🔺 Wye, Delta & the Copper Thermometer

A motor only gives you its leads. You can touch the three terminals A, B and C, but the per-phase coil you actually care about is buried inside, wired into either a wye or a delta. So every reading arrives scrambled by the connection, and shifted by temperature. To get back to the truth you have to undo both, in order: first work out what the wye or delta geometry did to your line-to-line number, then correct that number for the temperature the copper was sitting at when you measured it.

By the end, you can

1. Explain why a line-to-line measurement is not the per-phase resistance
2. Convert line-to-line resistance to phase resistance for both wye and delta windings
3. Calculate how copper resistance changes with temperature using the coefficient alpha
4. Infer winding temperature from a resistance reading, and record the reference temperature with every log entry

Intuition first

Think of the three windings as three identical lengths of copper hose. You are not allowed to look inside the machine. All you can do is stand at the outside fittings (the terminals A, B, C) and push water from one fitting to another, measuring how hard it is to push. The trouble is that the same three hoses can be plumbed two completely different ways, and the plumbing changes what your push feels like.

In a wye (also called a star), all three hoses meet at one hidden junction in the middle, the neutral. Push from A to B and the water has to travel out through hose A, across the junction, and back through hose B. You feel two hoses in series, so the resistance you read is twice the resistance of a single hose.

In a delta, the three hoses form a closed triangle. Push from A to B and the water splits: a little goes straight down the A-to-B hose, and the rest takes the long way around through the other two hoses in series. You feel one hose in parallel with two hoses in series, which works out lighter than a single hose.

Same three coils, two different answers, and the meter cannot tell you which plumbing it is looking at. You have to know. That is the first correction. The second correction is temperature: copper hoses that are warm push back a little harder, and that extra push is not damage, it is just heat. Once you can read both effects, an ohmmeter becomes a thermometer.

The connection correction

Let $R_{ph}$ be the resistance of one phase winding, the thing you actually want. Let $R_{LL}$ be what your meter reads between two line terminals. The relationship between them is pure series-parallel bookkeeping, and it is different for each topology.

Wye: two phases in series

In a wye the three coils share a common neutral point. A line-to-line probe enters one terminal, runs through that phase to the neutral, and comes back out through a second phase. The third phase dangles off the neutral and carries no current, so it does not count. You are measuring two equal windings end to end:

$$R_{LL} = 2\,R_{ph} \qquad\Longrightarrow\qquad R_{ph} = \frac{R_{LL}}{2}$$

So a wye motor reading $1.0\ \Omega$ line-to-line has a per-phase resistance of $0.5\ \Omega$. Halve the meter reading and you are done with the geometry.

Delta: one phase parallel with two in series

In a delta the coils form a triangle with no neutral. Probe two corners and the current splits into two paths: the single phase directly between those corners, and the series pair of the other two phases. One winding ($R_{ph}$) sits in parallel with two windings in series ($2\,R_{ph}$). Combine them:

$$R_{LL} = \frac{R_{ph}\cdot 2R_{ph}}{R_{ph} + 2R_{ph}} = \frac{2R_{ph}^2}{3R_{ph}} = \frac{2}{3}\,R_{ph} \qquad\Longrightarrow\qquad R_{ph} = \frac{3}{2}\,R_{LL}$$

So in a delta the meter reads only two-thirds of a phase, and the per-phase resistance is one and a half times the line-to-line value. Notice the trap: a wye and a delta wound from identical coils give different line-to-line readings, and if you assume the wrong topology you will be off by a factor of three between the two conversions.

The temperature correction

Copper is a metal, and a metal’s resistance climbs with temperature. Heat shakes the crystal lattice harder, the conduction electrons scatter more often, and the same wire resists more. Over the modest range a motor lives in, the rise is close to linear and is captured by a single number, the temperature coefficient of resistance, written $\alpha$. For copper,

$$\alpha \approx 0.00393\ \text{/}^\circ\text{C} \quad(\text{about } 0.39\%\ \text{per } ^\circ\text{C})$$

If you know the resistance $R(T_1)$ at one temperature $T_1$, the resistance at another temperature $T_2$ is

$$R(T_2) = R(T_1)\,\bigl[\,1 + \alpha\,(T_2 - T_1)\,\bigr]$$

Resistance rises with temperature, so $R(T_2) \gt R(T_1)$ whenever $T_2 \gt T_1$. A $0.5\ \Omega$ phase at $20\ ^\circ\text{C}$ becomes about $0.5\,(1 + 0.00393\cdot 40) \approx 0.579\ \Omega$ at $60\ ^\circ\text{C}$, a 16% jump from nothing but heat. That is why two honest people reading the same winding can disagree: they measured it warm and cold.

The same equation, run the other way, turns the disagreement into a measurement. Rearrange for the temperature difference:

$$T_2 - T_1 = \frac{1}{\alpha}\left(\frac{R(T_2)}{R(T_1)} - 1\right)$$

Take a cold reference reading $R(T_1)$ with the motor at a known, settled ambient temperature, run the motor, measure $R(T_2)$, and the ratio tells you how much hotter the copper got. This is the standard way to measure the average winding temperature of a motor without burying a sensor in it, because the winding is the sensor.

See it / Try it

The calculator below does both corrections so you can build intuition for the size of each. The top tool flips between wye and delta and between knowing the line-to-line or the phase value, so you can watch the factor of two and the factor of two-thirds appear. The bottom tool applies the copper temperature law. Try this: set the winding to wye with a line-to-line value of $1.0\ \Omega$, confirm the phase comes out at $0.5\ \Omega$, then feed that $0.5\ \Omega$ into the temperature tool as the value at $20\ ^\circ\text{C}$ and ask what it becomes at $80\ ^\circ\text{C}$.

You should see the phase resistance climb by roughly a quarter as the copper goes from room temperature to a hot running motor, and you should see the delta conversion move the phase number the opposite way from the wye conversion for the same line-to-line reading. Both effects are real, both are always present, and both have to be undone before a resistance number means anything you can compare.

?A motor is wye-connected. Your ohmmeter reads 1.2 ohm between two of its line terminals. What is the per-phase resistance?

• A 2.4 ohm, because line-to-line sees the phases in parallel
• B 0.6 ohm, because line-to-line sees two phases in series
• C 1.2 ohm, because the line-to-line reading is the phase resistance
• D 0.8 ohm, because delta divides by three-halves

?You record a winding at 0.50 ohm when the copper is 20 degC. After a run it reads 0.55 ohm. Using copper's alpha of about 0.00393 per degC, the winding is now roughly:

• A Cooler than 20 degC, since higher resistance means a colder metal
• B About 45 degC, a rise of roughly 25 degC inferred from the resistance ratio
• C Damaged, because winding resistance should never change with use
• D Unchanged in temperature, since 0.05 ohm is within meter noise

Lab

On the bench, always pair a winding-resistance reading with the copper’s temperature. Let the motor sit unpowered until it equilibrates with the room, measure each line-to-line resistance with a four-wire (Kelvin) connection so the lead and contact resistance does not swamp a sub-ohm winding, and write down the ambient temperature next to the number. That cold reading is your reference $R(T_1)$ at a known $T_1$. From it, decide the topology (count the leads, check the datasheet), convert to per-phase with the wye or delta factor, and optionally correct everything back to a standard $20\ ^\circ\text{C}$ so different days and different motors compare cleanly. A resistance with no temperature beside it is half a measurement, and the missing half is the half that starts arguments.

↘ Where alpha comes from, and why copper is such a clean thermometer

The temperature coefficient is not a property of a particular wire, it is a property of the material, inherited from how electrons scatter in the metal’s lattice. Resistivity $\rho$ (the intrinsic resistance of the material, independent of shape) follows the same linear law that the resistance does:

$$\rho(T) = \rho_0\,\bigl[\,1 + \alpha_0\,(T - T_0)\,\bigr]$$

where $\rho_0$ is the resistivity at the reference temperature $T_0$. Because resistance is just resistivity scaled by the wire’s geometry, $R = \rho\,\ell/A$, the geometry cancels out of the ratio $R(T_2)/R(T_1)$ and the same $\alpha$ governs both. That is why you can use a copper coefficient straight from a table on a winding whose exact length and cross-section you will never know.

Annealed copper sits at a resistivity near $1.72\times 10^{-8}\ \Omega\!\cdot\!\text{m}$ at $20\ ^\circ\text{C}$ with a temperature coefficient close to $0.00393\ \text{/}^\circ\text{C}$. Different references quote slightly different copper values (you will see numbers from about $0.0039$ to $0.0040\ \text{/}^\circ\text{C}$ depending on purity, anneal state, and the reference temperature the table was built around) so for careful work cite the figure with its reference temperature. We use $\alpha \approx 0.00393\ \text{/}^\circ\text{C}$ throughout, the annealed-copper value, which is why the worked numbers in this lesson assume $20\ ^\circ\text{C}$ references.

The linearity itself is an approximation. The honest law is exponential, $R(T) = R_0\,e^{\alpha(T-T_0)}$, and the familiar $R_0\,(1 + \alpha\,\Delta T)$ is just its first-order Taylor expansion, valid while $\alpha\,\Delta T \ll 1$. Over a motor’s range, even a $100\ ^\circ\text{C}$ rise gives $\alpha\,\Delta T \approx 0.39$, so the linear form is good to a few percent and the error is far smaller than the temperature swing you are trying to catch. Push to hundreds of degrees, or to cryogenic temperatures where the metal’s behavior changes character, and the straight line stops being a safe shortcut.

Grounded in Wikipedia: “Temperature coefficient”, “Electrical resistivity and conductivity”, “Synchronous motor” (CC BY-SA).

Key takeaways

• ▸A line-to-line reading is not the phase resistance; the winding connection always reshapes it.
• ▸Wye: two phases in series, so $R_{LL} = 2R_{ph}$ and $R_{ph} = R_{LL}/2$.
• ▸Delta: one phase parallel with two in series, so $R_{LL} = \tfrac{2}{3}R_{ph}$ and $R_{ph} = \tfrac{3}{2}R_{LL}$.
• ▸Copper resistance rises with temperature at about 0.39% per degC ($\alpha \approx 0.00393$/degC).
• ▸Run the law backwards and the winding becomes its own thermometer: $T_2 - T_1 = \tfrac{1}{\alpha}\!\left(\tfrac{R_2}{R_1} - 1\right)$.
• ▸Always record the winding temperature with the reading, ideally corrected to a standard 20 degC.

Practice 1 warm-up

A delta-connected motor reads $0.90\ \Omega$ line-to-line. What is its per-phase resistance?

↘ Show worked solution

For a delta, the line-to-line measurement sees one phase in parallel with the other two in series, so $R_{LL} = \tfrac{2}{3}R_{ph}$. Invert it: $R_{ph} = \tfrac{3}{2}\,R_{LL} = 1.5 \times 0.90\ \Omega = 1.35\ \Omega$. The per-phase resistance is larger than the line-to-line reading, the opposite of the wye case.

Practice 2 core

A wye motor reads $2.0\ \Omega$ line-to-line at $25\ ^\circ\text{C}$. First find the per-phase resistance, then find what that per-phase resistance becomes at $75\ ^\circ\text{C}$. Use $\alpha = 0.00393\ \text{/}^\circ\text{C}$ for copper.

↘ Show worked solution

Connection correction first: wye means $R_{ph} = R_{LL}/2 = 2.0/2 = 1.0\ \Omega$ at $25\ ^\circ\text{C}$.

Temperature correction next, with $T_2 - T_1 = 75 - 25 = 50\ ^\circ\text{C}$:

$$R(75^\circ) = R(25^\circ)\,[\,1 + \alpha\,(T_2 - T_1)\,] = 1.0\,[\,1 + 0.00393 \times 50\,] = 1.0 \times 1.1965 \approx 1.20\ \Omega.$$

So the phase climbs from $1.0\ \Omega$ to about $1.20\ \Omega$, a 20% rise from heat alone, with no change in the copper itself.

Practice 3 stretch

You log a wye motor’s winding at $0.80\ \Omega$ line-to-line with the copper settled at the room temperature of $22\ ^\circ\text{C}$. After a sustained run you measure $0.92\ \Omega$ line-to-line. Estimate the average winding temperature at the end of the run, and say why the per-phase versus line-to-line distinction does not change the answer.

↘ Show worked solution

Use the resistance ratio to find the temperature rise. Because both readings are taken the same way (line-to-line on the same wye motor), the connection factor of two appears in both $R_1$ and $R_2$ and cancels in the ratio, so you can work straight from the line-to-line numbers without converting to phase first:

$$T_2 - T_1 = \frac{1}{\alpha}\left(\frac{R_2}{R_1} - 1\right) = \frac{1}{0.00393}\left(\frac{0.92}{0.80} - 1\right) = \frac{0.15}{0.00393} \approx 38\ ^\circ\text{C}.$$

So the average winding temperature is about $22 + 38 = 60\ ^\circ\text{C}$. The wye-versus-delta correction only matters when you compare a resistance to an absolute per-phase spec; for a temperature rise inferred from two readings of the same winding, the geometry cancels and only the ratio matters.