Chapter 23, Problem x08 In a series RCa aircuit the dissipated power drops by a

Question

Chapter 23, Problem x08 In a series RCa aircuit the dissipated power drops by a factor of 9.0 when the frequency of the generator is changed from the resonant frequency to a nonresonant frequency, The peak voltage is held constant while this change is made. Determine the power factor of the carcuit at the nonresonant frequency. Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise. Units Number The number of significant digits is set to 2; the tolerance is +/2% Click if you would like to Show Work for this question: Open Show Work SHOW HENT LINK TO TEXT

Explanation / Answer

The current at resonance is:
es=V R , where V is the source voltage

At the other frequency (K), the current is:
=V ()=V (R² + X²)

The ratio of the power dissipations is (Pes) (P) = 9
Since:P = i²•R , then:

then:(es²•R) (i²•R)=(es)² (i)²=9
and:(es) (i)=9
(V R) (V )=9
() R=9

but:PF = R ...definition of power factor
PF = R ()
= 1 9

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