Question and Answer

Secondary burden=15VA, secondary winding current=5A= $I_{s}$

Secondary circuit impedance = $\frac{15}{5^{2}} = 0.6 Ω$

Secondary circuit reactance= $0.195 Ω$

Phase angle of secondary circuit $δ = \sin^{- 1} (\frac{0.195}{0.6}) = 0.325$

$∴ \sin δ = 0.00567, \cos δ = 0.9999$

Primary turns=40= $N_{P}$ , secondary turns =120= $N_{S}$

Turns ratio,n= $\frac{N_{S}}{N_{P}} = \frac{120}{40} = 3$ , Nominal ratio= $K_{n} = \frac{25}{5} = 5$

Magnetizing current $I_{m} = \frac{m a g n e t i z i n g m m f}{p r i m a r y w i n d i n g t u r n s} = \frac{13}{40} = 0.325$

Loss component $I_{e} = \frac{e x c i t a t i o n f o r l o s s}{p r i m a r y w i n d i n g t u r n s} = \frac{10.1}{40} = 0.2525$

Actual ratio $R = n + \frac{I_{e} \cos δ + I_{m} \sin δ}{I_{s}} = 3 + \frac{0.2525 * 0.9999 + 0.325 * 0.0567}{5}$

$= 3 + \frac{0.2524 + 0.001848}{5} = 3.0508$

Ratio error = $\frac{n o m i n a l r a t i o - a c t u a l r a t i o}{a c t u a l r a t i o} * 100 = \frac{5 - 3.0508}{3.0508} * 100 = + 63.889$

Phase angle $θ = \frac{180}{π} [\frac{I_{m} \cos δ - I_{e} \sin δ}{n I_{s}}] = 180 [\frac{0.325 * 0.9999 - 0.2525 * 0.0056}{3 * 5}]$

$= 180 [\frac{0.3249675 - 0.0014316}{15}] = 3.8824 °$

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