Determine the power factor and the true powder in the circuit of Figure 15-59.

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Accepted Answer

The capacitive reactance is

[latex]X_{C}= \frac{1}{2\pi fC} = \frac{1}{2\pi (10 \ kHz)(0.0047 \ \mu F)}= 3.39 \ k\Omega [/latex]

The total impedance of the circuit in rectangular form is

Z = R — j[latex]X_{C}[/latex] = 1.0 kΩ - j3.39 kΩ

Converting to polar form yields

[latex]Z =\sqrt{R^{2}+ X^{2}_{C}} \angle - an ^{-1} \left(\frac{X_{C}}{R} ight)[/latex]

[latex]\ \ \ \ \ = \sqrt{(1.0 \ k\Omega )^{2}+ (3.39 \ k\Omega )^{2}} \angle - an^{-1}\left(\frac{3.39 \ k\Omega}{1.0 \ k\Omega} ight)= 3.53 \angle -73.6° k\Omega[/latex]

The angle associated with the impedance is θ, the angle between the applied voltage and the total current; therefore, the power factor is

[latex]PF= \cos heta = \cos (-73.6°)= 0.282[/latex]

The current magnitude is

[latex]I=\frac{V_{s}}{Z}= \frac{15 \ V}{3.53 \ k\Omega }= 4.25 \ mA [/latex]

The true power is

[latex]P_{true}= V_{s}I\cos heta = (15 \ V)(4.25 \ mA)(0.282)= 18.0 \ mW[/latex]

Question 15.23: Determine the power factor and the true powder in the circui...

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