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Dran-View 6 User Guide

The following example illustrates what happens if you fail to exercise care when

doing power calculations. Given a capacitive load with a voltage signal expressed

as sin (ωt + .2°) and a current signal expressed as sin (ωt + 35°) we can compute

the watts as cos (.2 - 35° + 360°) = cos (325.2°) = .82 watts. The VAR is computed

as sin (325.2°) = -.57 VAR. By our convention, a negative sign on the VAR

indicates that the current leads the voltage and consistent with that convention, we

can say that it is capacitive. The positive sign on the watts indicates that the power

is flowing from the source to the load, as you would expect. If the sign of the watts

value was negative it would imply that the “load” was behaving as a generator. If

you were looking at the fundamental, it would indicate that you probably have your

current probe reversed. Now, if you use the trigonometric identities cos(θ) = cos(-

θ) and cos (90° - θ) = sin (θ) to express the signals in cos(ωt -δ) format you will get

very different results.

sin (ωt + .2°) = cos (90° -( ωt + .2°)) = cos (ωt - + 89.8°) = cos (ωt - 89.8°) (Volts)

sin (ωt + 35°) = cos (90° -( ωt + 35°)) = cos (-ωt + 55°) = cos (ωt - 55°) (Current)

Now, using the 89.8° and 55° values that would be presented in the un-normalized

cosine expansion phase table and using the same conventions as before to compute

watts and VAR you will get the correct value for watts (cos(89.8° - 55°) = .82

watts) but the value you get for VAR will be the negation of the previous

computation (sin(89.8° - 55°) = .57 VAR). If you had known to use (-89.8°-(-55°))

= -34.8° + 360° = 325.2° as your θ, then both your watts and VAR would have

worked out correctly. Dran-View always presents the phase angle θ in a manner

consistent with computing both watts and VAR correctly regardless of which

expansion form you choose.

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