## Expression for Back E.M.F or Induced E.M.F. per Phase in S.M.

**Case i)**Under excitation, E

_{bph }< V

_{ph }.

Z

_{s }= R_{a }+ j X_{s }=**|**Z_{s }**|**?? ? ? = tan

^{-1}(X_{s}/R_{a}) E

_{Rph }^ I_{aph }= ?, I_{a }lags always by angle ?. V

_{ph }= Phase voltage applied E

_{Rph }= Back e.m.f. induced per phase E

_{Rph }= I_{a }x Z_{s }V … per phase Let p.f. be cos?, lagging as under excited,

V

_{ph }^ I_{aph }= ? Phasor diagram is shown in the Fig. 1.

Fig. 1 Phasor diagram for under excited condition |

Applying cosine rule to ? OAB,

(E

_{bph})^{2}= (V_{ph})^{2}+ (E_{Rph})^{2}– 2V_{ph}E_{Rph }x (V_{ph }^ E_{Rph}) but V

_{ph }^ E_{Rph }= x = ? – ? (E

_{bph})^{2}= (V_{ph})^{2}+ (E_{Rph})^{2}– 2V_{ph}E_{Rph }x (? – ?) ……(1) where E

_{Rph }= I_{aph }x Z_{s } Applying sine rule to ? OAB,

E

_{bph}/sinx = E_{Rph}/sin? So once E

_{bph}is calculated, load angle ? can be determined by using sine rule.**Case ii)**Over excitation, E

_{bph }> V

_{ph }

p.f. is leading in nature.

E

_{Rph }^ I_{aph }= ? V

_{ph }^ I_{aph }= ? The phasor diagram is shown in the Fig. 2.

Fig.2 Phasor diagram for overexcited condition |

Applying cosine rule to ? OAB,

(E

_{bph})^{2}= (V_{ph})^{2}+ (E_{Rph})^{2}– 2V_{ph }E_{Rph}x cos(V_{ph }^ E_{Rph}) V

_{ph }^ E_{Rph }= ? + ?**.**(E

^{.}._{bph})

^{2}= (V

_{ph})

^{2}+ (E

_{Rph})

^{2}– 2 V

_{ph }E

_{Rph}cos(? + ?) …….(3)

But ? + ? is generally greater than 90

^{o}**.**cos (? + ?) becomes negative, hence for leading p.f., E

^{.}._{bph }>V

_{ph }.

Applying sine rule to ? OAB,

E

_{bph}/sin( E_{Rph }^V_{ph}) = E_{Rph}/sin? Hence load angle ? can be calculated once E

_{bph}is known.**Case iii)**Critical excitation

In this case E

_{bph ? }V_{ph}, but p.f. of synchronous motor is unity.**.**cos = 1

^{.}.**.**? = 0

^{.}.^{o}

i.e. V

_{ph }and I_{aph }are in phase and E

_{Rph }^I_{aph }= ? Phasor diagram is shown in the Fig. 3.

Fig. 3 Phasor diagram for unity p.f. condition |

Applying cosine rule to OAB,

(E

_{bph})^{2}= (V_{ph})^{2}+ (E_{Rph})^{2}– 2V_{ph }E_{Rph }cos ? …………(5) Applying sine rule to OAB,

E

_{bph}/sin? = E_{Rph}/sin? where E

_{Rph}= I_{aph }x Zs V

## Leave a comment