Q44GATE 2007MCQ

A single-phase, 50-kVA, 250 V/500 V two winding transformer has an efficiency of 95% at full load, unity power factor. If it is re-configured as a 500 V/750 V auto-transformer, its efficiency at its new rated load at unity power factor will be

🚀 Solve in Practice Mode

📖 Explanation

In two winding transformer

\begin{aligned} V_1 &=500V \\ V_2&= 250V\\ I_1 &=\frac{50 \times 10^3}{500}=100A \\ I_2&= \frac{50 \times 10^3}{250}=200A\\ \eta &= \frac{x S \cos \phi }{xS \cos \phi + \text{losses}}\\ \text{Full load}& \\ x&=1,\; \text{upf}, \; \cos \phi =1, \; S=50 \\ 0.95&= \frac{1 \times 50 \times 1}{1 \times 50 \times 1+ \text{losses}}\\ \text{losses} &= 2.63\\ \text{losses}&= \text{Iron loss}+ \text{Copper loss} \end{aligned}

When two winding transformer is connected as step up autotransformer $I_2=200,\;\; V_2=750$ kVA rating of auto transformer

\begin{aligned} S_{auto} &=V_2I_2 \\ &=750 \times 200 \\ &= 150kVA \end{aligned}

As current through windings and voltage across winding are equal in two-winding transformer and autotransformer. Losses remain same at full load efficiency at upf and full load

\begin{aligned} \%\eta &=\frac{xS_{auto} \cos \phi }{xS_{auto} \cos \phi +\text{ losses }}\times 100 \\ &= \frac{1 \times 150 \times 1}{1 \times 150 \times 1 +2.63} \times 100\\ &= 98.276\% \end{aligned}

Q55GATE 2003MCQ

A sinple phase transformer has a maximum efficiency of 90% at full load and unity power factor. Efficiency at half load at the same power factor is

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📖 Explanation

\begin{aligned} \text{Efficiency} &=\frac{\text{Output}}{\text{Input}} \\ &= \frac{\text{Output}}{\text{Output}+\text{Losses}}\\ \text{Output}&= x S \cos \phi \\ \text{Where,}\;\;S &=\text{Rating of the machine} \\ x&= \% \text{of the full load}\\ \cos \phi &= \text{Power factor} \\ P_i &= \text{Iron or core loss}\\ P_{cu} &=\text{Full load copper loss} \\ &\text{For maximum frequency}\\ P_i &=x^2P_{cu} \\ \text{Efficiency} =\eta &= \frac{xS \cos \phi }{xS \cos \phi +P_i+x^2 P_{cu}} \\ \text{Max efficiency}=\eta _{max}&=\frac{xS \cos \phi }{xS \cos \phi +2P_i}\\ \Rightarrow \;\;0.9&=\frac{1 \times S \times 1}{1 \times S \times 1+2P_i}\\ P_i&=0.055S \end{aligned}

Full load copper loss $= 1^2 P_{cu}=P_i$ $P_{cu}=0.055S$ Efficiency of halh load, upf $x=\frac{1}{2}$ $\eta =\frac{\frac{1}{2} \times S \times 1 }{\frac{1}{2} \times S \times 1 +0.055S+\left ( \frac{1}{2} \right )^2 \times 0.055S} \times 100$ $\;\;\;\approx 87.8\%$

Q56GATE 2003MCQ

Figure shows an ideal single-phase transformer. The primary and secondary coils are wound on the core as shown. Turns ratio $(\frac{N_{1}}{N_{2}}) = 2$ . The correct phasors of voltages $E_{1}$ , $E_{2}$ , currents $I_{1}$ , $I_{2}$ and core flux $\phi$ are as shown in

🚀 Solve in Practice Mode

📖 Explanation

As per lenz's law, $E_1=N_1\frac{d\phi }{dt}$ The positive direction of this emf opposes the positive current direction.]

\begin{aligned} \text{If,}\;\; \phi &=\phi _m\sin \omega t \\ E_1&= N_1 \frac{d}{dt}(\phi _m\sin \omega t )\\ &= N_1\phi _m \omega \cos \omega t \end{aligned}

$E_1$ leads the flux by $90^{\circ}$ Similarly, $E_2= N_2 \frac{d\phi }{dt}(\phi _m\sin \omega t )$ and $E_2$ leads the flux by $90^{\circ}$ So $E_1[/latex and [latex]E_2$ are in phase.

\begin{aligned} \text{But,}\;\;\frac{E_1}{E_2} &=\frac{N_1}{N_2}=2 \\ E_1&=2E_2 \;\text{or}\; E_1 \gt E_2 \\ I_2&=\frac{E_2}{R} \end{aligned}

Because of resistive load, $I_2$ is in phase with $E_2.$

\begin{aligned} \frac{I_1}{I_2} &=\frac{N_2}{N_1} =\frac{1}{2}\\ \rightarrow \;\;I_1 &=\frac{I_2}{2} \;\Rightarrow \; I_1 \lt I_2 \end{aligned}

$I_1$ and $I_2$ are in phase, because $I_1=I_2+ \text{Exciting current } (I_M)$ But, $I_m$ is very small. On the basis of above analysis, phasor diagrma is

Q57GATE 2003MCQ

Figure shows a $\Delta$ -Y connected, 3-phase distribution transformer used to step down the voltage from 11000 V to 415 V line-to-line. It has two switches $S_{1}$ and $S_{2}$ . Under normal conditions $S_{1}$ is closed and $S_{2}$ is open. Under certain special conditions $S_{1}$ is open and $S_{2}$ is closed. In such a case the magnitude of the voltage across the LV terminals a and c is

🚀 Solve in Practice Mode

📖 Explanation

As the transformer is $\Delta -Y$ connected, it means line-to-line voltage on primary side induces line to neutral for phase voltage on secondary side due to transformer.

\begin{aligned} \text{Turn ratio} &=\frac{N_1}{N_2}=a \\ &= \frac{11kV}{415\sqrt{3}}=\frac{11\sqrt{3}}{415} \times 10^3\\ V_{AB}&=aV_a \\ V_{BC}&=aV_b\\ V_{AC}&=aV_c \end{aligned}

When $S_2$ is closed, B and C are at same potential. So, $V_{BC}=V_B-V_C=0$ Applying KVL in, $\Delta$ connection

\begin{aligned} V_{AB}+V_{BC}+V_{AC} &=0 \\ \Rightarrow \;\; V_{AC} &=-V_{AB} \\ V_{AB}&= 11 \times 10^3 \angle 0^{\circ}\\ V_{AC}&= - 11 \times 10^3 \angle 0^{\circ}\\ &= 11 \times 10^3 \angle 180^{\circ} \end{aligned}

Voltage across terminals a and c,

\begin{aligned} V_{ac} &=V_a-V_c \\ &=\frac{V_{AB}}{a}-\frac{V_{AC}}{a} \\ &= \frac{1}{a}[V_{AB}-V_{AC}]\\ &=\frac{415}{11\sqrt{3} \times 10^3} \\ &\$[11 \times 10^3 \angle 0^{\circ} - 11 \times 10^3 \angle 180^{\circ} ]\$\\ &=\frac{415}{11\sqrt{3} \times 10^3} \$[11 \times 10^3 \times 2]\$\\ &\approx 480 V \end{aligned}

Q58GATE 2003MCQ

Figure shows an ideal three-winding transformer. The three windings 1, 2, 3 of the transformer are wound on the same core as shown. The turns ratio $N_{1}:N_{2}:N_{3}$ is 4:2:1. A resistor of 10 $\Omega$ is connected across winding-2. A capacitor of reactance 2.5 $\Omega$ is connected across winding-3. Winding-1 is connected across a 400 V, ac supply. If the supply voltage phasor $V_{1}=400\angle 0^{\circ}$ , the supply current phasor $I_1$ is given by

🚀 Solve in Practice Mode

📖 Explanation

Turn ratio, $N_1:N_2: N_3 is 4:2:1$ Induced emf in winding 1 $=E_1\approx V_1=400\angle 0^{\circ}V$ As we know,

\begin{aligned} \frac{E_1}{E_2} &=\frac{N_1}{N_2} \\ E_2&=\frac{N_2}{N_1}E_1=\left ( \frac{2}{4} \right ) 400\angle 0^{\circ}\\ &=200\angle 0^{\circ} \\ \text{and}\;\;\frac{E_1}{E_3} &=\frac{N_1}{N_3} \\ \Rightarrow \; E_3 &=\frac{N_3}{N_1}E_1 \\ &=\left ( \frac{1}{4} \right ) 400\angle 0^{\circ}= 100\angle 0^{\circ}V \end{aligned}

Current in secondary winding $I_2=\frac{E_2}{R}=\frac{200\angle 0^{\circ}}{10}=20\angle 0^{\circ}A$ Current in tertiary winding $I_3=\frac{E_3}{-j \times C}=\frac{100\angle 0^{\circ}}{-j2.5}=40\angle 90^{\circ}A$ $I_2$ referred to primary side

\begin{aligned} I'_2 &=\left ( \frac{N_2}{N_1} \right )I_2 \\ &= \left ( \frac{2}{4} \right ) \times (20\angle 0^{\circ})=0\angle 10^{\circ}A \end{aligned}

$I_3$ reffered to primary side

\begin{aligned} I'_3 &=\left ( \frac{N_3}{N_1} \right )I_3 \\ &= \left ( \frac{1}{4} \right ) \times (40\angle 90^{\circ})=10\angle 90^{\circ}A \end{aligned}

The suuply currrent

\begin{aligned} I_1 &=I'_2+I'_3 \\ &= 10\angle 0^{\circ}+10\angle 90^{\circ}=(10+j10)A \end{aligned}