Question

A current transformer with a bar primary has 300 turns in its secondary winding. The resistance and reactance of the secondary circuit are $1.5\Omega$ and $1.0\Omega$ respectively including the transformer winding with 5 A flowing in the secondary winding, the phase angle error is (assuming magnetizing mmf to be 100 AT and core loss to be 1.5 W)

• A. ${5.54}^o$
• B. ${2.12}^o$
• C. ${1.12}^o$
• D. ${6.28}^o$

Answer 1

The correct option is B ${2.12}^o$

Phase angle,
$\theta =\frac{180}{\pi }\left(\frac{I_m\cos \delta -I_c\sin \delta }{n{I}_s}\right)$

$\text{Given primary turns,}{N}_p=1$
$\text{and secondary turns,}{N}_s=300$

$\therefore \text{Turns ratio,}n=\frac{N_s}{N_p}=300$

Secondary burden impedance

$=\sqrt{(1.5{)}^2+(1.0{)}^2}=1.8\Omega$

$\therefore$ $cos\delta =\frac{R}{Z}=\frac{1.5}{1.8}=0.833$

$andsin\delta =\frac{X}{Z}=\frac{1.0}{1.8}=0.555$

Secondary voltage,

$V_s=I_s\times Z_2=5\times 1.8=9V$

Primary voltage,

$V_P=\frac{V_s}{300}=\frac{9}{300}=0.03V$

$I_C=\frac{coreloss}{V_p}=\frac{1.5}{0.03}=50A$

Phase angle error,

$\theta =\frac{180}{\pi }\left(\frac{100\times 0.833-50\times 0.555}{300\times 5}\right)={2.12}^o$