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The current coil in a wattmeter is connected in series with an ammeter & an inductive load. The ammeter reading is 5.5 A. A voltmeter & the voltage coil are connected across 200 Hz supply. The reading of voltmeter & wattmeter is 220V & 25 watt respectively. The inductance of the voltage circuit is 12 mH & its resistance R = 2400 Ω. If the voltage drop across the ammeter & the current coil are negligible, then what will be the approximate percentage error in the wattmeter reading.
  • a)
    26.12 %
  • b)
    29.35 %
  • c)
    36.37 %
  • d)
    43.37 %
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The current coil in a wattmeter is connected in series with an ammete...
Inductive reactance of pressure coil =2π × 200 × 12 × 10−13 = 15Ω
Resistance of pressure coil =2400Ω
Phase angle of pressure coil
Since,
tan⁡β = 152 / 400 = 6.25 × 10−3
So, β = tan−1⁡(6.25 × 10−3)
=0.36 or β = 0.36
We know that for inductive load,
Reading of the wattmeter αcos⁡β.cos⁡(ϕ−β)
True power αcos⁡ϕ
Where cos ϕ load p.f.
Therefore,
True power
= cos⁡ϕ / cos⁡β.cos⁡(ϕ − β) × Re ading of wattmeter
Here true power = 12R
= (5.5)2R
Which is equal to = 12Zcos⁡ϕ
Impedance
Z = 220 / 5.5 = 40Ω
So, true power = (5.5)2 × 40cos⁡ϕ
Reading of wattmeter is given as = 25 watt
So,
(5.5)2 × 40 cos⁡ϕ = cos⁡ϕ / cos⁡β⋅cos⁡(ϕ − β) × 5
Or,cos⁡β⋅cos⁡(ϕ − β) = 25(5.5)2 × 40 = 0.0206
since β = 0.36
So, cos⁡(0.36)⋅cos⁡(ϕ − 0.36) = 0.0206
or cos⁡(ϕ − β) ≃ 0.0206
(ϕ−β) ≃ cos−1⁡(0.0206)
ϕ−β ≃ 88.81
so, ϕ ≃ 88.81 + 0.36 ≃ 89.17
therefore, percentage error
=tan⁡ϕ⋅tan⁡β × 100
=tan⁡(89.17)⋅tan⁡(0.36) × 100
=43.37%
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Most Upvoted Answer
The current coil in a wattmeter is connected in series with an ammete...
And a load, while the potential coil is connected in parallel with the load. The current coil measures the current flowing through the circuit, while the potential coil measures the voltage across the load. By multiplying the readings of the two coils, the wattmeter is able to measure the power consumed by the load. The unit of measurement for power is watts (W).
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The current coil in a wattmeter is connected in series with an ammeter & an inductive load. The ammeter reading is 5.5 A. A voltmeter & the voltage coil are connected across 200 Hz supply. The reading of voltmeter & wattmeter is 220V & 25 watt respectively. The inductance of the voltage circuit is 12 mH & its resistance R = 2400 Ω. If the voltage drop across the ammeter & the current coil are negligible, then what will be the approximate percentage error in the wattmeter reading.a)26.12 %b)29.35 %c)36.37 %d)43.37 %Correct answer is option 'D'. Can you explain this answer?
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