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Part II.
4.3. Effective or RMS Value
The average power absorbed by the resistor in the
ac circuit is;
The power absorbed by the resistor in the dc circuit
is
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4.3. Effective or RMS Value
Equating these equations;
The effective value of the
voltage is found in the same
way as current;
This indicates that the effective value is the (square) root of the mean (or
average) of the square of the periodic signal. Thus, the effective value is
often known as the root-mean-square value, or rms value for short; and
we write
4.3. Effective or RMS Value
To find rms value of x(t), we first find its square
then find the mean of that.
and
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4.3. Effective or RMS Value
 The average power can be written in terms of the rms values;
 Similarly, the average power by a resistor R in;
Example 4.7.
Determine the rms value of the current waveform
in figure. If the current is passed through a 2 Ω
resistor, find the average power absorbed by the
resistor.
Solution:
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Example 4.7.
Example 4.8.
The waveform shown in figure is a half-wave
rectified sine wave. Find the rms value and the
amount of average power dissipated in a 10Ω
resistor.
Solution:
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Example 4.8.
Example 4.8.
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4.4. Apparent Power and Power Factor
İf the voltage and current at the terminals of a circuit
are;
4.4. Apparent Power and Power Factor
The power factor is;
The power factor angle;
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4.4. Apparent Power and Power Factor
 The value of pf ranges between zero to unity.
 For a purely resistive load, the voltage and current are in phase, so
that
0 and pf=1. This implies that apparent power is equal
to the average power.
 For a purely reactive load,
90 and pf=0.
 In this case the average power is zero.
 In between these two extreme cases pf is said to be leading or
lagging.
 Leading power factor means that current leads voltage, which
implies a capacitive load.
 Lagging power factor means that current lags voltage, implying an
inductive current.
Example 4.9.
Solution:
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Example 4.9.
Example 4.10.
Determine the power factor of
the entire circuit of Figure as
seen by the source. Calculate
the average power delivered by
the source.
Solution:
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Example 4.10.
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