Solved Problems · Set 42

Single-Phase Motor Equivalent Circuit

Part 6 · Single-Phase and Special Machines — halve the rotor and magnetising parameters, form two parallel blocks, and the whole machine becomes one series impedance.

Prof. Mithun Mondal 5 solved problems GATE · ESE · University

Set 42 — Single-Phase Motor Equivalent Circuit

Set 41 obtained the two revolving fields and the two slips. Turning that picture into numbers means drawing a circuit: the stator impedance in series with a forward block and a backward block, each built from half the magnetising reactance shunting half the rotor branch. Once \(Z_f\) and \(Z_b\) are on paper the machine is an ordinary series a.c. circuit.

The problems work outward from that circuit — mechanical power at a stated slip, torque when the magnetising branch is dropped, the various forward-to-backward ratios, a full efficiency calculation and a loss audit. The recurring lesson is that the backward field's small air-gap power hides a large rotor loss, because its slip is close to 2 and not close to 0.

Part 6 · Single-Phase Induction Motors · 5 solved problems

i Method Recap

  • Halve everything, then split it in two. The main-winding equivalent circuit puts the stator impedance \(R_1+jX_1\) in series with a forward block and a backward block, each built from half of the rotor and magnetising parameters:

    \[ x_m = \tfrac12X_m, \qquad r_2 = \tfrac12R_2', \qquad x_2 = \tfrac12X_2' \]

  • The two block impedances are parallel combinations, evaluated at the two slips:

    \[ Z_f = \frac{jx_m\left(\dfrac{r_2}{s}+jx_2\right)}{\dfrac{r_2}{s}+j(x_2+x_m)}, \qquad Z_b = \frac{jx_m\left(\dfrac{r_2}{2-s}+jx_2\right)}{\dfrac{r_2}{2-s}+j(x_2+x_m)} \]

  • One series impedance then gives the current and with it everything else:

    \[ Z_{01} = (R_1+jX_1) + Z_f + Z_b, \qquad I_1 = \frac{V}{Z_{01}} \]

  • Air-gap power is \(I_1^2\) times the real part of each block, because the magnetising branch is lossless:

    \[ P_{gf} = I_1^2\,\mathrm{Re}(Z_f), \qquad P_{gb} = I_1^2\,\mathrm{Re}(Z_b) \]

    Equivalently, split the voltage \(V_f = I_1Z_f\) across the forward block, find the rotor branch current, and use \(I_{2f}^2r_2/s\). Both routes are used on this page.

  • Torque, power and loss follow the double-field rules of Set 41:

    \[ T = \frac{P_{gf}-P_{gb}}{\omega_s}, \qquad P_m = (1-s)\left(P_{gf}-P_{gb}\right), \qquad P_{cu2} = sP_{gf}+(2-s)P_{gb} \]

  • When the magnetising branch is neglected the circuit collapses to a single series loop and the arithmetic becomes trivial:

    \[ Z = R_1 + \frac{r_2}{s} + \frac{r_2}{2-s} + j(X_1+X_2') \]

  • The forward and backward torque ratio is fixed by slip alone, a check worth applying to every answer:

    \[ \frac{T_f}{T_b} = \frac{2-s}{s} \quad\text{(equal branch currents)} \]
Problem 1Exam levelMechanical Power At 5% Slip

Find the gross mechanical power output at a slip of 0.05 for a 185 W, 4-pole, 110 V, 60 Hz single-phase induction motor whose main-winding constants are given below. Use the full double-revolving-field equivalent circuit.

ParameterSymbolValue
Resistance of the stator main winding\(R_1\)1.86 Ω
Leakage reactance of the stator main winding\(X_1\)2.56 Ω
Magnetising reactance of the main winding\(X_m\)53.5 Ω
Rotor resistance at standstill, referred to stator\(R_2'\)3.56 Ω
Rotor leakage reactance at standstill, referred to stator\(X_2'\)2.56 Ω
Solution

Halve the rotor and magnetising parameters before anything else, because each revolving field has half the amplitude and therefore sees half the machine:

\[ x_m = \frac{53.5}{2} = 26.75\ \Omega, \qquad r_2 = \frac{3.56}{2} = 1.78\ \Omega, \qquad x_2 = \frac{2.56}{2} = 1.28\ \Omega \]
Double-revolving-field equivalent circuit of a single-phase induction motor: stator resistance and leakage reactance in series with a forward block and a backward block, each formed by half the magnetising reactance in parallel with half the rotor branch
Main-winding equivalent circuit — forward and backward blocks in series with the stator impedance

The stator resistance and leakage reactance are not halved: they are traversed once by the whole stator current.

The forward block. Writing \(x_0 = x_2+x_m = 28.03\ \Omega\) and rationalising the parallel combination:

\[ Z_f = x_m\,\frac{\dfrac{r_2}{s}x_m + j\left[\left(\dfrac{r_2}{s}\right)^2 + x_2x_0\right]}{\left(\dfrac{r_2}{s}\right)^2 + x_0^2} \]
\[ Z_f = 26.75\,\frac{35.6\times26.75 + j\left[35.6^2 + 1.28\times28.03\right]}{35.6^2+28.03^2} = 12.41 + j16.98 = 21.03\angle53.8^\circ\ \Omega \]

Note \(r_2/s = 1.78/0.05 = 35.6\ \Omega\), comfortably larger than \(x_m\), so the forward block is dominated by the magnetising branch and is strongly inductive.

The backward block, identical in form but with \(r_2/(2-s) = 1.78/1.95 = 0.913\ \Omega\):

\[ Z_b = 26.75\,\frac{0.913\times26.75 + j\left[0.913^2 + 1.28\times28.03\right]}{0.913^2+28.03^2} = 0.83 + j1.25 = 1.50\angle56.4^\circ\ \Omega \]

Fourteen times smaller than \(Z_f\). The backward field is nearly a short circuit across the air gap, which is exactly why it contributes little voltage but a great deal of current.

Total impedance and stator current:

\[ \begin{aligned} Z_{01} &= (1.86+j2.56) + (12.41+j16.98) + (0.83+j1.25) \\ &= 15.10 + j20.79 = 25.69\angle54.0^\circ\ \Omega \\ I_1 &= \frac{110}{25.69} = 4.28\ \text{A} \end{aligned} \]

Divide the terminal voltage between the two fields, then find the current in each rotor branch:

\[ \begin{aligned} V_f &= I_1|Z_f| = 4.28\times21.03 = 90.0\ \text{V}, & V_b &= I_1|Z_b| = 4.28\times1.50 = 6.42\ \text{V} \\ Z_{2f} &= \left|\frac{r_2}{s}+jx_2\right| = |35.6+j1.28| = 35.62\ \Omega, & Z_{2b} &= \left|\frac{r_2}{2-s}+jx_2\right| = 1.572\ \Omega \\ I_{2f} &= \frac{90.0}{35.62} = 2.53\ \text{A}, & I_{2b} &= \frac{6.42}{1.572} = 4.08\ \text{A} \end{aligned} \]

The backward rotor current is larger than the forward one, even though the backward field is given only 6.4 V. This is the single most counter-intuitive number in the whole subject.

Air-gap power for each field, which in this convention is also the torque in synchronous watts:

\[ \begin{aligned} P_{gf} &= I_{2f}^2\frac{r_2}{s} = 2.53^2\times35.6 = 227.4\ \text{W} \\ P_{gb} &= I_{2b}^2\frac{r_2}{2-s} = 4.08^2\times0.913 = 15.2\ \text{W} \\ P_g &= 227.4 - 15.2 = 212.2\ \text{W} \end{aligned} \]

The gross mechanical power is the net air-gap power reduced by the slip:

\[ P_m = (1-s)\left(P_{gf}-P_{gb}\right) = 0.95\times212.2 = 202\ \text{W} \]

The motor is rated 185 W at the shaft, and 202 W of gross mechanical power leaves about 17 W for core loss, friction and windage — a sensible figure for a machine this size, which is the sanity check worth doing here. The corresponding developed torque is \(212.2/188.5 = 1.13\) N·m at \(\omega_s = 188.5\) rad/s.

Two impedances, one series circuit, and the rest is bookkeeping. Once \(Z_f\) and \(Z_b\) are on paper the machine behaves like any series a.c. circuit; the only single-phase-specific steps are halving the parameters at the start and subtracting the backward air-gap power at the end.
Answer\(Z_f = 12.41+j16.98\ \Omega,\ Z_b = 0.83+j1.25\ \Omega,\ I_1 = 4.28\ \text{A},\ P_m = 202\ \text{W}\)
Problem 2CoreTorque, Magnetising Branch Ignored

A 120 V, 60 Hz, 4-pole single-phase induction motor has the approximate equivalent circuit shown, in which the magnetising impedance has been omitted so that the forward and backward rotor branches carry the same current. The parameters are

ElementValue
Stator main-winding resistance \(R_1\)0.15 Ω
Half the rotor resistance \(\tfrac12R_2'\)0.1 Ω
Total series leakage reactance \(X_1+X_2'\)1.2 Ω

Estimate the torque developed when the rotor runs at 1764 rpm.

Approximate single-phase induction motor equivalent circuit with the magnetising branch omitted: stator resistance, total leakage reactance, and the two rotor resistances 0.5R2/s and 0.5R2/(2-s) all in one series loop
Approximate circuit — with \(X_m\) removed the whole machine is a single series loop
Solution

Find the slip from the speed:

\[ N_s = \frac{120\times60}{4} = 1800\ \text{rpm}, \qquad s = \frac{1800-1764}{1800} = 0.02 \]

Assemble the series impedance. With no magnetising branch the two rotor resistances simply add in series with the stator:

\[ \begin{aligned} Z &= R_1 + \frac{0.5R_2'}{s} + \frac{0.5R_2'}{2-s} + j\left(X_1+X_2'\right) \\ &= 0.15 + \frac{0.1}{0.02} + \frac{0.1}{1.98} + j1.2 \\ &= 0.15 + 5.0 + 0.0505 + j1.2 = 5.20 + j1.2 = 5.34\angle13^\circ\ \Omega \end{aligned} \]

The forward resistance is a hundred times the backward one — the ratio \((2-s)/s = 99\). Almost the entire series resistance belongs to the forward field.

The main-winding current:

\[ I_m = \frac{V}{Z} = \frac{120\angle0^\circ}{5.34\angle13^\circ} = 22.48\angle-13^\circ\ \text{A} \]

A power factor of \(\cos13^\circ = 0.974\), which is unrealistically high precisely because the magnetising reactance has been thrown away. That is the price of the approximation, and it does not affect the torque much because the magnetising branch carries no real power.

Torque is the difference of the two air-gap powers divided by synchronous speed. The same current flows in both branches, so:

\[ T = \frac{I_m^2}{\omega_s}\left(\frac{0.5R_2'}{s} - \frac{0.5R_2'}{2-s}\right) = \frac{22.48^2}{188.5}\left(\frac{0.1}{0.02} - \frac{0.1}{1.98}\right) \]
\[ T = 2.682\left(5.0 - 0.0505\right) = 13.41 - 0.14 = 13.27\ \text{N·m} \]

The backward field removes only 0.14 N·m — about 1% — at this slip. Written out this way the two contributions are visible separately, which is worth doing every time.

Cross-check through mechanical power. The gross output is

\[ P_m = T\,\omega_m = 13.27\times\frac{2\pi\times1764}{60} = 13.27\times184.7 = 2452\ \text{W} \]

Consistent with \((1-s)(P_{gf}-P_{gb}) = 0.98\times2502 = 2452\) W. A 2.4 kW machine drawing 22.5 A at 120 V is a large single-phase motor, but the numbers hang together.

Torque uses the difference of the resistances; current uses their sum. Both appear in the same expression here, and mixing them up — taking \(I^2\) times the total resistance for torque — would give 13.55 N·m instead of 13.27 N·m, and would predict torque at standstill where there is none.
Answer\(s = 0.02,\ I_m = 22.48\ \text{A},\ T = 13.27\ \text{N·m}\)
Problem 3ChallengeForward And Backward Ratios

A 220 V, 6-pole, 50 Hz single-winding single-phase induction motor has the following equivalent-circuit parameters referred to the stator:

\[ R_{1m} = 3.0\ \Omega, \quad X_{1m} = 5.0\ \Omega, \quad R_2' = 1.5\ \Omega, \quad X_2' = 2.0\ \Omega \]

Neglect the magnetising current. When the motor runs at 97% of synchronous speed, compute

  1. the ratio \(E_{mf}/E_{mb}\)
  2. the ratio \(V_f/V_b\)
  3. the ratio \(T_f/T_b\)
  4. the gross total torque
  5. the ratios \(T_f/T\) and \(T_b/T\).
Solution

The slip. Running at 97% of synchronous speed means the rotor is 3% behind:

\[ s = 1 - 0.97 = 0.03, \qquad 2-s = 1.97 \]
Equivalent circuit of a single-winding single-phase induction motor showing the forward rotor branch R2/s in parallel with jX and the backward branch R2/(2-s) in parallel with jX
The forward and backward rotor branches, each shunted by the magnetising reactance

(a) The ratio of the two air-gap e.m.f.s. Each equals the current times the corresponding block impedance; with the magnetising current neglected, \(X\to\infty\) and the parallel combination reduces to the rotor branch alone:

\[ \frac{E_{mf}}{E_{mb}} = \frac{\left|\dfrac{R_2'}{s}+jX_2'\right|}{\left|\dfrac{R_2'}{2-s}+jX_2'\right|} = \frac{|50+j2|}{|0.761+j2|} = \frac{50.04}{2.140} = 23.38 \]

(b) The ratio of the two applied voltages. Redrawing the circuit in symmetrical form, the stator impedance is shared equally between the two halves, so each half carries \(\tfrac12\left(R_{1m}+jX_{1m}\right)\) in series with its own rotor branch:

Symmetrical form of the single-phase induction motor circuit with the stator impedance divided equally between the forward half and the backward half, each half driven by its own voltage Vf and Vb
Symmetrical form — each half of the machine is driven by its own voltage
\[ \begin{aligned} \text{impedance to } V_f &= \tfrac12\left[\left(3+\frac{1.5}{0.03}\right) + j(5+2)\right] = \tfrac12(53+j7) \\ \text{impedance to } V_b &= \tfrac12\left[\left(3+\frac{1.5}{1.97}\right) + j(5+2)\right] = \tfrac12(3.76+j7) \end{aligned} \]
\[ \frac{V_f}{V_b} = \frac{|53+j7|}{|3.76+j7|} = \frac{53.46}{7.95} = 6.73 \]

Note this is far smaller than the e.m.f. ratio of 23.38, because the backward half must also carry the stator drop, which is the same in both halves.

(c) The torque ratio. Both branches carry the same current, so everything cancels except the slips:

\[ \frac{T_f}{T_b} = \frac{P_{gf}}{P_{gb}} = \frac{\tfrac12I_m^2R_2'/s}{\tfrac12I_m^2R_2'/(2-s)} = \frac{2-s}{s} = \frac{1.97}{0.03} = 65.7 \]

(d) The gross total torque. First the impedance seen at the stator terminals, which is the sum of the two halves:

\[ Z = \tfrac12\left[(53+j7)+(3.76+j7)\right] = 28.38+j7 = 29.23\angle13.9^\circ\ \Omega \]
\[ I_m = \frac{220}{29.23} = 7.53\ \text{A} \]
\[ N_s = \frac{120\times50}{6} = 1000\ \text{rpm}, \qquad \omega_s = \frac{2\pi\times1000}{60} = 104.72\ \text{rad/s} \]
\[ T = T_f - T_b = \frac{I_m^2R_2'}{2\omega_s}\left(\frac1s - \frac1{2-s}\right) = \frac{7.53^2\times1.5}{2\times104.72}\left(33.33 - 0.508\right) = 13.31\ \text{N·m} \]

Cross-check: the gross mechanical power is \(T\omega_m = 13.31\times104.72\times0.97 = 1352\) W, against an input of \(I_m^2\times28.38 = 1607\) W — a plausible efficiency before core, friction and windage losses.

(e) Each torque as a fraction of the total. Divide the two components by their difference:

\[ \begin{aligned} \frac{T_f}{T} &= \frac{1/s}{\dfrac1s-\dfrac1{2-s}} = \frac{1}{1-\dfrac{s}{2-s}} = 1.015 \\ \frac{T_b}{T} &= \frac{1/(2-s)}{\dfrac1s-\dfrac1{2-s}} = \frac{1}{\dfrac{2-s}{s}-1} = 0.015 \end{aligned} \]

The forward field must produce 101.5% of the useful torque so that the backward field can take 1.5% away. The two fractions always differ by exactly 1.

Every ratio on this page is a different question about the same two numbers. The e.m.f. ratio (23.4) compares the rotor branches alone; the voltage ratio (6.73) includes the shared stator drop; the torque ratio (65.7) compares the real powers. Quoting one when the question asked for another is the most frequent error in this problem type.
Answera\(23.38\) b\(6.73\) c\(65.7\) d\(T = 13.31\ \text{N·m}\) e\(T_f/T = 1.015,\ T_b/T = 0.015\)
Problem 4Exam levelEfficiency From The Full Circuit

A 230 V, 50 Hz, 4-pole single-phase induction motor runs on its main winding alone with the following constants referred to the stator:

\[ R_1 = 2.0\ \Omega, \quad X_1 = 2.8\ \Omega, \quad X_m = 60\ \Omega, \quad R_2' = 4.0\ \Omega, \quad X_2' = 2.8\ \Omega \]

The combined core, friction and windage loss is 40 W. At a slip of 0.05 determine

  1. the forward and backward block impedances
  2. the stator current, power factor and input power
  3. the two air-gap powers and the rotor copper loss
  4. the gross mechanical power, shaft output and efficiency.
Solution

Halve the rotor and magnetising parameters:

\[ x_m = 30\ \Omega, \quad r_2 = 2.0\ \Omega, \quad x_2 = 1.4\ \Omega, \quad x_0 = x_2+x_m = 31.4\ \Omega \]
\[ \frac{r_2}{s} = \frac{2.0}{0.05} = 40\ \Omega, \qquad \frac{r_2}{2-s} = \frac{2.0}{1.95} = 1.026\ \Omega \]

(a) The two block impedances:

\[ Z_f = 30\,\frac{40\times30 + j\left[40^2+1.4\times31.4\right]}{40^2+31.4^2} = 13.92 + j19.07\ \Omega \]
\[ Z_b = 30\,\frac{1.026\times30 + j\left[1.026^2+1.4\times31.4\right]}{1.026^2+31.4^2} = 0.935 + j1.368\ \Omega \]

(b) One series circuit gives the current and the input power:

\[ Z_{01} = (2.0+j2.8) + (13.92+j19.07) + (0.935+j1.368) = 16.86 + j23.24 = 28.71\angle54.0^\circ\ \Omega \]
\[ I_1 = \frac{230}{28.71} = 8.01\ \text{A}, \qquad \cos\phi = \cos54.0^\circ = 0.587\ \text{lagging} \]
\[ P_{in} = VI_1\cos\phi = 230\times8.01\times0.587 = 1082\ \text{W} \]

Equivalently \(P_{in} = I_1^2\,\mathrm{Re}(Z_{01}) = 64.19\times16.86 = 1082\) W. A power factor below 0.6 is normal for a single-phase motor running on the main winding alone.

(c) Air-gap powers come from the real parts of the blocks, since the magnetising reactance absorbs no watts:

\[ \begin{aligned} P_{gf} &= I_1^2\,\mathrm{Re}(Z_f) = 64.19\times13.92 = 893.5\ \text{W} \\ P_{gb} &= I_1^2\,\mathrm{Re}(Z_b) = 64.19\times0.935 = 60.0\ \text{W} \end{aligned} \]
\[ P_{cu2} = sP_{gf} + (2-s)P_{gb} = 0.05\times893.5 + 1.95\times60.0 = 44.7 + 117.0 = 161.7\ \text{W} \]

Check the stator: \(P_{cu1} = I_1^2R_1 = 128.4\) W, and \(1082 - 128.4 = 953.6\ \text{W} = P_{gf}+P_{gb}\). Everything crossing the air gap is accounted for.

(d) Mechanical power, shaft output and efficiency:

\[ P_m = (1-s)\left(P_{gf}-P_{gb}\right) = 0.95\times833.5 = 791.8\ \text{W} \]
\[ P_{shaft} = 791.8 - 40 = 751.8\ \text{W} \]
\[ \eta = \frac{751.8}{1082} = 0.695 = 69.5\% \]

Cross-check the loss ledger: \(128.4+161.7+40 = 330.1\) W of loss, and \(1082-330.1 = 751.9\) W of output. The shaft torque is \(751.8/149.2 = 5.04\) N·m at 1425 rpm, against a developed torque of \(833.5/157.1 = 5.31\) N·m.

Under 70% efficiency for a 750 W machine, and the backward field is most of the reason. It removes 60 W of air-gap power and then burns 117 W in the rotor — together 16% of the input, spent achieving nothing. A three-phase machine of the same rating would be around 80% efficient.
Answera\(Z_f = 13.92+j19.07\ \Omega,\ Z_b = 0.935+j1.368\ \Omega\) b\(I_1 = 8.01\ \text{A},\ \cos\phi = 0.587,\ P_{in} = 1082\ \text{W}\) c\(P_{gf}=893.5\ \text{W},\ P_{gb}=60.0\ \text{W},\ P_{cu2}=161.7\ \text{W}\) d\(P_m = 792\ \text{W},\ P_{shaft} = 752\ \text{W},\ \eta = 69.5\%\)
Problem 5CoreRotor Copper-Loss Split

A 220 V, 50 Hz, 6-pole single-phase induction motor runs at 940 rpm. From its equivalent circuit the main-winding current is 6.0 A, the forward block has a resistive part of 18.5 Ω and the backward block 1.1 Ω. The stator main-winding resistance is 2.5 Ω. Determine

  1. the two air-gap powers
  2. the rotor copper loss attributable to each field, and the total
  3. the fraction of the rotor loss caused by the backward field
  4. the gross mechanical power, and the rotor loss as a percentage of the net air-gap power.
Solution

The slip:

\[ N_s = \frac{120\times50}{6} = 1000\ \text{rpm}, \qquad s = \frac{1000-940}{1000} = 0.06, \qquad 2-s = 1.94 \]

(a) The air-gap powers are the stator current squared times the resistive part of each block:

\[ P_{gf} = 6.0^2\times18.5 = 666\ \text{W}, \qquad P_{gb} = 6.0^2\times1.1 = 39.6\ \text{W} \]

The backward field takes only 5.9% of what the forward field takes.

(b) The rotor copper loss belonging to each field. Each field's rotor loss is its own slip times its own air-gap power — and the backward field's slip is 1.94, not 0.06:

\[ \begin{aligned} P_{cu2,f} &= sP_{gf} = 0.06\times666 = 40.0\ \text{W} \\ P_{cu2,b} &= (2-s)P_{gb} = 1.94\times39.6 = 76.8\ \text{W} \\ P_{cu2} &= 40.0 + 76.8 = 116.8\ \text{W} \end{aligned} \]

(c) The backward field dominates the rotor heating:

\[ \frac{P_{cu2,b}}{P_{cu2}} = \frac{76.8}{116.8} = 0.658 = 65.8\% \]

Two thirds of the rotor loss is produced by a field that supplies only 6% of the air-gap power — and that removes torque rather than adding it. Almost all of that 76.8 W is at \((2-s)f = 97\) Hz, so it appears in the cage as a nearly-supply-frequency current.

(d) Mechanical power and the effective loss fraction:

\[ P_g = 666 - 39.6 = 626.4\ \text{W}, \qquad P_m = (1-s)P_g = 0.94\times626.4 = 588.8\ \text{W} \]
\[ \frac{P_{cu2}}{P_g} = \frac{116.8}{626.4} = 0.186 = 18.6\% \]

Check the balance: \(P_{gf}+P_{gb} = 705.6\) W enters the rotor, \(116.8\) W is dissipated, \(588.8\) W leaves as mechanical power. In a three-phase machine at 6% slip the rotor loss would be exactly 6% of the air-gap power; here it is 18.6%.

For completeness, the input power and stator copper loss:

\[ P_{in} = I_m^2\left(R_1+18.5+1.1\right) = 36\times22.1 = 795.6\ \text{W}, \qquad P_{cu1} = 36\times2.5 = 90\ \text{W} \]
"Rotor copper loss equals slip times air-gap power" is still true — twice. The whole trap is remembering that the backward field's slip is \(2-s\), a number close to 2, so its small air-gap power is multiplied by a large factor. Applying \(s\) to both fields would give 42.4 W instead of 116.8 W and understate the rotor heating by a factor of nearly three.
Answera\(P_{gf}=666\ \text{W},\ P_{gb}=39.6\ \text{W}\) b\(40.0\ \text{W}\) forward, \(76.8\ \text{W}\) backward, \(116.8\ \text{W}\) total c\(65.8\%\) d\(P_m = 588.8\ \text{W},\ 18.6\%\)
Formulas

Key Formulas

QuantityRelationNotes
Halved parameters\(x_m=\tfrac12X_m,\ r_2=\tfrac12R_2',\ x_2=\tfrac12X_2'\)\(R_1,X_1\) are not halved — Problems 1, 4
Forward block\(Z_f = jx_m\left(\dfrac{r_2}{s}+jx_2\right)\Big/\left[\dfrac{r_2}{s}+j(x_2+x_m)\right]\)Problems 1, 4
Backward block\(Z_b = jx_m\left(\dfrac{r_2}{2-s}+jx_2\right)\Big/\left[\dfrac{r_2}{2-s}+j(x_2+x_m)\right]\)Typically \(|Z_b|\approx|Z_f|/14\) — Problem 1
Rationalised form\(Z = x_m\dfrac{ax_m + j\left(a^2+x_2x_0\right)}{a^2+x_0^2},\ x_0=x_2+x_m\)\(a = r_2/s\) or \(r_2/(2-s)\)
Total impedance\(Z_{01} = R_1+jX_1+Z_f+Z_b\)Problems 1, 4
Approximate circuit\(Z = R_1+\dfrac{0.5R_2'}{s}+\dfrac{0.5R_2'}{2-s}+j(X_1+X_2')\)\(X_m\) omitted — Problems 2, 3
Air-gap powers\(P_{gf}=I_1^2\mathrm{Re}(Z_f),\ P_{gb}=I_1^2\mathrm{Re}(Z_b)\)Problems 4, 5
Branch voltages and currents\(V_f=I_1Z_f,\ I_{2f}=V_f/\left|\dfrac{r_2}{s}+jx_2\right|\)Problem 1
Developed torque\(T = \left(P_{gf}-P_{gb}\right)/\omega_s\)13.27 N·m — Problem 2
Torque ratio\(T_f/T_b = (2-s)/s\)65.7 at 3% slip — Problem 3
Torque fractions\(T_f/T = \dfrac{1}{1-s/(2-s)},\quad T_b/T = \dfrac{1}{(2-s)/s-1}\)They differ by 1 — Problem 3
Rotor copper loss\(P_{cu2} = sP_{gf}+(2-s)P_{gb}\)65.8% from the backward field — Problem 5
Mechanical power\(P_m = (1-s)\left(P_{gf}-P_{gb}\right)\)Problems 1, 4, 5
Shaft output and efficiency\(P_{shaft} = P_m - P_{rot},\quad \eta = P_{shaft}/P_{in}\)69.5% — Problem 4
E.m.f. ratio\(E_{mf}/E_{mb} = \left|\dfrac{R_2'}{s}+jX_2'\right|\Big/\left|\dfrac{R_2'}{2-s}+jX_2'\right|\)23.38 — Problem 3
Pitfalls

Common Mistakes

  1. Halving the stator impedance along with everything else. Only \(X_m\), \(R_2'\) and \(X_2'\) are split between the two fields; the whole stator current flows through the whole of \(R_1+jX_1\) — Problems 1 and 4.

  2. Halving \(R_2'\) and then also writing \(R_2'/2s\). Once \(r_2 = R_2'/2\) has been defined, the forward branch is \(r_2/s\), not \(r_2/2s\). In Problem 1 the double halving would give 17.8 Ω instead of 35.6 Ω.

  3. Using \(s\) for the backward rotor loss. The backward field's loss is \((2-s)P_{gb} = 76.8\) W, not \(sP_{gb} = 2.4\) W — Problem 5(b).

  4. Multiplying each air-gap power by \((1-s)\) separately. Mechanical power uses the net air-gap power once: \((1-s)(P_{gf}-P_{gb})\) — Problems 1, 4 and 5.

  5. Taking \(I^2\) times the total resistance for torque. Torque needs the difference of the two branch resistances; the sum gives the input power. In Problem 2 the confusion turns 13.27 N·m into 13.55 N·m and predicts a non-zero torque at standstill.

  6. Assuming the backward rotor current is small because \(V_b\) is small. In Problem 1 the backward branch sees only 6.42 V but its impedance is 1.57 Ω, so it carries 4.08 A — more than the forward branch's 2.53 A.

  7. Confusing the e.m.f. ratio with the voltage ratio. Problem 3 gives 23.38 for \(E_{mf}/E_{mb}\) but only 6.73 for \(V_f/V_b\), because the latter includes the stator drop shared by both halves.

  8. Reporting the stator current with a misplaced decimal. In Problem 3, \(220/29.23\) is 7.53 A; a 753 A reading in a 220 V fractional-kW machine should be rejected on sight, and the torque formula in fact uses 7.53.

  9. Forgetting the rotational loss when efficiency is asked for. Problem 4's 791.8 W is gross mechanical power; the shaft delivers 40 W less, and quoting \(791.8/1082 = 73.2\%\) overstates the efficiency by nearly four points.

  10. Trusting the power factor from the approximate circuit. Dropping \(X_m\) in Problem 2 gives \(\cos13^\circ = 0.974\), which no single-phase induction motor achieves. The approximation is acceptable for torque, not for current or power factor.

→ Looking Ahead

Every problem on this page began by being handed \(R_1\), \(X_1\), \(X_m\), \(R_2'\) and \(X_2'\). Nothing in the equivalent circuit is useful until those five numbers exist, and no manufacturer supplies them. They have to be measured, and measured on a machine that cannot be opened up or driven at synchronous speed.

The measurement uses the same two tests as the transformer and the three-phase induction motor — one at rated voltage with the rotor free, one at reduced voltage with the rotor locked — but with two single-phase complications. The auxiliary winding must be disconnected so that only the main winding is under test, and the no-load reading contains a stubborn backward-field term that will not go away because the backward slip is 2, not 0.

Next: Set 43 — Testing of Single-Phase Induction Motors, where the blocked-rotor test yields \(R_2'\) and the leakage reactances, the no-load test yields \(X_m\), and the rotational loss is separated into its core and friction-and-windage parts.