Question 3.28: A 6000 V/100 V, 50 Hz potential transformer has the followin...

A 6000 V/100 V, 50 Hz potential transformer has the following parameters as seen from HV side.
R_{1}= 780\Omega X_{1} = 975\Omega     X_{m} = 443 k\Omega
\acute{R_{2} } = 907\Omega    \acute{X_{2} } = 1075\Omega
(a) The primary is excited at 6500 V and the secondary is left open. Calculate the secondary voltage,
magnitude and phase.
(b) The secondary is loaded with 1 k\Omega resistance, repeat part (a)
(c) The secondary is loaded with 1 k\Omega reactance, repeat part (a)

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The potentiometer equivalent circuit as seen from HV side is drawn on Fig.3.76.
Turn ratio, \frac{N_{1} }{N_{2} } =\frac{6000}{1000} =60
(a)   Secondary open; Z_{b}=\infty
V_{1} = 6500 V
\acute{V_{2} } =\left\lgroup\frac{jX_{m} }{R_{1}+ jX_{1} } \right\rgroup V_{1}
\acute{V_{2} } =\frac{J443\times 10^{3} }{780+ j443\times 10^{3} } \times 6500
\acute{V_{2} }=1 \angle 0.1°\times 6500 = 6500 \angle 0.1° V
V_{2} =6500\times \left\lgroup\frac{N_{2} }{N_{1} } \right\rgroup = 108 V, \angle 0.1°
(b)   Z_{b} =R_{b} =1 kW,\acute{R_{b} } =\left(60\right)^{2} \times 1=3600 k\Omega
As \acute{R_{b} } is far larger than \acute{R_{2} }  and  \acute{X_{2} }, we can ignore \acute{R_{2} }, \acute{X_{2} }
Then
\bar{Z_{m} } =jX_{m} \parallel \acute{R_{b} }
\bar{Z_{m} }=\left\lgroup\frac{j443\times 3600}{3600+ j443} \right\rgroup= 439.7 \angle 83° = 53.6 + j 436.4 k\Omega
\left(R_{1}+ jX_{1} \right) + \bar{Z_{m} } =\left(0.78 + j 0.975\right) + \left(53.6 + j 436.4\right) =54.38 + j 473.4 = 440.77 \angle82.9° k\Omega
\bar{V_{m} } =\left[\frac{\bar{Z_{m} } }{\left(R_{1}+jX_{1} \right)+ \bar{Z_{m} } } \right]V_{1}
\bar{V_{m} }=\left[\frac{439.7 \angle83°}{440.77 \angle82.9°} \right] \times 6500= 6484 \angle 0.1°
\bar{\acute{V_{2} } } =\bar{V_{m} } =6484 \angle 0.1° V
V_{2} =\frac{6484}{60} =108.07 V ; phase 0.1°
Exact value should be\frac{6500}{60} = 108.33
Error =\frac{108.33-108.07}{108 .33} = 0.26\%
(c)   \bar{Z_{b} } =jXb ; Xb = 1 k\Omega
\bar{\acute{Z_{b} } }= j 3600 k\Omega
Ignoring \acute{R_{2} },\acute{X_{2} } in comparison
\bar{Z_{m} } =j 443\parallel j 3600 = j\frac{443\times 3600}{443+3600} =j 394.45 k\Omega
\left(R_{1}+ jX_{1} \right) +\bar{Z_{m} }=\left(0.78 + j 0.975\right) + j 394.45 = 0.78 + j 395.425 = 395.426 \angle 89.89°
\bar{\acute{V_{2} } } =\bar{V_{m} }=\frac{394.45 \angle 90°}{395.426 \angle 89.89°} \times 6500=6484\angle 0.01°
V_{2} =\frac{6484}{60}= 108.07, phase 0.01°
V_{2} is same as in resistive load (part (b) except for change in phase. In any case phase is almost zero

3.76

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