Q2GATE 2023NAT

When the winding $c-d$ of the singlephase, 50 $\mathrm{Hz}$ , two winding transformer is supplied from an AC current source of frequency $50 \mathrm{~Hz}$ , the rated voltage of $200 \mathrm{~V}$ (rms), $50 \mathrm{~Hz}$ . is obtained at the open-circuited terminals a-b. The cross sectional area of the core $5000 \mathrm{~mm}^{2}$ and the average core length traversed by the mutual flux is $500 \mathrm{~mm}$ . The maximum allowable flux density in the core is $B_{\max }=1 \mathrm{~Wb} / \mathrm{m}^{2}$ and the relative permeability of the core material is $5000$ . The leakage impedance of the winding $a-b$ and winding c-d at $50 \mathrm{~Hz}$ are $(5+j 100 \pi \times 0.16) \Omega$ and $(11.5+j 100 \pi \times 0.36) \Omega$ , respectively. Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding $a-b$ in millihenry is ____ (Round off to 1 decimal place).

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \mathrm{A} & =5000 \mathrm{~mm}^{2} \\ \mathrm{I} & =500 \mathrm{~mm} \\ \mathrm{~B}_{\max } & =1 \mathrm{~Wb} / \mathrm{m}^{2} \\ \mathrm{~V}_{\mathrm{ab}} & =200 \mathrm{~V} \\ \sqrt{2} \pi \mathrm{fN}_{1} \phi_{\max } & =200 \\ \sqrt{2 \pi} \times 50 \times \mathrm{N}_{1} \times 1 & \times 5000 \times 10^{-6}=200 \quad[\because \phi=\beta \mathrm{A}] \\ \Rightarrow \quad \mathrm{N}_{1} & =181 \end{aligned}

Now, mutual inductance,

\begin{aligned} M&=\frac{N^{2}}{I / \mu A} \\ & =\frac{4 \pi \times 10^{-7} \times 5000 \times 5000 \times 10^{-6} \times (181)^{2}}{0.5} \\ & =2.05843 \mathrm{H} \end{aligned}

Therefore, self inductance of winding.

\begin{aligned} L & =\text { Mutual + leakage } \\ & =2.05843+0.16 \\ & =2.21843 \mathrm{H} \text { or } 2218.43 \mathrm{mH} \end{aligned}

Q3GATE 2021MCQ

In a single-phase transformer, the total iron loss is $2500\:W$ at nominal voltage of $440\:V$ and frequency $50\:Hz$ . The total iron loss is $850\:W$ at $220\:V$ and $25\:Hz$ . Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} W_{i 1} &=2500 \mathrm{~W} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ W_{i 2} &=850 \mathrm{~W} \text { at } 220 \mathrm{~V}, 25 \mathrm{~Hz} \\ W_{i 3} &=R_{e_{3}}+P_{H_{3}} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ R_{\theta_{3}} &=?, \quad P_{h_{3}}=?, \quad \frac{v}{f}=\mathrm{constant} \\ \Rightarrow \qquad \qquad\quad 2500 &=A f+B f^{2} &\left\{\frac{400}{50}=\frac{220}{25}=\text { Constant }\right\} \\ \text { or, }\qquad \qquad \quad \frac{2500}{f} &=A+B f \\ \text{or},\qquad \qquad \frac{2500}{50}&=A+B(50) &\ldots(i)\\ \text{and}\qquad \qquad \frac{850}{25}&=\mathrm{A}+\mathrm{B}(25)&\ldots(ii) \end{aligned}

Solving (i) and (ii), we get

\begin{aligned} 25 B &=\frac{2500}{50}-\frac{850}{25}=\frac{2500-1700}{50} \\ &=\frac{800}{50}=16 \\ B &=\frac{16}{25} \\ \text{and from (i)},\qquad A &=50-\frac{16}{25} \times 50=50-32=18\\ \text{So, at} 50 \mathrm{~Hz}\\ P_{h}&=A f=18 \times 50=900 \mathrm{~W} \\ P_{e}&=B f=\left(\frac{16}{25}\right) \times (50)^{2}=1600 \mathrm{~W} \end{aligned}

Q13GATE 2017NAT

A three-phase, three winding $\Delta / \Delta / Y$ (1.1kV/6.6kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is _______.

🚀 Solve in Practice Mode

📖 Explanation

Load on 6.6 kV winding = 900 kVA @0.8 pf lag $\therefore \;\; I_2=\frac{900 \times 10^3}{\sqrt{3} \times 6.6 \times 10^3}$ $\;\;\;\;=78.73\angle -36.86^{\circ} A$ Transforamtion Ratio

\begin{aligned} k &=\frac{1.1}{6.6}=\frac{1}{6} \\ \therefore \;\; I'_2&=\frac{I_2}{k}=472.4\angle -36.86^{\circ} A \end{aligned}

Load on 400 V winding = 300 kVA at 0.6 p.f. $\therefore \;\; I_3=\frac{300 \times 10^3}{\sqrt{3} \times 400}$ $=433.025\angle -53.13^{\circ}A$ Transformation ratio $= \frac{1.1 \times 1000}{\frac{400}{\sqrt{3}}}=4.76$

\begin{aligned} I'_3 &=\frac{I_3}{k}=\frac{4333.025}{4.76} \\ &= 90.97\angle -53.13^{\circ}A \end{aligned}

So, line value of current

\begin{aligned} &=\sqrt{3} \times 90.97\angle -53.13^{\circ} \\ &= 157.56 \angle -53.13^{\circ} \end{aligned}

Current drawn by 1.1 kV winding will be

\begin{aligned} I_1&=I'_2+I'_3 \\ &=472.4\angle -36.86 +157.56\angle -53.13^{\circ} \\ &=377.96 -j283.37 +94.53 -126.04 \\ &= 472.49 -j409.41\\ &= 625.19\angle -40.9^{\circ}A \end{aligned}

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