The series capacitance to be added to increase the power factor is : 345 μF
Given data :
Power line voltage ( V[tex]_{rms}[/tex] ) = 120 V
Frequency = 60 Hz
Current drawn by power line ( I[tex]_{rms}[/tex] ) = 8.40 A
Average energy dissipation ( P ) = 850 W
power factor = 1
Determine the value of series capacitance needed to increase the power factor to 1
we will apply the formulae below
first step : calculate the value of the phase angle
[tex]\beta[/tex] = cos⁻¹ [tex]( \frac{P}{V_{rms}* I_{rms} } )[/tex]
= cos⁻¹ ( 850 / ( 120 * 8.4 ) )
= 32.5°
Final step : Determine the series Capacitance
Series Capacitance ( C ) = [tex](\frac{1}{(2\pi f)(\frac{V_{rms} }{Ix_{rms} }) (sin\beta -cos\beta tan\beta _{t} } )[/tex]
Therefore C = [tex](\frac{1}{(2\pi *60)\frac{120}{8.4} * sin32.5-cos32.5*tan0} )[/tex]
= 345 * 10⁻⁶ F = 345 μF
Hence we can conclude that The series capacitance to be added to increase the power factor is : 345 μF
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