- #1

Martin Harris

- 103

- 6

- Homework Statement
- Find currents Is (stator current), Ir (rotor current) and Im (magnetic current)

- Relevant Equations
- Kirchoff's Law

Ohm's Law

Hi,

Please find attached below the Induction Generator circuit.

I have the following parameters:

I am requested to find the currents.

Kirchoff's Law I

$$Is = \frac{Vsl} { Zoverall} $$

$$Zoverall = Zs+Zparallel$$

$$Zs = Rs+iXs = 0.2+1.8849 Ω $$

$$Zm = iXm = (0+37.6991i) Ω $$

$$Zr = \frac {Rr} {s} + iXr = (-10 +1.8849i)Ω $$

$$Zparallel = \frac {Zm*Zr} {Zm+Zr} = \frac { (0+37.6991i)*(-10 +1.8849i)} {(0+37.6991i)+(-10 +1.8849i)} $$

$$Zparallel = (-8.5261+3.9490i) Ω $$

Hence $$Zoverall = Zs+Zparallel = (0.2+1.8849 Ω) + (-8.5261+3.9490i) Ω $$

$$Zoverall =(-8.3261+5.8339i)$$

$$Is = \frac{220V} {-8.3261+5.8339i } $$

$$Is = (-17.7222 -12.4175i) A = 21.6395 A$$

Now we have to find currents Ir and Im

$$Vs = Is*Zs = 21.6395 A *1.8954 Ω = 41.0172 V$$

$$Vm=220V-Vs = 220V-41.0172V = 178.9828 V$$

$$Im = \frac {Vm} {Zm} = \frac {178.9828 V} {(0+37.6991i) Ω}$$

$$Im = (0-4.7476i)A = 4.7466 A $$

$$Ir = I am - Is = (0-4.7476i)A - (-17.7222 -12.4175i) A$$

$$Ir = (17.7222-7.6699i)A$$

$$Ir = 19.3107 A$$

Do these calculations for I am and Ir make any sense? Current through the stator (Is) seems to be all right, but I'm not sure if what I did to calculate I am and Ir makes any sense.

Can we say m and r branches are in parallel?

I would be more than grateful if someone can check the calculation for I am and Ir.

Many thanks!

Please find attached below the Induction Generator circuit.

I have the following parameters:

Parameter | Value |

R_{s} = R_{r} (Resistance on stator and rotor) | 0.2 Ω |

L_{s} = L_{r} (inductance on stator and rotor) | 5 mH = 0.005 H |

p (number of pairs of poles) | 2 |

f (frequency) | 60 Hz |

Vsl (stator line voltage) - delta connection | 220 V |

s (slip of the generator) | -0.02 |

I am requested to find the currents.

Kirchoff's Law I

_{m}= I_{s}+ I_{r}$$Is = \frac{Vsl} { Zoverall} $$

$$Zoverall = Zs+Zparallel$$

$$Zs = Rs+iXs = 0.2+1.8849 Ω $$

$$Zm = iXm = (0+37.6991i) Ω $$

$$Zr = \frac {Rr} {s} + iXr = (-10 +1.8849i)Ω $$

$$Zparallel = \frac {Zm*Zr} {Zm+Zr} = \frac { (0+37.6991i)*(-10 +1.8849i)} {(0+37.6991i)+(-10 +1.8849i)} $$

$$Zparallel = (-8.5261+3.9490i) Ω $$

Hence $$Zoverall = Zs+Zparallel = (0.2+1.8849 Ω) + (-8.5261+3.9490i) Ω $$

$$Zoverall =(-8.3261+5.8339i)$$

$$Is = \frac{220V} {-8.3261+5.8339i } $$

$$Is = (-17.7222 -12.4175i) A = 21.6395 A$$

Now we have to find currents Ir and Im

$$Vs = Is*Zs = 21.6395 A *1.8954 Ω = 41.0172 V$$

$$Vm=220V-Vs = 220V-41.0172V = 178.9828 V$$

$$Im = \frac {Vm} {Zm} = \frac {178.9828 V} {(0+37.6991i) Ω}$$

$$Im = (0-4.7476i)A = 4.7466 A $$

$$Ir = I am - Is = (0-4.7476i)A - (-17.7222 -12.4175i) A$$

$$Ir = (17.7222-7.6699i)A$$

$$Ir = 19.3107 A$$

Do these calculations for I am and Ir make any sense? Current through the stator (Is) seems to be all right, but I'm not sure if what I did to calculate I am and Ir makes any sense.

Can we say m and r branches are in parallel?

I would be more than grateful if someone can check the calculation for I am and Ir.

Many thanks!

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