Q2GATE 2024NAT

In the circuit shown, $Z_{1}=50 \angle-90^{\circ}$ and $Z_{2}=200 \angle-30^{\circ} \Omega$ . It is supplied by a three phase $400 \mathrm{~V}$ source with the phase sequence being R-Y-B. Assume the watt meters $W_{1}$ and $W_{2}$ to be ideal. The magnitude of the difference between the readings of $W_{1}$ and $W_{2}$ in watts is ______ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_{R Y} & =400 \angle 0^{\circ} \\ V_{Y B} & =400 \angle-120^{\circ} \\ V_{B R} & =400 \angle-240^{\circ} \\ V_{P C} & =V_{R B}=400 \angle-60^{\circ} \\ I_{1} & =\frac{V_{R B}}{Z_{1}}=\frac{400 \angle-60^{\circ}}{50 \angle-90^{\circ}}=8 \angle 30^{\circ} \\ I_{2} & =\frac{V_{R Y}}{Z_{2}}=\frac{400 \angle 0^{\circ}}{200 \angle-30^{\circ}}=2 \angle 30^{\circ} \Rightarrow I_{C C(2)} \\ I_{L 1} & =I_{C C(1)}=I_{1}+I_{2}=8 \angle 30^{\circ}+2 \angle 30^{\circ} \\ & =10 \angle 30^{\circ} \Rightarrow I_{C C(1)} \\ W & =V_{R B} \times I_{2 \theta} \cos \left(\angle V_{R B} \text { and } I_{L O}\right) \\ W_{1} & =400 \times 10 \times \cos \left(-60^{\circ}-30^{\circ}\right)=0 \text { Watt } \\ W_{2} & =V_{Y B} \times I_{2} \cos \left(\angle V_{Y B} \text { and } I_{2}\right) \\ & =400 \times 2 \times \cos \left(-120^{\circ}-30^{\circ}\right)=-692.8 \text { Watt } \\ W_{2} & =0-(-692.8)=692.8 \text { Watt } \end{aligned}

Q6GATE 2017MCQ

The load shown in the figure is supplied by a 400 V (line to line) 3-phase source (RYB sequence). The load is balanced and inductive, drawing 3464 VA. When the switch S is in position N, the three watt-meters $W_1,W_2 \; and \; W_3$ read 577.35W each. If the switch is moved to position Y, the readings of the watt-meters in watts will be:

🚀 Solve in Practice Mode

📖 Explanation

When switch is on Neutral side

\begin{aligned} W &= W_1+W_2+W_3\\ &=577.35+577.35+577.35 \\ &=1732 \; Watts \\ \text{VA load} &= 3464 =3 V_{ph}I_{ph}\\ \Rightarrow \; V_{ph}I_{ph} &=1154.66 \end{aligned}

Each $\Omega$ meter reading

\begin{aligned} &= V_{RN}I_{R} \cos (\angle V_{RN} \; \text{and}\; I_R)\\ 577.35&= V_{ph}I_{ph} \cos (\phi )\\ \cos (\phi ) &=\frac{577.35}{1154.66}=05 \\ \phi &= \cos ^{-1} (0.5)=60^{\circ} \end{aligned}

When switch is on Y-phase side,

\begin{aligned} W_1&= V_{RY}I_R \cos (\angle V_{RY} \; \text{and}\; I_R)\\ &= \sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}+\phi ) \\ &= \sqrt{3} \times 1154.66 \times \cos (30^{\circ}+ 60^{\circ})\\ &= 0 \text{Watt}\\ W_3 &= V_{BY}I_R \cos (\angle V_{BY} \; \text{and}\; I_R)\\ &=\sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}-\phi )\\ &=\sqrt{3} \times 1154.66 \times \cos (30^{\circ}- 60^{\circ}) \\ &= 1732 \text{Watt}\\ W_1&=0\\ W_2&=0\\ W_3&=1732 \text{Watt} \end{aligned}

Q20GATE 2004MCQ

A single-phase load is connected between R and Y terminals of a 415 V, symmetrical, 3-phase, 4-wire system with phase sequence RYB. A wattmeter is connected in the system as shown in figure. The power factor of the load is 0.8 lagging. The wattmeter will read

🚀 Solve in Practice Mode

📖 Explanation

Line of line voltage $=450V$ Phase to neutral voltage $=\frac{415}{\sqrt{3}}V$ Taking $V_R$ as the reference

\begin{aligned} V_{RN} &=\frac{415}{\sqrt{3}} \angle 0^{\circ}V\\ V_{yN} &=\frac{415}{\sqrt{3}} \angle -120^{\circ}V\\ \text{and } V_{BN} &=\frac{415}{\sqrt{3}} \angle -240^{\circ}\\ V_{Ry} &=V_{RN}-V_{yN} \\ &=\frac{415}{\sqrt{3}} \angle 0^{\circ}- \frac{415}{\sqrt{3}} \angle -120^{\circ} \\ &=415 \angle 30^{\circ}V \\ & \text{Load current,}\\ I_L&= \frac{V_{Ry}}{Z}=\frac{415 \angle 30^{\circ}}{100 \angle 36.87^{\circ}}\\ &= 4.15 \angle -6.87^{\circ} A \end{aligned}

Current through current coil, $I_{cc}=I_L=4.15\angle -6.87^{\circ}A$ Voltage across pressure coil $V_{BN}=\frac{415}{\sqrt{3}}\angle -240^{\circ}V$ Phase angle between $V_{BN} \text{and} I_{cc}$ $\phi =-240^{\circ}-(-6.87^{\circ})=233.13^{\circ}$ Wattmeter reading $=|V_{BN}||I_L| \cos \phi$ $\;\;\; =\frac{415}{\sqrt{3}} \times 4.15 \times \cos 233.13$ $\;\;\; \approx -597W$

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