Set 41 — Double-Revolving-Field Theory
A single-phase winding cannot produce a rotating field. What it produces is a field that pulses along one axis, and the whole of this set follows from a single trigonometric identity that rewrites that pulsation as two constant-amplitude fields of half the size, turning in opposite directions at synchronous speed. The rotor then belongs to two machines at once: one at slip \(s\), one at slip \(2-s\).
The problems drill the consequences in order — the two rotor frequencies, the two branch resistances with their easily lost factor of one half, the two air-gap powers and the torque that is their difference. The last problem closes the loop by putting \(s=1\) into that difference and getting zero, which is why Set 44 exists at all.
A single-phase winding produces a pulsating field, not a rotating one. Resolve it into two fields of half amplitude turning in opposite directions at synchronous speed:
\[ \mathcal{F} = \mathcal{F}_{max}\sin\omega t\,\cos\theta = \tfrac12\mathcal{F}_{max}\sin(\omega t-\theta) + \tfrac12\mathcal{F}_{max}\sin(\omega t+\theta) \]The rotor sees two slips. Running at slip \(s\) with respect to the forward field, it runs at \(2-s\) with respect to the backward field, so two rotor frequencies exist together:
\[ N_s = \frac{120f}{P}, \qquad s = \frac{N_s-N}{N_s}, \qquad f_{2f} = sf, \qquad f_{2b} = (2-s)f \]Each field gets half the rotor. The standstill rotor resistance referred to the main stator winding, \(R_2'\), is split between the two branches and then scaled by the branch slip:
\[ R_f = \frac{R_2'}{2s}, \qquad R_b = \frac{R_2'}{2(2-s)} \]The factor 2 in the denominator is the halving of the field; the \(s\) and \(2-s\) are the two slips. Losing either factor is the commonest error on this page.
Air-gap power and torque follow separately for each field, and the two torques oppose:
\[ P_{gf} = I^2R_f, \qquad P_{gb} = I^2R_b, \qquad T = \frac{P_{gf}-P_{gb}}{\omega_s} \]At standstill the two are equal, so the motor will not start. Putting \(s=1\) gives \(R_f = R_b = R_2'/2\), hence \(P_{gf} = P_{gb}\) and \(T = 0\) whatever the current. Every starting method in Set 44 exists to break this symmetry.
Rotor loss and mechanical power collect the contributions of both fields:
\[ P_{cu2} = sP_{gf} + (2-s)P_{gb}, \qquad P_m = (1-s)\left(P_{gf}-P_{gb}\right) \]The forward field still obeys the ordinary maximum-torque condition, so the slip at which torque peaks is fixed by the rotor time constant alone:
\[ s_{mT} = \frac{R_2'}{X_2'} \]
A 6-pole, 50 Hz single-phase induction motor runs at 900 rpm. Determine the frequencies of the currents induced in the cage rotor.
Synchronous speed first, since both revolving fields turn at it — one forward, one backward:
The slip with respect to the forward field:
The rotor is simultaneously a member of two circuits. Relative to the backward field the rotor is running against the rotation, so the relative speed is \(N_s+N\) and the slip is \(2-s\):
Check the second one from the geometry: the backward field passes the rotor at \(1000+900 = 1900\) rpm, and a 6-pole machine at 1900 rpm gives \(1900\times6/120 = 95\) Hz.
Both currents flow in the same bars at the same time. The cage therefore carries a 5 Hz component that produces useful torque and a 95 Hz component that produces braking torque and a great deal of heat — the near-supply-frequency current sees a high rotor reactance and contributes almost nothing but loss.
A single-phase motor rated 2 HP is supplied at 240 V a.c. Its efficiency is 70% and its power factor 0.8 lagging. Find the input current.
Convert the rating to watts. The horsepower rating is the shaft output, never the input:
Work back through the efficiency to get the electrical input:
A single-phase supply has no \(\sqrt3\), so the input power is simply \(VI\cos\phi\):
The apparent power drawn is \(240\times11.1 = 2664\) VA for 1492 W of shaft work — the product \(\eta\cos\phi = 0.56\) is what makes small single-phase motors such poor citizens on a distribution circuit.
A 4-pole single-phase induction motor supplied at 100 V, 50 Hz rotates clockwise at 1000 rpm. Its standstill rotor resistance referred to the main stator winding is 1.7 Ω. Find the effective rotor resistance of the backward branch, and compare it with that of the forward branch.
The two branch resistances are what the equivalent circuit needs. Each revolving field has half the amplitude, so each branch carries half the rotor resistance, and that half is then divided by the slip its own field sees:
Here \(R_2'\) is the standstill rotor resistance referred to the main stator winding.
Synchronous speed and slip:
The backward branch, with \(2-s = 5/3\):
The forward branch, for contrast:
The ratio is always \((2-s)/s\) and never depends on \(R_2'\). At 33% slip the forward branch is only five times the backward one; at a normal running slip of 4% it would be 49 times, which is why the backward field is usually ignored in a rough estimate but never in a loss calculation.
A 6-pole, 50 Hz induction motor has a rotor resistance of 0.01 Ω per phase referred to the stator. Its stalling (pull-out) speed is 900 rpm. Determine the resistance that must be inserted in each rotor phase so that maximum torque occurs at starting.
Stalling speed identifies the slip at which torque peaks. Pull-out is by definition the maximum-torque point:
The maximum-torque condition gives the standstill rotor reactance, which is the one machine constant not supplied:
Adding external resistance changes \(R_2'\) but leaves \(X_2'\) untouched, so this number is fixed for the rest of the problem.
Maximum torque at starting means \(s_{mT} = 1\), because standstill is \(s = 1\). The condition therefore demands that the total rotor-circuit resistance equal the reactance:
The rheostat supplies the difference, not the whole of it — the winding already contributes 0.01 Ω:
Ten times the winding's own resistance, which is typical: a machine designed for a good running efficiency has a rotor resistance far below the value that would give the best starting torque.
A 230 V, 50 Hz, 4-pole single-phase induction motor runs at 1440 rpm with only its main winding energised. Its standstill rotor resistance referred to the main winding is 2.5 Ω, and at this speed the main-winding current is 4.0 A. Neglecting the magnetising branch, so that the same current flows through both rotor branches, determine
- the forward and backward slips
- the effective resistance of each branch
- the forward and backward air-gap powers
- the rotor copper loss and the gross mechanical power
- the developed torque.
The two slips. The forward slip comes from the speed; the backward slip is its complement about 2:
The branch resistances, each carrying half of \(R_2'\):
A factor of 49 between them — exactly \((2-s)/s\), as Problem 3 showed.
The air-gap powers. With the magnetising branch neglected the same 4 A flows through both branches, so each air-gap power is just \(I^2\) times its own resistance:
The rotor copper loss takes a fraction \(s\) of the forward air-gap power and a fraction \(2-s\) of the backward one:
The two halves are identical, and that is no accident: both equal \(\tfrac12I^2R_2' = \tfrac12(16)(2.5) = 20\) W. Whenever the same current flows in both branches, the forward and backward fields dissipate exactly the same rotor copper loss — even though one delivers 500 W across the air gap and the other only 10 W.
The gross mechanical power:
Check the balance: \(P_{gf}+P_{gb} = 510.2 = P_m + P_{cu2} = 470.2 + 40.0\) W. Every watt that crosses the air gap in either direction is accounted for.
The developed torque, from the net air-gap power at synchronous speed:
The same number follows from \(P_m/\omega_m = 470.2/150.80 = 3.12\) N·m, as it must.
The main winding of a single-phase induction motor carries a current \(i = \sqrt2\,I\sin\omega t\) and produces a pulsating mmf \(\mathcal{F} = \mathcal{F}_{max}\sin\omega t\,\cos\theta\) along its own axis.
- Resolve this mmf into two constant-amplitude revolving fields.
- With the magnetising branch neglected, so that the same current flows in both rotor branches, obtain the ratio \(T_f/T_b\) as a function of slip alone.
- Evaluate that ratio at standstill and at 4% slip, and hence explain why the machine develops no starting torque.
- For a 4-pole, 50 Hz machine, find the speed at which the backward torque has fallen to 5% of the forward torque.
Split the pulsating mmf with a product formula. Using \(\sin A\cos B = \tfrac12\sin(A-B)+\tfrac12\sin(A+B)\):
Each term has a constant amplitude \(\tfrac12\mathcal{F}_{max}\) and a crest that moves at \(\mathrm{d}\theta/\mathrm{d}t = \pm\omega\), i.e. at synchronous speed in opposite directions. The half-amplitude is the origin of every factor of \(\tfrac12\) in the equivalent circuit.
Each wave acts on the rotor like an independent machine. The forward wave sees slip \(s\), the backward wave slip \(2-s\), so the two branch resistances are \(R_2'/2s\) and \(R_2'/2(2-s)\). Torque is air-gap power divided by \(\omega_s\), and with the same current \(I\) in both branches:
The current, the voltage and the rotor resistance all cancel. The balance between the two fields is set by the speed and by nothing else.
At standstill the two fields are exactly matched. Put \(s=1\):
Note what this argument does not depend on: the supply voltage, the rotor current, or the rotor resistance. A stationary single-phase motor develops zero net torque no matter how hard it is driven, because the two fields are mirror images of one another.
Once turning, the symmetry is broken and stays broken. At 4% slip:
This is the crucial asymmetry: the machine has no starting torque but a perfectly good running torque, so if it is given a push in either direction it accelerates in that direction and keeps going. The direction of rotation is decided entirely by whichever way the rotor first moves.
The 5% point. Set the torque ratio to 0.05 and solve for the slip:
Alternative method — the torque–speed curves. Draw the ordinary three-phase torque–slip curve for the forward field, then draw the same curve reflected about both axes for the backward field, which is at slip \(2-s\). Adding the two ordinates gives the resultant. The sum passes through the origin of the torque axis at \(N = 0\) — the graphical statement of part (c) — and is antisymmetric in speed, which is the graphical statement that the machine will run equally well either way.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Field decomposition | \(\mathcal{F}_{max}\sin\omega t\cos\theta = \tfrac12\mathcal{F}_{max}\left[\sin(\omega t-\theta)+\sin(\omega t+\theta)\right]\) | Two half-amplitude waves — Problem 6 |
| Synchronous speed | \(N_s = 120f/P\) | Both fields turn at it — Problems 1, 3, 4, 5 |
| Forward and backward slip | \(s\) and \(2-s\) | They sum to 2 at every speed |
| Rotor frequencies | \(f_{2f}=sf,\quad f_{2b}=(2-s)f\) | 5 Hz and 95 Hz — Problem 1 |
| Forward branch resistance | \(R_f = R_2'/(2s)\) | Problems 3, 5 |
| Backward branch resistance | \(R_b = R_2'/\left[2(2-s)\right]\) | Problems 3, 5 |
| Branch ratio | \(R_f/R_b = (2-s)/s\) | Independent of \(R_2'\) — Problems 3, 6 |
| Air-gap powers | \(P_{gf}=I^2R_f,\quad P_{gb}=I^2R_b\) | Equal currents only if \(X_m\) is ignored — Problem 5 |
| Net torque | \(T = \left(P_{gf}-P_{gb}\right)/\omega_s\) | 3.12 N·m — Problem 5 |
| Rotor copper loss | \(P_{cu2} = sP_{gf} + (2-s)P_{gb}\) | Split 20 W / 20 W — Problem 5 |
| Mechanical power | \(P_m = (1-s)\left(P_{gf}-P_{gb}\right)\) | Problem 5 |
| Zero starting torque | \(s=1 \Rightarrow R_f=R_b \Rightarrow T=0\) | Independent of \(V\) and \(I\) — Problem 6 |
| Slip for maximum torque | \(s_{mT} = R_2'/X_2'\) | Forward field — Problem 4 |
| Rheostat for starting peak | \(R_{ext} = X_2' - R_2'\) | 0.09 Ω/phase — Problem 4 |
| Single-phase input power | \(P_{in} = VI\cos\phi = P_{out}/\eta\) | No \(\sqrt3\) — Problem 2 |
Common Mistakes
Dropping the factor of one half in the branch resistances. Writing \(R_b = R_2'/(2-s)\) instead of \(R_2'/\left[2(2-s)\right]\) doubles the answer in Problem 3, from 0.51 Ω to 1.02 Ω. The half comes from the field, not from the slip.
Using \(1-s\) for the backward slip. The backward field passes the rotor at \(N_s+N\), giving \(2-s\). In Problem 1 the wrong choice turns 95 Hz into 45 Hz.
Quoting a single rotor frequency. A single-phase cage carries two frequencies at once — 5 Hz and 95 Hz in Problem 1 — and the high-frequency one is responsible for most of the rotor heating.
Assuming the backward field is negligible because its air-gap power is small. In Problem 5 it takes only 10.2 W across the air gap yet dissipates 20 W in the rotor, exactly as much as the forward field does.
Applying \(P_m = (1-s)P_g\) to each field separately and adding. The correct form uses the net air-gap power once: \(P_m = (1-s)(P_{gf}-P_{gb})\) — Problem 5(d).
Subtracting the rheostat resistance from the reactance instead of adding the winding resistance to it. The condition \(s_{mT}=1\) fixes the total rotor resistance at \(X_2'\), so \(R_{ext} = X_2'-R_2' = 0.09\) Ω, not 0.1 Ω — Problem 4.
Believing the rotor rheostat raises the peak torque. It only moves the peak to a different slip; the maximum value is fixed by \(V\) and \(X_2'\) — Problem 4.
Inserting a \(\sqrt3\) into the single-phase power equation. Problem 2 needs \(P = VI\cos\phi\); the three-phase form would give 6.4 A instead of 11.1 A.
Treating the horsepower rating as input power. The 2 HP of Problem 2 is shaft output, so it must be divided by \(\eta\) before the power factor is applied.
Explaining zero starting torque by "no rotating field". There are two rotating fields at standstill, not none; they simply produce equal and opposite torques — Problem 6(c).
The double-revolving-field picture has now done everything it can do on its own: two slips, two branch resistances, two air-gap powers and a torque that is their difference. Problem 5 got its numbers by assuming the same current flows in both rotor branches, which is only true if the magnetising reactance is ignored. That assumption is comfortable but it is worth several percent on every answer, and it hides the fact that the forward and backward branches present very different impedances to the stator.
Removing the assumption means drawing the circuit properly: the stator impedance in series with two parallel combinations, each consisting of half the magnetising reactance shunting half the rotor branch. Once that circuit is on paper the same quantities come out again, but exactly, and the stator current, power factor and efficiency come with them.
Next: Set 42 — Single-Phase Motor Equivalent Circuit, where \(Z_f\) and \(Z_b\) are computed as complex parallel combinations and the whole performance of the machine is read off one series impedance.