the sum of changes around a loop must be 0 (which is KVL).

However, the assumption that there exists no time-varying magnetic field is violated in the circuit consisting of the voltmeters, their probe wires, and the mains conductor. Accordingly, MIT Professor Lewin has argued that the voltage between the two lead points of the Volt-meter is undefined. In a class demonstration, Lewin shows two voltmeters connected to the same test points showing different voltages. Lewin's position is further explained in this blog.

(Lewin's demonstration has is discussed here, but I am not satisfied with the explanation given.)

Others, such as Electroboom/Mehdi have argued (see also here) that the problem with the example Lewin provides lies in measurement technique. Volt-meter probe wires should not arranged to form a loop cut by a changing magnetic field. If the probe wires are re-arranged, consistent voltage measurements are observed in Lewin's experiment. If we take that point of view, however, and re-arrange the test leads to avoid loops cut by changing magnetic fields in order to get a consistent/proper voltage measurement, then one of the test leads needs to pass through the CT core. This, however, will result in a voltage drop consisting only of the conductor resistance.

Hence, the conundrum. How should the test leads be arranged? If they form a loop cut by a changing magnetic field, is the voltage well defined? If they are not cut by that changing magnetic field, where is the voltage drop created by the CT?

*The equations called "Maxwell's Equations" were first formulated in their modern form by the electrical engineer Oliver Heaviside.

What is the relationship with the dimensions of the CT?

Very little, provided the core is not saturated. If the core is not saturated, most of the magnetic lines of force will be confined to the core, and any wire loop around those lines of force will have approximately the same induced emf.

What is the effect of the angle between the axis of the CT and that of the cable. (CT)

Almost none, for the reasons explained above.

The leakage inductance of the CT adds a voltage drop in quadrature to the primary current, aka an inductance, in series with the primary.

The reflected secondary load, if resistive, adds a voltage drop in phase with the primary current, aka a resistance, in series with the primary.

Consider that another way of looking at this mechanism is that the CT secondary voltage is transformed into the primary. This induction is seen across closed turns around the core, so the trajectory of your meter leads, VM1 and VM2, will be significant in the actual voltages you measure. As you move the meter leads along the wire to investigate 'where' the drop is happening, you may get unexpected results. This works for pure IR resistive drop. It doesn't work (without an ice bag for the forehead and a lot of care) for induced voltages.

• Re this: a resistance, in series with the primary - I think you mean a resistor in parallel with the primary. Jan 26 at 17:42
• @Andyaka Let me think about that. If the secondary resistance of an ideal transformer is shorted, the primary will appear short. If we add leakage idncutance, that appears in series. I did specify leakage, not primary inductance. Jan 27 at 7:28
• @Andyaka Of course, the primary inductance is in parallel with the leakage and secondary in series. Now I need to bend my head around what resonating Lpri does for power output, what magnetising current is actually trying to saturate the core, but that was a different question! Jan 27 at 8:34