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what is pf. how to calculate capacitor bank value to
maintain unity pf.

Answer Posted / ezzuddin

The correct answer and below is the proof of the formula:

KVAR= KW X [&#8730;(1–(COS&#1060;)^2)]/COS&#1060;

The proof:
Pythagoras Theorem states that in a right-angled triangle
the square of the hypotenuse is equal to the sum of the
squares of the other two sides.
so that:
Q is the hypotenuse.
S is the opposite.
P is the adjacent.
Tan&#1060;= opposite/adjacent = S/P -----> S=P x(Tan&#1060;)
Pf=COS&#1060;= adjacent/hypotenuse = P/Q ---->Q=P/COS&#1060;
Pythagoras said that:
Q^2= P^2 + S^2
P^2/(COS&#1060;)^2= P^2 + (P^2 x (Tan&#1060;)^2)
by divide P^2:
1/(COS&#1060;)^2 = 1 + (Tan&#1060;)^2
by changing the both sides:
(Tan&#1060;)^2 = (1/(COS&#1060;)^2) - 1
(Tan&#1060;)^2 = (1/(COS&#1060;)^2) - ((COS&#1060;)^2/(COS&#1060;)^2)
by Taking the (1/(COS&#1060;)^2) Common factor:
(Tan&#1060;)^2 = {1/(COS&#1060;)^2)} x ( 1 - (COS&#1060;)^2)
by taking the square root:

Tan&#1060; = [&#8730;(1–(COS&#1060;)^2)]/COS&#1060;

so that:
KVAR = KW x [&#8730;(1–(COS&#1060;)^2)]/COS&#1060;

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