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Consider a transformer shown in Fig. 10.13 having primary and secondary windings of resistances R_{1} and R_{2} and reactance X_{1} and X_{2} respectively. The impedance of primary winding is given by Z_{1} = R_{1} + j X_{1} and impedance of secondary winding is given by Z_{2} = R_{2} + jX_{2}.

**The phasor diagrams of above transformer on: **

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(i) Pure resistive,

(ii) Resistive-inductive, and

(iii) Resistive-capacitive loads are shown in Fig. 10.14(a), (b) and (c) respectively.

Draw OA representing secondary terminal voltage V_{2} and OI_{2} representing secondary current I_{2} in phase as well as magnitude. Since voltage drops due to secondary winding resistance and reactance are I_{2} R_{2} in phase with current I_{2} and I_{2} X_{2} leading current I_{2} by π/2 respectively, so draw AB parallel to OI_{2} and equal to I_{2} R_{2} in magnitude representing resistive drop in secondary winding and draw BC perpendicular to AB and equal to I_{2} X_{2} in magnitude representing reactive drop of secondary winding. Since phasor sum of terminal voltage V_{2}, secondary resistive drop I_{2} R_{2} and secondary reactive drop I_{2} X_{2} is equal to induced emf E_{2} in secondary winding so phasor OC represents secondary induced emf E_{2}.

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**Hence we have: **

E_{2} = v_{2} + I_{2} (R_{2} + j X_{2}) = V_{2} + I_{2} Z_{2} … (10.15)

The induced emf E_{1} in primary winding is in phase with E_{2} and equal to N_{1}/N_{2} E_{2} in magnitude, so take OD = N_{1}/N_{2} OC representing E_{1}. Produce DO to D’ taking OD’ = OD hence representing (-E_{1}).

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The induced primary current I’_{1} is equal to – I_{2} N_{2}/N_{1}so draw OI’_{1}, equal to OI_{2} × N_{2}/N_{1} by producing line I_{2} O. Draw line OI_{0} representing no-load current in magnitude as well as in phase. The phasor sum of induced primary current I’_{1} and no-load current I_{0} gives primary current represented by phasor OI_{1}, in Fig. 10.14.

Since voltage drop due to primary winding resistance and reactance are I_{1} R_{1} in phase with primary current I_{1} and I_{1} X_{1} leading current I_{1} by π/2 respectively, so draw D’ F parallel to OI_{1}, and equal to I_{1} R_{1} in magnitude representing resistive drop in primary winding and draw FG perpendicular to D’ F equal to I_{1} X_{1} in magnitude representing reactive drop in primary winding. As the phasor sum of (- E_{1}), primary resistive drop and primary reactive drop gives the applied voltage V_{1} to primary winding, hence phasor OG represents the applied voltage V_{1} in magnitude as well as in phase.

i.e. V_{1} = – E_{1} + I_{1} (R_{1} + j X_{1}) = – E_{1} + I_{1} Z_{1} … (10.16)

The phase angle φ_{1} between V_{1} and I_{1} gives the power factor angle of the transformer.

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Since no-load current I_{0}, resistive drops I_{1} R_{1} and I_{2} R_{2} and reactive drops, I_{1} X_{1} and I_{2} X_{2} are very small, so neglecting these we have φ_{2} = φ’_{1} = φ_{1} = φ, the phase angle of the load. In Fig. no-load current, resistive drops and reactive drops are shown, for clarity, on exaggerated scales.

**From phasor diagrams we have: **

(a) For pure resistive load [phasor diagram 10.14 (a)]

(b) For resistive-inductive load [phasor diagram 10.14 (b)]

(c) For resistive-capacitive load [phasor diagram 110.14 (c)]