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V = V rms ∠ f v and  I



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Bog'liq
Electric Circuit Analysis by K. S. Suresh Kumar

V
=
V
rms

f
v
and 
I
=
I
rms

f
i
are the voltage phasor and current phasor at load terminals as per 
passive sign convention, then the Q delivered to the load circuit is V
rms
I
rms 
sin(
f
v
-
f
i
) VAr. The other 
expressions are relevant when the voltage phasor and/or current phasor across a pure reactive element 
is known. In this case (
f
v
-
f
i
) is assured to be 90
°
if the element is an inductor and –90
°
if the element 
is a capacitor. Then, the reactive power delivered to that element, i.e., Q is given by


Complex Power under Sinusoidal Steady-State Condition 
7.49
Q V I
I
X
V
X
=
=
=
rms rms
rms
rms
2
2
, where V
rms
is the rms value of voltage across the element, I
rms 
is the 
rms value of current through the element and X is the reactance of that element. Note that reactance 
of an inductor is a positive value and that of a capacitor is a negative value.
7.10 
complex power under sInusoIdal steady-state condItIon
Can we get these P and Q values from the voltage phasor and current phasor straightaway 
by multiplying them together? We will try. Let 
V
=
V
rms

f
v
V rms be the voltage phasor 
and 
I
=
I
rms

f
i
A rms be the current phasor; both specified as rms quantities. Then, 
VI
=

+
=
+
+
+
V
V I
jV I
v
i
v
i
v
i
rms rms
rms rms
rms rms
I
(
)
cos(
)
sin(
)
f f
f f
f f
..
We are not able to identify P or Q in the real and imaginary parts of this quantity since we know 
that 
P V I
and Q V I
v
i
v
i
=

=

rms rms
rms rms
cos(
)
sin(
),
f f
f f
But this observation prompts us to try 

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