Q21GATE 2004MCQ

The circuit in figure is used to measure the power consumed by the load. The current coil and the voltage coil of the wattmeter have 0.02 $\Omega$ and 1000 $\Omega$ resistances respectively. The measured power compared to the load power will be

🚀 Solve in Practice Mode

📖 Explanation

Load Power (true power) $=VI \cos \phi$ $=200 \times 20 \times 1=4000W$ Resistance of current coil $=R_{cc}=0.02$ Current through CC $=I_{c}=20A$ Power consumed by current coil $=I_c^2 R_{cc}$ $=20^2 \times 0.02=8W$ Measured power =Power consumed by load + Power consumed by current coil Measured power =4000+8=4008W $\% \text{error} =\frac{\text{Measured power}-\text{True power}}{\text{True power}} \times 100 =\frac{4008-4000}{4000} \times 100 =0.2\% \;\;(\text{more})$

Q23GATE 2003MCQ

A wattmeter reads 400 W when its current coil is connected in the R-phase and its pressure coil is connected between this phase and the neutral of a symmetrical 3-phase system supplying a balanced star connected 0.8 p.f. inductive load. This phase sequence is RYB. What will be the reading of this wattmeter if its pressure coil alone is reconnected between the B and Y phases, all other connections remaining as before ?

🚀 Solve in Practice Mode

📖 Explanation

Taking $V_R$ as the reference and assuming phase to neutral voltage $=V$

\begin{aligned} V_R &=V\angle 0^{\circ} \\ V_Y &= V\angle -120^{\circ}\\ and \;\; V_B &=V\angle -240^{\circ} \end{aligned}

\begin{aligned} V_RI_R \cos \phi &=400 W \\ VI_R \times 0.8&= 400W\\ VI_R&=400\;\;...(i) \\ I_R&=I_R\angle -36.87^{\circ} \end{aligned}

[p.f. =0.8 lag, inductive load]

\begin{aligned} V_{YB} &=V_{YB}-V_B \\ &= V\angle -120^{\circ}- V \angle -240^{\circ}\\ &=\sqrt{3}V\angle -90^{\circ} V\\ I_R&=I_R\angle -36.87^{\circ}A \end{aligned}

Angle between $V_{YB}$ and $I_{R}$ $\theta =-90^{\circ}-(-36.87^{\circ})=53.13^{\circ}$ As pressure coil connected between Y and B phase. Reading of wattmeter $=V_{YB}I_R\cos \theta$ From equation (i) $=\sqrt{3}V \times I_R \times \cos (-53.13) =\sqrt{3}\times 500 \times 0.6=519.6W$

Q24GATE 2003MCQ

The voltage-flux adjustment of a certain 1-phase 220 V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it is 85 $^{\circ}$ (instead of 90 $^{\circ}$ ). The errors introduced in the reading of this meter when the current is 5 A at power factor of unity and 0.5 lagging are respectively

🚀 Solve in Practice Mode

📖 Explanation

Measured value $=VI \sin (\Delta -\phi )$ where, $\Delta =$ Phase angle between voltage and flux $\cos \phi$ = Poer factor True value= $VI \cos \phi$ Error=Measured value - True value CASE-I:

\begin{aligned} \Delta &= 80^{\circ}\\ pf&= \cos \phi =1\\ \phi &= 0^{\circ}\\ V&= 220V\\ I&= 5A \\ \text{Error}&=VI \sin (\Delta -\phi )-VI \cos \phi \\ &=220 \times 5 \sin(85-0^{\circ})-220 \times 1 \times 1\\ &\approx -4.2W \end{aligned}

CASE-II:

\begin{aligned} \Delta &= 85^{\circ}\\ pf&= \cos \phi =0.5\\ \phi &= 60^{\circ}\\ V&= 220V\\ I&= 5A \\ \text{Error}&=VI \sin (\Delta -\phi )-VI \cos \phi \\ &=220 \times 5 \sin(85-60^{\circ})-220 \times 5 \times 0.5\\ &= -85.1W \end{aligned}

Q25GATE 2002MCQ

The line-to-line input voltage to the 3 phase, 50 Hz, ac circuit shown in figure is 100 V rms. Assuming that the phase sequence is RYB, the wattmeters would read.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_L=&= 100V\\ I_L &=I_p =\frac{V_p}{Z}\\ &=\frac{100}{\sqrt{3} \times 5\angle 60^{\circ}} \\ &= \frac{20}{\sqrt{3}} \angle -60^{\circ}\\ W_1&= V_LI_L \cos (30^{\circ} -\phi )\\ &=100 \times \frac{20}{\sqrt{3}} \cos (30^{\circ}-(-60^{\circ}))\\ &= \frac{2000}{\sqrt{3}} \cos (90^{\circ})\\ W_2&= V_LI_L \cos (30^{\circ} +\phi )\\ &=100 \times \frac{20}{\sqrt{3}} \cos (30^{\circ}-60^{\circ})\\ &=1000W \end{aligned}

Q26GATE 2001MCQ

If an energy meter disc makes 10 revolutions in 100 seconds when a load of 450 W is connected to it, the meter constant (in rev/kWh) is

🚀 Solve in Practice Mode

Q27GATE 2001MCQ

The minimum number of wattmeter (s) required to measure 3-phase, 3-wire balanced or unbalanced power is

🚀 Solve in Practice Mode