Equivalent Circuit Problems

Set up the series branch. Take the phase voltage as reference and put the slip inside the rotor resistance:

In the approximate circuit this branch hangs directly across the supply, so it can be solved without knowing the air-gap e.m.f.

The referred rotor current:

Its power factor of \(\cos7.91^\circ = 0.990\) is excellent, because at 3% slip \(R_2'/s\) dominates the leakage reactance seven times over. This is why an induction motor's rotor behaves almost resistively at running speed.

The exciting current is now trivial, since the branch sits at the terminals:

Almost purely reactive, as an air-gapped magnetic circuit must be: the real part carries only the core loss.

Add the two currents to get the stator current:

Note how badly the magnetising current damages the power factor: the rotor draws its current at 0.99, and the machine as a whole manages only 0.875.

Air-gap power and torque. All the power entering \(R_2'/s\) crosses the gap:

Mechanical power and gross efficiency:

Confirm the input by adding the losses, which is the check that catches almost every slip in this kind of work:

In the approximate circuit the stator copper loss uses \(I_2'\), not \(I_1\), because the exciting branch has been moved in front of the stator impedance. That is exactly the approximation being made.

Split the rotor resistance into its two parts. The mechanical-load resistance carries the gross output directly:

Equivalently \(R_2'/s = 4\ \Omega\), of which 0.08 Ω is rotor copper and 3.92 Ω is the mechanical load. At 2% slip 98% of the air-gap power is mechanical.

The impedance of the series branch, stator plus rotor plus load resistance:

The secondary current, with the phase voltage as reference:

The no-load current from the admittances:

10.6 A of exciting current against 16.1 A of load current — a very large ratio, typical of a small low-voltage machine, and the reason its power factor is poor.

The primary current and power factor:

The gross power output is the power dissipated in \(R_L'\) — that resistance is not a real resistance at all but the mechanical load:

The air-gap power is larger: \(P_g = 3I_2'^2(R_2'/s) = 3\times16.14^2\times4 = 3126\ \text{W}\), and the 63 W difference is the rotor copper loss, \(sP_g\).

Speeds and the gross torque:

The same number comes from \(P_g/\omega_s = 3126/125.66 = 24.87\ \text{N·m}\), as it must.

The input power, taken from the line quantities:

Equivalently \(3V_1\,\mathrm{Re}(I_1) = 3\times66.40\times17.43 = 3471\ \text{W}\) — the same calculation with the \(\sqrt3\) undone.

Check the input against the losses before dividing:

The gross efficiency:

It is called gross because friction and windage have not been deducted; the shaft efficiency would be a little lower.

Draw the exact circuit and identify the parallel section. Here the magnetising branch stays where it belongs, across the air-gap e.m.f., so the rotor branch and \(jX_m\) must be combined before anything else:

Combine the two branches at A–B:

Because \(jX_m\) is lossless, the real part 8.58 Ω represents only the power delivered to the rotor. That fact is used two steps below.

Add the stator impedance to get the whole circuit as seen by the supply:

The input current and power factor:

The air-gap power is the real power absorbed by \(Z_{AB}\):

Check it against the rotor current. Current division gives \(I_2' = I_1\dfrac{jX_m}{(R_2'/s)+j(X_2'+X_m)} = 11.73\ \text{A}\), and \(3\times11.73^2\times10 = 4127\ \text{W}\) ✓.

Mechanical power, speed and electromagnetic torque:

The output power, after the rotational loss:

Comfortably above the 3.73 kW rating, so 4% slip is a shade beyond full load for this machine.

The loss ledger and the efficiency. The core loss is not represented in this circuit — only \(B_0\) was given — so it must be added to the input by hand:

Convert the exciting admittance into a branch impedance, since the exact circuit needs it in series form:

So the magnetising reactance is 19.8 Ω against a stator leakage of 0.6 Ω — a ratio of 33, which is where the size of the eventual error comes from.

Combine the magnetising and rotor branches:

Add the stator impedance and solve:

Find the air-gap e.m.f. and the true rotor current:

Here is the whole difference in one line: the rotor sees 236.5 V, not the terminal 254 V, because the stator impedance drops 17.5 V on the way in. The approximate circuit hands the rotor the full terminal voltage.

Air-gap power, torque and efficiency on the exact circuit:

Ledger check: stator copper \(3I_1^2R_1 = 952\ \text{W}\), core loss \(3E_1^2G_m = 839\ \text{W}\), air-gap 20 037 W — total 21 828 W ✓.

The comparison:

Why the approximation always errs upwards. Moving the exciting branch to the terminals removes the stator impedance from in front of the magnetising branch and from in front of the rotor. Both currents are therefore driven by the full terminal voltage instead of by the smaller internal e.m.f.:

Torque goes as \(I_2'^2\), so a 2.9% current error becomes a 6.0% torque error. The approximation is optimistic about every performance figure except the power factor.

When the approximation is safe. The error scales with the ratio of the stator impedance to the magnetising impedance:

The rule of thumb is that the approximate circuit is acceptable for machines above about 10 kW at normal supply voltages, and unreliable for fractional-kilowatt machines — which is precisely where the magnetising current is largest.

Recognise the condition for maximum torque as a maximum-power-transfer problem. The torque follows the air-gap power, which is the power delivered to the single resistance \(R_2'/s\). Since \(s\) is the only thing that varies, the torque peaks when that resistance matches the magnitude of the impedance looking back towards the source:

Substitute the data. The rotor impedance is \((0.6+j1)\ \Omega\), so \(R_2' = 0.6\ \Omega\) and \(X_2' = 1\ \Omega\):

A very high pull-out slip — this rotor has an unusually large resistance relative to its leakage reactance, which is what a high-starting-torque design looks like.

The referred rotor current at that slip, which is where the calculation is most often got wrong. The resistance in the loop is \(R_1+R_2'/s_{maxT}\), not \(R_1+R_2'\):

Using \(R_1+R_2' = 1\ \Omega\) instead would give 103 A — that is the standstill current, and it belongs to \(s=1\), not to \(s = 0.294\).

The maximum torque:

The closed form contains no \(R_2'\) at all, which is the standard result: rotor resistance moves the peak along the slip axis but does not change its height.

The power factor at 5% slip. Now \(R_2'/s = 0.6/0.05 = 12\ \Omega\), and the two parallel branches must be added:

How far below pull-out this operating point is:

The current at standstill is only 41% larger than at pull-out, but the torque is lower, because at \(s=1\) most of that current is reactive. Torque follows \(I_2'^2R_2'/s\), and the resistance term collapses as the slip rises.

Why a Thevenin reduction is needed at all. The maximum-power-transfer argument of Problem 5 requires a single source in series with a single impedance feeding \(R_2'/s\). The exact circuit does not have that form — \(jX_m\) sits in parallel. Removing the rotor branch and looking back gives exactly the required form.

The Thevenin voltage is the open-circuit voltage at the rotor terminals, a simple divider between \(jX_m\) and the stator impedance:

A 3% reduction on the terminal voltage — small, but it enters the torque squared.

The Thevenin impedance, with the source shorted:

The familiar shortcuts \(R_{Th}\approx R_1\left(\dfrac{X_m}{X_1+X_m}\right)^2 = 0.471\ \Omega\) and \(X_{Th}\approx X_1 = 1.2\ \Omega\) reproduce these to within 2.5%: \(R_{Th}\) is right to three figures and \(X_{Th}\) a shade high.

The slip at maximum torque is now the matching condition applied to the reduced circuit:

The maximum torque from the closed form:

Check from the current: at \(s_{maxT}\), \(R_2'/s = 2.417\ \Omega\) and \(I_2' = 224.2/\sqrt{(0.471+2.417)^2+2.371^2} = 60.0\ \text{A}\), giving \(3\times60.0^2\times2.417/157.08 = 166.2\ \text{N·m}\) ✓.

The starting torque, the same expression at \(s = 1\):

A cage machine of this design starts at just over a third of its pull-out torque while drawing 88.8 A against the 60.0 A at pull-out — the classic argument for a starter.

What the approximate circuit would have said. Replacing \(V_{Th}, R_{Th}, X_{Th}\) by \(V_1, R_1, X_1\):

The pull-out slip barely moves, because it depends on a ratio in which the two errors partly cancel; the pull-out torque moves by 3.8%, because \(V^2\) is not forgiving. As always the approximation is optimistic.