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A 3 Phase, 10 KVA load has power factor of 0.342. The power is measured by two wattmeter method. Find reading of each wattmeter when i) Power Factor is leading ii) Power Factor is logging
written 8.3 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 4.6 years ago
basic electrical engineering
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written 8.3 years ago by gravatar for teamques10 teamques10 ★ 68k

Given

$10\times10^3 VA \\ Pf=0.342 \\ P_{apperent}=10kvk \\ \sqrt3V_LI_L=10^4 \\ V_LI_L=\dfrac{10^4}{\sqrt3}=5773.5VA \\ pf=0.342 \emptyset=\cos_{-1}(0.342)$

1. Pf is lending, then

$\hspace{0.5cm}$ $w_1=V_LI_L \cos(30-\emptyset)=5773 \cos(30-70) \\=4422.76watt \\=4.42kw \\ w_2=V_LI_L \cos(30+\emptyset)=-1002.55watt \\ =-1.0025kw$

2. pf is logging, then

$\hspace{0.5cm}$ $w_1=V_LI_L \cos(30+\emptyset)=5773\cos(30+70) \\ =-1.0025kw \\w_2=V_LI_L\cos(30-\emptyset)=4.42kw$
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