Question

A current transformer has a single turn primary and $400$ turns on secondary winding. The resistance annd reactance of the secondary circuit are $2\Omega$ and $3\Omega$ respectively including transformer winding. When $6A$ current is flowing in the secondary winding, the magnetizing mmf is $100AT$ and iron loss is $2W$. The value of ratio error will be

• A. $3.42\%$
• B. $-4.14\%$
• C. $-2.24\%$
• D. $1.62\%$

Answer 1

The correct option is B $-4.14\%$

Primary winding turns,
$N_P=1$

Secondary winding turns,
$N_s=400$

Turn ratio,
$n=\frac{N_S}{N_P}=400$

Seconday circuit burden impedance
$=\sqrt{(2{)}^2+(3{)}^2}=3.606\Omega$

For secondary winding circuit.
$cos\delta =\frac{2}{3.606}=0.554$
$and,sin\delta =\frac{3}{3.606}=0.832$

Secondary induced voltage
$E_S=6\times 3.606$
$=21.64V$

Primary inuced voltage
$E_P=\frac{E_S}{n}=\frac{21.64}{400}=0.054V$

Loss component of current referred to primary winding
$I_c=\frac{\text{Iron loss}}{E}=\frac{2}{0.054}=37.04A$

Magnetizing current,
$I_m=\frac{\text{Magnetizing mmf}}{\text{Primary winding turns}}=\frac{100}{1}=100A$

Acutal ratio,
$R=n+\frac{I_{C}cos\delta +I_{m}sin\delta }{I_S}$

$=400+\frac{37.04cos\delta +100sin\delta }{I_s}$

$=400+\frac{103.72}{6}=400+17.28$

$=417.28$

Percentage ratio error

$=\frac{K_n-R}{R}\times 100;$ $Where{K}_a=nominalratio$

$=\frac{400-417.28}{417.28}\times 100=-4.14\%$