Q1GATE 2024MSQ

For a two-phase network, the phase voltages $V_{p}$ and $V_{q}$ are to be expressed in terms of sequence voltages $V_{\alpha}$ and $V_{\beta}$ as

\left[\begin{array}{l}V_{p} \\ V_{q}\end{array}\right]=S\left[\begin{array}{l}V_{\alpha} \\ V_{\beta}\end{array}\right]

. The possible option(s) for matrix $S$ is/are

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Q2GATE 2022MCQ

In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced three-phase supply of $100\sqrt{3}$ with phase sequence $ABC$ . The star connected load has $Z_A=10\Omega$ and $Z_B=20\angle 60^{\circ}$ . The value of $Z_C$ in $\Omega$ , for which the voltage difference across the nodes $n$ and $n'$ is zero, is

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📖 Explanation

Given: n & n' are at same potential, therefore, $I_{nn'}=0$ $I_{A}+I_{B}+I_{C}=0$ $\frac{E_A}{Z_A}+\frac{E_B}{Z_B}+\frac{E_C}{Z_C}=0$ where , $E_A=100\angle 0^{\circ}V \text{ and }Z_A=10\Omega$ $E_B=100\angle -120^{\circ}V \text{ and }Z_B=20\angle 60^{\circ}$ $E_C=100\angle 120^{\circ}V \text{ and }Z_C=?$ $\therefore \frac{100\angle 0^{\circ}}{10}+\frac{100\angle -120^{\circ}}{20\angle 60^{\circ} }+\frac{100\angle 120^{\circ}}{Z_C}=0$ $\Rightarrow Z_C=20\angle -60^{\circ} \Omega$

Q3GATE 2021NAT

A three-phase balanced voltage is applied to the load shown. The phase sequence is $\text{RYB}$ . The ratio $\frac{\left | I_{B} \right |}{|I_{R}|}$ is ____________.

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📖 Explanation

\begin{aligned} I_{R} &=\frac{V_{R B}}{-j 10}=\frac{V_{L} \angle-60^{\circ}}{-j 10}=\frac{V_{L}}{10} \angle 30^{\circ} \\ I_{Y} &=\frac{V_{Y B}}{j 10}=\frac{V_{L} \angle-120^{\circ}}{-j 10}=\frac{V_{L}}{10} \angle 150^{\circ} \\ \text { and }\qquad \qquad I_{B}&=-\left(I_{R}+I_{Y}\right)=\frac{V_{L}}{10} \angle-90^{\circ} \\ \therefore \qquad \qquad \left|\frac{I_{B}}{I_{R}}\right|&=1 \end{aligned}

Q4GATE 2015MCQ

In the given network $V_{1}=100\angle 0^{\circ}$ V, $V_{2}=-120\angle 0^{\circ}$ V, $V_{3}=+120\angle 0^{\circ}$ V. The phasor current i (in Ampere) is

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📖 Explanation

$'i'$ is taken as input, Apply KCL at node Node voltage is $100\angle 120^{\circ}$ $i=\frac{100\angle 120^{\circ}-100\angle 0^{\circ}}{-j1}$ $+\frac{100\angle 120^{\circ}-100\angle -120^{\circ}}{j1}$ $i=173.2\angle -60^{\circ}A$

Q5GATE 2014MCQ

The line A to neutral voltage is $10 \angle 15^{\circ}$ V for a balanced three phase star connected load with phase sequence ABC . The voltage of line B with respect to line C is given by

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📖 Explanation

Given, $V_{AN}=10\angle 15^{\circ}volt$ As the system is balanced and phase sequence is ABC, therefore, $V_{AN}=10\angle 15^{\circ}$ $V_{BN}=10\angle 135^{\circ}V$ and $V_{CN}=10\angle 255^{\circ}V$ $\therefore$ voltage of line w.r.t. line C is $V_{BC}=V_{BN}-V_{CN}$ $V_{BC}=10\angle 255^{\circ}-10\angle 135^{\circ}$ $V_{BC}=10\sqrt{3}\angle -75^{\circ} Volt$

Q6GATE 2012MCQ

A two-phase load draws the following phase currents : $i_{1}(t)=I_{m}sin(\omega t-\phi _{1})$ , $i_{2}(t)=I_{m}cos(\omega t-\phi _{2})$ . These currents are balanced if $\phi _{1}$ is equal to.

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📖 Explanation

In two phase, current are balanced if phase difference is $90^{\circ}$ . $i_1(t)=I_m \sin (\omega t-\phi _1)$ $i_2(t)=I_m \sin (\omega t-\phi _2+90^{\circ})$ Phase difference is $90^{\circ},$ $\omega t-\phi _2+90^{\circ}-(\omega t-\phi _1)=90^{\circ}$ $\phi _1-\phi _2=0$ $\phi _1=\phi _2$

Q7GATE 2005MCQ

For the three-phase circuit shown in the figure, the ratio of the currents $I_{R}$ : $I_{Y}$ : $I_{B}$ is given by

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📖 Explanation

Assuming phase sequence to be RYB Taking $V_{RY}$ as the reference, $V_{RY}=V\angle 0^{\circ}$ $V_{YB}=V\angle -120^{\circ}$ $V_{BR}=V\angle -240^{\circ}$ $I_R=\frac{V_{RB}}{R_1}=-\frac{V_{BR}}{R_1} =-\frac{V\angle -240^{\circ}}{R_1}=\frac{V}{R_1}\angle -60^{\circ}$ $I_Y=\frac{V_{YB}}{R_1} =-\frac{V\angle -120^{\circ}}{R_1}$ Using KCL, $I_R+I_Y+I_B=0$ $\frac{V}{R_1}\angle -60^{\circ}+\frac{V}{R_1}\angle -120^{\circ} I_B=0$ $\Rightarrow \; I_B=\sqrt{3}\frac{V}{R_1}\angle 90^{\circ}$ So, $I_R:I_Y:I_B=\frac{V}{R_1}:\frac{V}{R_1}:\sqrt{3}\frac{V}{R_1}$ $\;\;\; =1:1:\sqrt{3}$

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