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.

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.
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.

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