WEBVTT

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Now, we are going to discuss about power factor

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For that, we will first consider a simple
circuit consisting of an alternating voltage

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source across which purely resistive load is connected. For
this voltage source, a current will flow through the load

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As the load is purely resistive, there will
be no phase difference between the voltage

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and the current in the circuit. For a purely sinusoidal
voltage supplied by the source, let's draw the waveform of both

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current and voltage. Here in the picture, the black waveform
represents the voltage and red waveform represents the current

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Now, if the load is purely inductive instead
of resistive, the current lags behind the

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source voltage by 90 degrees.
It is shown here in this figure

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The black waveform here also represents the
voltage and red waveform represents the current

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Now if the load is purely capacitive instead
of resistive, the current leads ahead the

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source voltage by 90 degrees.
It is shown here in this figure

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The black waveform here also represents the
voltage and red waveform represents the current

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Now imagine, the load has all, resistance, inductance and capacitance. And due
to collective effect of these three parameters, the current shifts behind the

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source voltage by an angle theta.
It is shown here in this figure

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The black waveform here also represents the
voltage and red waveform represents the current

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Now come to the resistive load circuit. Here, if we multiply voltage
waveform with current waveform, we will get the power waveform

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Let's do that. For that, we will consider the voltage
waveform as Vmax sin omega t and current waveform as

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Imax sin omega t and if we multiply them,
we will get power P equals Vmax sin omega

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t into Imax sin omega t. This may be rewritten as P
equals Vmax into Imax by 2 into 2 sin omega t sin omega t

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This implies P equals Vmax into
Imax by 2 into 2 sin square omega t

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This can be rewritten as sin square omega t
plus cos square omega t minus within bracket

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cos square omega t minus sin square omega t. Or, P equals
Vmax into Imax by 2 into within bracket 1 minus cos 2 omega t

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And this is power waveform. Now let's plot this expression
of power waveform along with voltage and current waveform in

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same figure. Here you see, the power waveform does
not have any negative part, although the corresponding

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voltage and current waveform do have equal positive and negative part. Now let's
find the average value of this power and for that, we have to integrate the power

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waveform equation 0 to 2
pi and divide it by 2 pi

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Look here, this will ultimately
come as Vmax into Imax by 2

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Now come back to the circuit where the load has
all, resistance, capacitance and inductance values

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And due to the collective effect of these
three parameters, the current shifts behind

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the source voltage by an angle
theta as we have already shown

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Here, as the current shifts behind the source
voltage by an angle theta, the waveform equation

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of that can be represented as
Imax sin omega t minus theta

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By multiplying voltage waveform Vmax sin
omega t and current waveform Imax sin omega

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t minus theta, we get instantaneous power
P equals to Vmax sin omega t into Imax sin

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omega t minus theta. After simplifying this equation,
we get the final power waveform here as P equals Vmax

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into Imax by 2 into within bracket cos
theta minus cos 2 omega t minus theta

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Now let's plot this expression of power waveform
along with voltage and current waveform in

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same figure. Here you see, the
power waveform has a negative part

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Now let's find the average value of this power
and for that, we have to integrate the power

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waveform equation 0 to 2
pi and divide it by 2 pi

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Look here, this will ultimately come
as Vmax into Imax by 2 into cos theta

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So far, we have calculated only average power
of a purely resistive load and an arbitrary

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load with same sinusoidal voltage and current with same magnitude
peak. Now let's come to the actual definition of power factor

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Power factor is a measure to the degree to which
a given load matches that of a pure resistance

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That means it can be considered as the ratio
of average power of an arbitrary load to the

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average power of a purely resistive load for
same sinusoidal voltage and current with same

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magnitude peak. Hence, here the power factor would
be Vmax into Imax by 2 into cos theta by Vmax into

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Imax by 2 which is nothing but simple cos theta. Hence, power factor
can be defined as cosine of the angle between voltage and current in

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any load. But it is true only for sinusoidal alternating signal. But for other
form of signals, it is determined by the ratio of average power of an arbitrary

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load to the average power of a purely resistive
load for same sinusoidal voltage and current

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with same magnitude peak. Hope we could
give you a basic idea of power factor

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Thank you