Q2GATE 2006MCQ

A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be -0.5% and 30 minutes respectively. If the number of secondary turns is reduced by 1 new ratio-error(%) and phaseerror( min) will be respectively

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📖 Explanation

Nominal ratio $=k_n=\frac{200}{1}=200$ Primary current $=I_P=160A$ Actual ratio $= R_1=n_1\left ( 1+\frac{I_e}{I_P} \right )\;\;...(i)$ $\% \text{Ratio error}$ $=\frac{k_n - R_1}{R_1}\times 100$ $-0.5=\frac{200-R_1}{R_1} \times 100$ Actual ratio $=R_1=201$ $n_1=\text{turn ratio}=200$ $R_1=200\left ( 1+\frac{I_e}{I_P} \right )\;\;...(ii)$ when number of secondary is reduced by 1

\begin{aligned} N_s &= 199\\ n_2&=199 \\ R_2&=n_2\left ( 1+\frac{I_e}{I_P} \right ) \\ &= 199\left ( 1+\frac{I_e}{I_P} \right )\;\;...(iii) \end{aligned}

Dividing equation (ii) and (iii),

\begin{aligned} \frac{R_1}{R_2} &=\frac{200}{199} \\ \Rightarrow \;R_2 &= R_1 \times \left ( \frac{200}{199} \right )\\ &=201 \times \frac{200}{199}\approx 200 \\ \% \text{Ratio error} &=\frac{k_n-R_2}{R_2} \times 100\\ &= \frac{200-200}{200}\times 100=0 \end{aligned}

Phase angle error $=\theta =\frac{180}{\pi}\left ( \frac{I_m}{I_P} \right )$ Reduction of one or two turns of the secondary winding, no doubt, reduces the ration error, but it has no effect on the phase angle error.

Q3GATE 2004MCQ

A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 $\Omega$ . The magnetizing ampere-turns is 200. The phase angle between the primary and second current is $175.4^{\circ}$ . The core flux in the CT under the given operating conditions is

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📖 Explanation

$Z_s=\text{Burden}=1\Omega$ Voltage induced in the secondary

\begin{aligned} E_s&=I_S \times Z_S=5 \times 1=5V \\ E_s&=4.44 f\phi _mN_s \\ \therefore \;\; \phi _m&=\frac{E_s}{1.44fN_s} \\ &= \frac{5}{4.44 \times 50 \times 500}=45\mu Wb \end{aligned}

Q4GATE 2003MCQ

A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure resistance of 1 $\Omega$ and it draws a current of 5 A. If the magnetic core requires 250 AT for magnetization, the percentage ratio error is

🚀 Solve in Practice Mode

📖 Explanation

Nominal ratio $=\frac{500}{5}=100$ Number of turn in primary N=1 (primary bar) Magnetizing current, $I_m=\frac{\text{mmf}}{\text{No. of turns}}$ $\;\;=\frac{250}{1}=250A$ Secondary current, $I_s=5A n=\frac{500}{5}=100$ Primary current, $I_P=\sqrt{(nI_s)^{2+}I_m^2}$ $\;\;=\sqrt{(100\times 5)^2 +(250)^2}$ $\Rightarrow \; I_P=559.017A$ Actual ratio $\;\;=\frac{I_P}{5}=\frac{559.017}{5}=111.803$ Ratio error $\;\;=\frac{\text{Nominal Ratio} - \text{Actual Ratio}}{\text{Actual Ratio}} \times 100$ $\;\;=\frac{100-111.803}{111.803}\times 100=-10.56\%$

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