Question 3.27: A 250 A/5 A,50 Hz current transformer has the following para...

(a) The equivalent circuit with secondary shorted is drawn in Fig. 3.75.
By current division
\bar{\acute{I_{2} } } =\left\lgroup\frac{jX_{m} }{\acute{R_{2} }+j\acute{X_{2} }+jX_{m} } \right\rgroup I_{1}
\bar{I_{1} } = 250\angle 0° A
\bar{\acute{I_{2} } }=\frac{j256\times 10^{3} }{109+ j\left(551+256\times 10^{3} \right) } \times 256
\bar{\acute{I_{2} } } =\left\lgroup\frac{j256}{256.51\angle 89.975} \right\rgroup \times 250A
\bar{I_{2} } =\frac{j256}{256.5\angle 89.975°} \times 5; I_{2}=\left\lgroup\frac{N_{1} }{N_{2} } \right\rgroup \acute{I_{2} } =\frac{1}{50} \times 250°=5 A
I_{2}= 4.989 A phase = 0.025° (negligible)
Error =\frac{5-4.989}{5} \times 100=0.22\%
(b)  \acute{R_{b} } =200\mu \Omega in series with \acute{R_{2} },\acute{X_{2} },R_{b} =\left\lgroup\frac{1}{50} \right\rgroup ^{2} \times 200=0.08\mu \Omega
\bar{\acute{I_{2} } }=\frac{j256\times 10^{3} }{\left(109+ 0. 08\right)+j\left(551 +256 \times 10^{3} \right) } \times 250A
\bar{\acute{I_{2} } }=\frac{256\angle 90°}{256.551\angle 89.975°} \times 5=4.989 \angle 0.025°
No change as \acute{R_{b} } =0.08\mu \Omega is negligible