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A 3 phase, 10 KVA lead has power factor of 0.342.The power is measured by two wattmeter method. Find the reading of each wattmeter when. (i) Power factor is leading
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Given:

S = 10 kva

P.f = 0.342

W1 = ?

W2 = ?

Formula:

P=Scosϕ

P.f=cosϕ

ϕ=cos1(P.f)=70

Case 1: Lagging P.f

$W_1 +W_2 =P =S.(P.f)\\ =10\times 0.342\\ =3.4kW ...(1)\\ \tan \phi = \left [ \sqrt{3}\dfrac{(W_1-W_2)}{W_1+W_2} \right ]\\ \therefore W_1-W_2 =(W_1+W_2)\dfrac{\tan \phi}{\sqrt{3}}\\ …

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