Set 25 — Induction Motor Fundamentals — Slip and Rotor Quantities
An induction motor is a transformer whose secondary is free to turn. The stator sets up a field rotating at synchronous speed; the rotor runs a little slower, and the difference — expressed as a fraction and called the slip — is what induces anything in the rotor at all. This set establishes the three consequences of that statement: the rotor frequency is \(sf\), the rotor e.m.f. is \(sE_2\) and the rotor reactance is \(sX_2\), while the rotor resistance is untouched by speed.
The problems then use those relations in the direction that matters for starting: with the slip rings shorted the rotor is a low-resistance, high-reactance circuit that draws an enormous current at a wretched power factor, and adding external resistance improves the power factor faster than it reduces the current. Getting the standstill rotor e.m.f. right — which means reading the stator connection and the per-phase turns ratio correctly — is the step on which every number here depends.
Synchronous speed and slip. The stator field rotates at \(N_s\), fixed by the supply frequency and the pole number alone; the rotor lags it by the slip:
\[ N_s = \frac{120f}{P}, \qquad s = \frac{N_s-N}{N_s}, \qquad N = N_s(1-s) \]Slip scales three rotor quantities and leaves one alone. The rotor conductors are cut at the slip speed, so
\[ f_2 = sf, \qquad E_{2s} = sE_2, \qquad X_{2s} = sX_2, \qquad R_2 \text{ unchanged} \]Here \(E_2\) and \(X_2\) are the standstill (\(s = 1\)) values.
Rotor current and rotor power factor then follow from one impedance triangle:
\[ I_2 = \frac{sE_2}{\sqrt{R_2^2+(sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2+(sX_2)^2}} \]The standstill rotor e.m.f. comes from the stator phase voltage, multiplied by the per-phase turns ratio \(K = N_2/N_1\). Read the stator connection first: a delta stator puts the full line voltage on each stator winding, a star stator only \(V_L/\sqrt3\).
Torque follows the rotor's real power, \(T \propto E_2I_2\cos\phi_2\), so improving the rotor power factor can raise the torque even while the current falls.
Maximum torque occurs where the rotor resistance equals the rotor reactance at that slip:
\[ R_2 = sX_2 \;\Longrightarrow\; s_b = \frac{R_2}{X_2}; \qquad \text{at starting } (s=1):\; R_2 = X_2 \]Currents refer across in the inverse turns ratio, exactly as in a transformer, and the stator connection must be undone before the ratio is applied:
\[ \frac{I_2}{I_{1,ph}} = \frac{N_1}{N_2} = \frac{1}{K} \]
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 50 Hz. Calculate
- the speed at which the magnetic field of the stator rotates
- the speed of the rotor when the slip is 0.04
- the frequency of the rotor currents when the slip is 0.03
- the frequency of the rotor currents at standstill.
The stator field turns at synchronous speed, which depends on nothing but the supply frequency and the number of poles:
No property of the rotor, and no amount of load, can change this number.
The rotor speed at 4% slip:
The rotor is 60 rpm behind the field; that 60 rpm of relative motion is the entire source of the rotor e.m.f.
The rotor frequency at 3% slip. The rotor conductors are cut by the field at the slip speed, so their frequency is the supply frequency scaled by \(s\):
In mechanical terms the field overtakes the rotor at \(sN_s = 0.03\times1500 = 45\) rpm, and a 4-pole machine turning at 45 rpm generates \(45\times4/120 = 1.5\) Hz — the same answer, arrived at from the geometry.
At standstill the slip is unity, so the rotor is simply a short-circuited transformer secondary:
The range the rotor frequency lives in. Between standstill and synchronism the rotor frequency falls from the full supply frequency to zero:
| Condition | Slip \(s\) | Speed \(N\) | Rotor frequency \(f_2\) |
|---|---|---|---|
| Standstill | 1 | 0 | 50 Hz |
| Running, 4% slip | 0.04 | 1440 rpm | 2 Hz |
| Running, 3% slip | 0.03 | 1455 rpm | 1.5 Hz |
| Synchronism (unreachable) | 0 | 1500 rpm | 0 |
A 3-phase induction motor having a star-connected rotor has an induced e.m.f. of 80 V between the slip rings at standstill on open circuit. The rotor has a resistance and a standstill reactance per phase of 1 Ω and 4 Ω respectively. Calculate the rotor current per phase and the rotor power factor when
- the slip rings are short-circuited
- the slip rings are connected to a star-connected rheostat of 3 Ω per phase.
Convert the slip-ring voltage to a phase value. The 80 V is measured between rings, that is between lines of a star-connected rotor:
Both parts are at standstill, \(s = 1\), so this full e.m.f. and the full standstill reactance apply throughout.
(a) Slip rings short-circuited. The rotor circuit is then 1 Ω of resistance against 4 Ω of reactance:
A large current at a power factor of a quarter: nearly all of it is reactive and contributes nothing to torque. This is why a plain short-circuited rotor starts badly.
(b) With 3 Ω per phase in the rheostat the resistance becomes \(3+1 = 4\ \Omega\) while the reactance is unchanged:
Why the starting torque rises even though the current falls. Torque is set by the in-phase component of the rotor current, \(T \propto E_2I_2\cos\phi_2\):
| Case | \(I_2\) | \(\cos\phi_2\) | \(I_2\cos\phi_2\) |
|---|---|---|---|
| Rings shorted | 11.2 A | 0.243 | 2.72 A |
| 3 Ω rheostat | 8.16 A | 0.707 | 5.77 A |
The current drops by 27% but the power factor nearly triples, so the torque-producing component — and with it the starting torque — rises by a factor of 2.1.
A 3-phase, 400 V, star-connected induction motor has a star-connected rotor with a stator-to-rotor turn ratio of 6.5. The rotor resistance and standstill reactance per phase are 0.05 Ω and 0.25 Ω respectively. What value of external resistance per phase must be inserted in the rotor circuit to obtain maximum torque at starting, and what is the rotor starting current with that resistance in place?
The standstill rotor e.m.f. per phase. The stator is in star, so each stator winding sees \(400/\sqrt3\), and the rotor winding on the same pair of poles has \(1/6.5\) of the turns:
The condition for maximum torque is that the rotor resistance equals the rotor reactance at the slip concerned. Starting means \(s = 1\), so the reactance is the full standstill value:
The external resistance is the difference between what is required and what the rotor already has:
Five times the rotor's own resistance — a large addition, but the machine's own \(R_2/X_2\) is only 0.2, so without it the peak torque would occur at 20% slip and the starting torque would be far below the maximum.
The rotor current with that resistance. Resistance and reactance are now equal, so the impedance is \(\sqrt2\) times either:
Equivalently \(I_2 = E_2/(\sqrt2 X_2)\), and the rotor power factor at maximum torque is always \(1/\sqrt2 = 0.707\) whatever the machine.
A 1100 V, 50 Hz delta-connected induction motor has a star-connected slip-ring rotor with a phase transformation ratio of 3.8 (stator to rotor). The rotor resistance and standstill leakage reactance are 0.012 Ω and 0.25 Ω per phase respectively. Neglecting stator impedance and magnetising current, determine
- the rotor current at start with the slip rings shorted
- the rotor power factor at start with the slip rings shorted
- the rotor current at 4% slip with the slip rings shorted
- the rotor power factor at 4% slip with the slip rings shorted
- the external rotor resistance per phase required to obtain a starting current of 100 A in the stator supply lines.
Establish the rotor phase voltage, which is where the connection matters. In a \(\Delta/Y\) machine the stator phase voltage is the same as the line voltage, so each stator winding is across the full 1100 V:
Had the stator been in star, each winding would see only 635 V and every current below would be smaller by \(\sqrt3\).
(i) and (ii) Starting, slip rings shorted — full e.m.f. against full standstill reactance:
A brutal current at a power factor of five per cent. Almost the whole 1157 A is magnetising the leakage paths and doing no work.
(iii) and (iv) At 4% slip. Three quantities scale with \(s\) and the resistance does not:
The e.m.f. has fallen by a factor of 25 but the impedance has fallen by 16, so the current is still 741 A — and now at 0.768 power factor, which is why the running torque at 4% slip vastly exceeds the starting torque despite the smaller current.
(v) Work back from the stator line current. The limit is 100 A in the supply lines, and the stator is in delta, so each stator winding carries only
Magnetising current is to be neglected, so this winding current is entirely the reflection of the rotor current.
Refer it to the rotor through the same per-phase ratio, currents transforming inversely to turns:
The rotor impedance that produces that current at standstill, and hence the resistance it must contain:
The trap in this part is to use the 100 A directly as the stator winding current. That omits the delta's \(\sqrt3\), gives \(I_2 = 380\) A and an external resistance of 0.707 Ω — which would in fact let \(\sqrt3\times100 = 173\) A into the supply lines.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Synchronous speed | \(N_s = 120f/P\) | rpm; independent of load — Problem 1 |
| Slip | \(s = (N_s-N)/N_s\) | \(N = N_s(1-s)\) — Problem 1 |
| Rotor frequency | \(f_2 = sf\) | 50 Hz at standstill, ~2 Hz running |
| Rotor e.m.f. | \(E_{2s} = sE_2\) | \(E_2\) is the standstill value |
| Rotor reactance | \(X_{2s} = sX_2\) | Falls with slip; \(R_2\) does not |
| Rotor impedance | \(Z_2 = \sqrt{R_2^2+(sX_2)^2}\) | Problems 2, 4 |
| Rotor current | \(I_2 = sE_2/Z_2\) | Per phase — Problems 2, 3, 4 |
| Rotor power factor | \(\cos\phi_2 = R_2/Z_2\) | 0.048 at start, 0.768 at 4% — Problem 4 |
| Standstill rotor e.m.f. | \(E_2 = K\,V_{1,ph}\) | \(K = N_2/N_1\), stator phase voltage |
| Star rotor e.m.f. | \(E_{2,ph} = E_{\text{slip rings}}/\sqrt3\) | Problem 2 |
| Current referral | \(I_2 = I_{1,ph}/K\) | Stator phase current — Problem 4 |
| Delta stator | \(V_{1,ph} = V_L,\; I_{1,ph} = I_L/\sqrt3\) | Problem 4 |
| Torque-producing current | \(T \propto E_2I_2\cos\phi_2\) | Why a rheostat helps — Problem 2 |
| Maximum-torque slip | \(s_b = R_2/X_2\) | At starting requires \(R_2 = X_2\) — Problem 3 |
| Current at max torque | \(I_2 = E_2/(\sqrt2\,X_2)\), \(\cos\phi_2 = 0.707\) | Problem 3 |
Common Mistakes
Scaling the rotor resistance with slip. Only \(E_2\), \(X_2\) and \(f_2\) carry a factor \(s\); \(R_2\) is a property of the copper and never moves — Problem 4(iii).
Quoting the rotor frequency in rpm. \(f_2 = sf\) is a frequency in hertz. If a speed is wanted it is the slip speed \(sN_s = 45\) rpm, which is not numerically 60 times the frequency unless the machine has two poles — Problem 1(c).
Using the slip-ring voltage as the phase e.m.f. The 80 V of Problem 2 is between rings, so a star rotor has only \(80/\sqrt3 = 46.2\) V per phase.
Using the stator line voltage in the turns-ratio calculation. It must be the stator phase voltage: \(400/\sqrt3\) for the star stator of Problem 3, but the full 1100 V for the delta stator of Problem 4.
Forgetting the delta on the current side. A 100 A line current in Problem 4(v) means 57.7 A in each stator winding; using 100 A gives an external resistance almost half the correct value.
Subtracting the external resistance from the wrong quantity. The condition \(R_2 = X_2\) fixes the total rotor-circuit resistance, so the rheostat must supply \(X_2 - R_2\) — Problem 3.
Assuming a lower rotor current means a lower torque. Adding resistance in Problem 2 cut the current from 11.2 A to 8.16 A and more than doubled the torque, because torque follows \(I_2\cos\phi_2\).
Applying the maximum-torque condition at the running slip when the question says starting. \(R_2 = sX_2\) becomes \(R_2 = X_2\) only because \(s = 1\) at standstill — Problem 3.
Confusing rotor-referred and stator-referred values. Every number in Problems 2 to 4 is an actual rotor quantity; multiplying by \(1/K^2\) to get \(R_2'\) belongs to the equivalent-circuit work of Set 26 onwards, not here.
The rotor has now been described completely at any speed: an e.m.f. \(sE_2\) driving a current through \(R_2 + jsX_2\), with a power factor that improves as the machine runs up. Problems 2 and 3 have already shown, informally, that torque follows the product \(E_2I_2\cos\phi_2\) and that rotor resistance decides where along the speed range the best torque appears.
That informal statement now has to become an equation. Writing the torque out in full turns it into a function of slip with a single maximum, and the position and height of that maximum answer almost every question an examiner can ask about starting, pull-out and the effect of a rotor rheostat.
Next: Set 26 — Torque and Torque–Slip Relations, where the torque expression is written explicitly, the slip for maximum torque is derived, and the ratios \(T/T_{max}\) and \(T_{st}/T_{max}\) do the work.