Amotor attached to a 120 v/60 hz power line draws an 8.40 a current. its average energy dissipation is 850 w. how much series capacitance needs to be added to increase the power factor to 1.0?

The series capacitance to be added to increase the power factor is : 345 μF

Given data :

Power line voltage ( V ) = 120 V

Frequency = 60 Hz

Current drawn by power line ( I ) = 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

= cos⁻¹  

   = cos⁻¹ ( 850 / ( 120 * 8.4 ) )

   = 32.5°

Final step : Determine the series Capacitance

Series Capacitance ( C ) =

Therefore C =

                    = 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

Learn more about series capacitance : link

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

the answer would be   dignified - apex