Question

A $\text{20VA, 100/2 A}$ current transformer has an iron loss of 0.5W and magnetizing current of 1.2A. If the rated output is supplied to a motor having a ratio of resistance to reactance of 4, the value of phase angle (in degree) will be

• A. ${0.23}^o$
• B. ${0.32}^o$
• C. ${0.50}^o$
• D. ${1.0}^o$

Answer 1

The correct option is B ${0.32}^o$

We know,
In the absence of data the turn ratio is equal to the nominal ratio.

$\text{Nominal ratio,}{K}_a=\frac{100}{2}=50$

$\text{Total burden on transformer = Burden of meter}=20VA$

$E_P=\frac{VA}{I_P}=\frac{20}{100}=0.2V$

$\text{Loss component of currnet,}$

$I_c=\frac{\text{Iron loss}}{E_P}$

$I_c=\frac{0.5}{0.20}=2.5A$

$\text{Mangetizing component,}$

$I_m=1.2A$

$\text{Secondary circuit phase angle}$

$\delta =ta{n}^{-1}\frac{X}{R}=ta{n}^{-1}\frac{1}{4}={14.036}^o$

$cos\delta =0.970andsin\delta =0.242$

$\text{Phase angle,}$

$\theta =\frac{180}{\pi }\left[\frac{I_{m}cos\delta -I_{s}sin\delta }{n{I}_s}\right]$


$\theta =\frac{180}{\pi }\times \left[\frac{1.2\times 0.970-2.5\times 0.242}{50\times 2}\right]$


$\theta =\frac{180}{\pi }\times \left[\frac{1.164-0.605}{100}\right]$


$\theta ={0.320}^o$