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

Welcome back.

and 2 others joined a min ago.

5.3kviews

written 8.7 years ago by

teamques10 ★ 69k

• modified 4.9 years ago

basic electrical engineering

ADD COMMENT EDIT

1 Answer

1.1kviews

written 8.7 years ago by

teamques10 ★ 69k

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

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

Create a free account to keep reading this post.

and 4 others joined a min ago.

ADD COMMENT EDIT

Community

Content

Company