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,
tanβ = 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 αcosβ.cos(ϕ−β)
True power αcosϕ
Where cos ϕ load p.f.
Therefore,
True power
= cosϕ / cosβ.cos(ϕ − β) × Re ading of wattmeter
Here true power = 12R
= (5.5)2R
Which is equal to = 12Zcosϕ
Impedance
Z = 220 / 5.5 = 40Ω
So, true power = (5.5)2 × 40cosϕ
Reading of wattmeter is given as = 25 watt
So,
(5.5)2 × 40 cosϕ = cosϕ / cosβ⋅cos(ϕ − β) × 5
Or,cosβ⋅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
=tanϕ⋅tanβ × 100
=tan(89.17)⋅tan(0.36) × 100
=43.37%