Induction motor starters

Switch on a large induction motor directly across the line and it draws a violent gulp of current — five to seven times its rated value — for the second or two it takes to spin up. At standstill the rotor is not yet generating any back-EMF to oppose the flow, so the machine behaves almost like a short-circuited transformer. Left unchecked, that inrush dims lights, trips breakers, and stresses windings every time the motor starts. Starters exist to tame this surge without giving up too much starting torque in exchange. This article works through why the surge happens, how DOL, star-delta, autotransformer and rotor-resistance starters each attack it differently, and the one trade-off — current falls with voltage, torque falls with voltage squared — that governs every reduced-voltage method.

The core insight

At standstill an induction motor's rotor resistance is at its lowest effective value, so starting current is large; reduced-voltage starters (star-delta, autotransformer) cut that current by lowering stator voltage, but because torque depends on voltage squared, torque falls even faster, while a rotor-resistance starter (wound-rotor motors only) is the one method that can cut current and raise starting torque at once.

The core idea

Why starting current is so large. At standstill, slip $s=1$ , so the rotor-branch resistance term $R_2'/s$ collapses to its minimum value $R_2'$ . With almost no opposing resistance, the equivalent circuit looks close to a transformer with its secondary shorted, and both stator and rotor current shoot up to five to seven times the full-load value for a cage motor started directly on the line.

The universal trade-off. Because developed torque is proportional to the square of the applied stator voltage while current is (roughly) proportional to voltage itself,

T \propto V_1^2, \qquad I \propto V_1,

where: $T$ is developed torque, $I$ is stator/line current, and $V_1$ is the stator (phase) voltage applied at starting. Any method that reduces $V_1$ to cut current pays for it with a disproportionately larger drop in torque — this single relation explains why every reduced-voltage starter is a compromise, never a free lunch.

The four methods, in one line each.

Direct-On-Line (DOL): full line voltage from the start; no reduction, so full starting torque but also the full 5-7x inrush. Reserved for small motors or a supply that can absorb the surge.
Star-delta: windings normally run in delta are switched to star for starting, then back to delta once the motor has picked up speed.
Autotransformer: a three-phase autotransformer taps a reduced voltage (commonly 50-80% of line) for starting, then switches to full voltage; the tap can be chosen, unlike star-delta's fixed ratio.
Rotor-resistance starter (wound-rotor only): external resistance is inserted in the rotor circuit via slip rings and cut out in steps as the rotor speeds up. This raises $R_2$ directly rather than lowering $V_1$ , so it escapes the $T \propto V^2$ penalty entirely.

Figure

Bar chart comparing starting current and starting torque as multiples of full load for DOL star delta and autotransformer starting

Derivation AWhy star-delta cuts starting current to about one third of DOL

The goal is to show, using only the relations between phase and line quantities in star and delta connections, that switching a motor's stator windings from their normal delta connection to star for starting reduces the line starting current to about one third of the DOL value — a result that is stated but never proved in most first courses, and that examiners like to test by asking a student to reconstruct it from scratch rather than just quote it.

Step 1. Fix the notation. Let $V_L$ be the fixed line (supply) voltage and let $Z$ be the fixed standstill impedance per phase of one stator winding (the same physical winding, same $Z$ , whichever way it is connected — only the connection changes). In a delta connection each winding is directly across the line, so the phase voltage equals the line voltage: $V_{ph,\Delta} = V_L$ . In a star connection each winding sees only the line-to-neutral voltage: $V_{ph,Y} = V_L/\sqrt3$ .

Step 2. Find the phase current in each connection. Phase current is simply phase voltage over the fixed winding impedance $Z$ : in delta, $I_{ph,\Delta} = V_L/Z$ ; in star, $I_{ph,Y} = (V_L/\sqrt3)/Z = V_L/(\sqrt3 Z)$ . The star phase current is already $1/\sqrt3$ of the delta phase current, because the winding sees a smaller voltage.

Step 3. Convert phase current to line current for each connection — this is the step students most often skip. In delta, line current is $\sqrt3$ times phase current: $I_{L,\Delta} = \sqrt3\, I_{ph,\Delta} = \sqrt3\, V_L/Z$ . In star, line current equals phase current (each winding carries the full line current): $I_{L,Y} = I_{ph,Y} = V_L/(\sqrt3 Z)$ .

Step 4. Take the ratio of star line current to delta line current. $\dfrac{I_{L,Y}}{I_{L,\Delta}} = \dfrac{V_L/(\sqrt3 Z)}{\sqrt3\, V_L/Z} = \dfrac{1}{\sqrt3}\times\dfrac{1}{\sqrt3} = \dfrac{1}{3}$ . The factor of $1/\sqrt3$ from the reduced phase voltage and a second factor of $1/\sqrt3$ from the star-to-delta line-current relation multiply together, giving exactly $1/3$ — not $1/\sqrt3$ , a common error.

Step 5. Read off the torque consequence. Since $T \propto V_1^2$ and the phase (and hence effective) voltage across each winding in star is $1/\sqrt3$ of its delta value, torque in star is $(1/\sqrt3)^2 = 1/3$ of the DOL torque — the same factor of $1/3$ appears for both current and torque here only because the star connection changes both quantities through the identical $1/\sqrt3$ voltage step and (for current alone) one extra $1/\sqrt3$ from the line-current relation, whereas torque has no such extra factor. This is why star-delta happens to give current and torque both cut to about one third, a coincidence of this particular method rather than the general rule ( $I \propto V$ , $T \propto V^2$ ) that governs autotransformer starting. Result: $I_{L,Y} = \tfrac13 I_{L,\Delta}$ and $T_{Y} = \tfrac13 T_{\Delta}$ , matching the standard star-delta rule used in Worked Example 1 below.

Worked example

Problem. A cage induction motor draws a DOL starting current of $6\times$ full-load current (FLC) and a DOL starting torque of $1.5\times$ full-load torque (FLT). Find the starting line current and starting torque, as multiples of full-load values, (a) with a star-delta starter, and (b) with an autotransformer starter tapped at 60% of line voltage (assume the autotransformer reduces line-side starting current by the square of the tap ratio, since motor current falls by the tap ratio and the transformer action supplies a further tap-ratio reduction on the line side).

Step 1. Star-delta line current. From the derivation above, $I_{start,Y} = \tfrac13 I_{start,DOL} = \tfrac13 \times 6 = 2\times$ FLC.

Step 2. Star-delta starting torque. $T_{start,Y} = \tfrac13 T_{start,DOL} = \tfrac13 \times 1.5 = 0.5\times$ FLT.

Step 3. Autotransformer, tap ratio $k = 0.6$ . Torque scales as $k^2$ : $T_{start} = k^2 T_{start,DOL} = 0.36 \times 1.5 = 0.54\times$ FLT.

Step 4. Autotransformer line current also scales as $k^2$ (motor current falls by $k$ , and the autotransformer's turns ratio contributes a further factor of $k$ when referring that current to the line side): $I_{start} = k^2 I_{start,DOL} = 0.36 \times 6 = 2.16\times$ FLC.

Answer: Star-delta gives $2\times$ FLC and $0.5\times$ FLT; the 60% autotransformer gives $2.16\times$ FLC and $0.54\times$ FLT — a marginally better torque-per-ampere trade than star-delta here, and adjustable simply by choosing a different tap, unlike star-delta's fixed $1/3$ reduction.

Competitive stretch

If the same motor instead used a wound-rotor rotor-resistance starter and the added rotor resistance were chosen to shift the pull-out slip to exactly $s_{max}=1$ (i.e. $R_2 = X_2$ at standstill), starting torque would rise to the machine's maximum (pull-out) torque — larger than the $1.5\times$ FLT DOL value — while starting current, which depends mainly on total rotor-circuit impedance, would fall well below $6\times$ FLC. No stator-voltage method can improve current and torque simultaneously this way, which is exactly why rotor-resistance starting is prized wherever slip rings are available.

Common traps

Treating star-delta's current and torque reduction as proof that reduced-voltage methods always cut both by the same factor — it is a coincidence of star-delta's particular $1/\sqrt3$ -then- $1/\sqrt3$ structure; for an autotransformer, current and torque both scale as $k^2$ , not as $k$ and $k^2$ separately mixed up.
Using $1/\sqrt3$ instead of $1/3$ for the star-delta current ratio — forgetting that the phase-voltage reduction ( $1/\sqrt3$ ) and the star-to-delta line-current conversion ( $1/\sqrt3$ ) both apply, and they multiply.
Assuming a rotor-resistance starter can be fitted to a cage motor — the cage rotor bars are short-circuited internally with no external terminals; only wound-rotor (slip-ring) motors allow external rotor resistance.
Dismissing DOL as inferior in every case — for small motors, or a supply stiff enough to absorb the inrush, DOL remains standard because it delivers full starting torque with the least equipment.

Ask the tutor: want to work through how the pull-out slip $s_{max}=R_2/X_2$ from the torque-speed curve pins down exactly how much rotor resistance to add for maximum starting torque?

Why does star-delta reduce line current to exactly one third and not one half or 1 over root 3?

Two separate $1/\sqrt3$ factors are at work and they multiply. Switching from delta to star drops the voltage across each winding to $1/\sqrt3$ of the line voltage, which alone would cut phase current to $1/\sqrt3$ . But in star the line current equals the phase current, whereas in delta the line current is $\sqrt3$ times the phase current — so converting phase current back to line current brings in a second factor of $1/\sqrt3$ . Multiplying $1/\sqrt3 \times 1/\sqrt3$ gives exactly $1/3$ , which is why the standard rule of thumb quotes one third, not $1/\sqrt3$ (about $0.577$ , a common wrong answer).

If autotransformer starting can use any tap, why is star-delta still common?

Star-delta needs no extra transformer hardware — just a switch that reconnects the same windings, so it is cheap, compact and low-maintenance, which matters for the very common case of small-to-medium cage motors where the fixed $1/3$ current reduction is good enough. An autotransformer starter is bulkier and costlier because it adds a genuine three-phase transformer, but it earns its keep when the fixed star-delta ratio does not give enough starting torque for the load, since the tap can be raised (say to 70-80%) to trade some current reduction for extra torque.

Why does adding rotor resistance help a wound-rotor motor start, when adding series impedance anywhere else in a circuit usually only limits current?

Because rotor resistance does two things at once in an induction motor, not one. It limits starting current the way any added resistance would, by increasing total rotor-circuit impedance. But it also shifts the slip at which maximum torque occurs, $s_{max} = R_2/X_2$ , since a larger $R_2$ pushes $s_{max}$ closer to $s=1$ (standstill). Choosing $R_2$ so that $s_{max}$ lands near $s=1$ means the motor's maximum possible torque is delivered right at starting, instead of at some higher speed the motor never gets to use during the starting interval — that is the second effect no stator-voltage method can offer.

Induction motor starters

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads