sequence of solving these 2 questions for power in AC circuit?

Question

i need to know the sequence or steps of answering these 2 questions
q1.A two branch parallel circuit has branch impedance z1=2-j5 , z2=1+j1
obtain the complete power Triangle for the circuit if the 2 ohm resistor consumes 20 watt.
also calculate the total apparent power,total active power, total reactive power,total power factor

q2.An inductive load draws 600 watt, 0.769 pf (power factor) is connected across a voltage source 200 volt, 50 hz.
it is required to improve the pf to 0.98 lagging, what is the value of the capacitor to be connected in parallel.
i need just the sequence of using the right rules to get the correct answer for both questions, no need for final answer or calculations, and of coarse explanation excepted

Answer 1

Just the steps, right?



q1. The current in the 2 ohm branch is found from the wattage it dissipates. The voltage across both branches is found by summing the resistive drop and the (90 degree displaced) capacitive voltage drop. The current in the 1 ohm branch can be found from the voltage across it and its impedance. The remaining values are defined by simple equations.



q2. To improve the power factor, draw the phase diagram, draw the effect of the parallel capacitor on incoming I-V phase relation, adjust C to get the .98 pf.



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Answer 2

Question 1



V of 2 Ohm resistor= sq. rt. (2 Ohms x 20 Watts) = 6.325 Volts

i of 2 Ohm resistor = 6.325V / 2 Ohms = 3.162 A

V of capacitor = (Xc) x (i of c) = 5 Ohm x 3.162A = 15.81 Volts

V Total = Square root [(6.325V squared) + ( 15.81V squared)] = 17.03V

Capacitor Volt-Amps = 15.81V x 3.162A = 50 VA



Z of 1 Ohm resistor in series with inductor = square root [(1 Ohm squared) + (1 Ohm squared)] = 1.414 Ohms.

i of 1 ohm resistor and inductor = 17.03V / 1.414 Ohm = 12.04 Amps

V of inductor = 12.04 A X 1 Ohm = 12.04 Volts

Watts dissipated in 1 Ohm resistor = 12.04 Amps squared x 1 Ohm = 145 Watts

Inductor Volt-Amps = 12.04V x 12.04A = 145 VA



Adjacent side of power triangle = 20 Watts + 145 Watts = 165 Watts

Opposite side of power triangle = 145VA - 50 VA = 95 Volt-Amps

Hypotenuse of power triangle = square root of [(165 squared) + (95 squared)] = 190.39 Volt-Amps



Total apparent power = 190.39 Watts

Total real power = 165 Watts

Total reactive power = 95 Watts

Total power factor = 165 / 190.39 = .87



Question 2

Load current x 200V x .769 = 600 Watts

Load current = 3.9 Amps

3.9 x 3.9 x load resistance = 600 Watts

Load resistance = 600/15.21 = 39.45 Ohms

Angle theta = arc cos .769 = 39.74 degrees

tan theta = XL / 39.45 Ohms

XL = 32.79 Ohms



Improved angle theta = arc cos .98 = 11.48 degrees

tan 11.48 = XL/39.45

XL = 8.01 Ohms

old XL - new XL = needed value of Xc

Xc = 32.79 Ohms - 8.01 Ohms = 24.78 Ohms

Xc = 1 / 2pi f c

24.78 = 1 / (6.28) x (50) x c

c = 128 uF

Answer 3

All you could desire to do is combine like words. So in case you combine the x's first, 1x + 2x + ...+ 21x would provide you 231x (you in simple terms upload the coefficients). next you will sum up the a million's. 21 a million's would in simple terms provide you 21. So, your answer would be 231x + 21.