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Determining single phase IM equivalent circuit parameters

Short answer

DC test gives stator resistance R1 (corrected for skin effect). Blocked-rotor test with locked rotor and reduced voltage at rated current yields R2', X1, X2' via measured Vsc, Isc, Psc. No-load test at rated voltage gives magnetising reactance Xm and rotational losses.

Discuss the procedure to determine the parameters of equivalent circuit of one phase induction motor.

The parameters of the single-phase induction motor equivalent circuit (R1R_1, X1X_1, R2′R_2', X2′X_2', XmX_m and rotational loss) are found from three simple tests on the main winding (auxiliary winding left open): a DC resistance test, a blocked-rotor test and a no-load test.

     A        W
 o--(A)--+--[W]--+-------+
         |       |       |
 1-ph    (V)   Main     rotor
 variac  |    winding  (free or
 supply  |       |      blocked)
 o-------+-------+-------+

1. DC resistance test

Pass DC through the main winding and measure VdcV_{dc} and IdcI_{dc}:

R1=VdcIdc(multiply by about 1.1–1.2 for AC/skin effect)R_1 = \frac{V_{dc}}{I_{dc}} \quad(\text{multiply by about } 1.1\text{–}1.2 \text{ for AC/skin effect})

2. Blocked-rotor (short-circuit) test

Procedure: hold the rotor stationary, apply a reduced voltage through a variac until rated current flows. Record VscV_{sc}, IscI_{sc}, PscP_{sc}.

At s=1s = 1 the forward and backward halves are equal, and since Xm≫X_m \gg rotor impedance, the magnetising branch is neglected. The circuit becomes R1+jX1+R2′+jX2′R_1 + jX_1 + R_2' + jX_2'.

Zsc=VscIsc,Rsc=PscIsc2Xsc=Zsc2−Rsc2R2′=Rsc−R1X1=X2′=Xsc2\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}}, \qquad R_{sc} = \frac{P_{sc}}{I_{sc}^2} \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} \\ R_2' &= R_{sc} - R_1 \\ X_1 &= X_2' = \frac{X_{sc}}{2} \end{aligned}

3. No-load test

Procedure: run the motor at rated voltage with no load (start it with the auxiliary winding, then open it). Record V0V_0, I0I_0, P0P_0.

At no load s≈0s \approx 0:

  • Forward branch: 0.5R2′/s→∞0.5R_2'/s \to \infty, so only j0.5Xmj0.5X_m remains.
  • Backward branch: 0.5R2′/(2−s)≈R2′/40.5R_2'/(2-s) \approx R_2'/4; since 0.5Xm≫0.5X_m \gg this, only R2′/4+j0.5X2′R_2'/4 + j0.5X_2' remains.
 R1   X1   0.5Xm     R2'/4   0.5X2'
o-/\/-mmm---mmm---+--/\/\---mmm---o
Z0=V0I0,cos⁡ϕ0=P0V0I0X0=Z0sin⁡ϕ0=X1+0.5Xm+0.5X2′Xm=2(X0−X1−0.5X2′)=2(X0−1.5X1)\begin{aligned} Z_0 &= \frac{V_0}{I_0}, \qquad \cos\phi_0 = \frac{P_0}{V_0 I_0} \\ X_0 &= Z_0 \sin\phi_0 = X_1 + 0.5X_m + 0.5X_2' \\ X_m &= 2\left(X_0 - X_1 - 0.5X_2'\right) = 2(X_0 - 1.5X_1) \end{aligned}

Rotational (core + friction + windage) loss

Prot=P0−I02(R1+R2′4)P_{rot} = P_0 - I_0^2\left(R_1 + \frac{R_2'}{4}\right)

This loss is assumed constant and is subtracted from the mechanical power when finding output.

Summary

TestConditionGives
DC testDC supplyR1R_1
Blocked rotorRotor locked, reduced V, rated IR2′R_2', X1X_1, X2′X_2'
No loadRated V, no loadXmX_m, rotational loss

With these values, the full forward/backward equivalent circuit is drawn, and current, power factor, torque and efficiency at any slip can be calculated.

✓ Answer checked by Claude Sonnet 5.5

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