Fig Two shows two Cartesian coordinate diagram
(Apparent Power)2 = (True Power)2 + (Reactive Power)2
Reactive Power = inductive reactive power - nhl jerseys from chinacapacitive reactive power
REACTIVE POWER
Fig Two shows two Cartesian coordinate diagrams. The upper nba jerseys from chinadiagram shows Xc with R = 0.
Since R = 0,
Xc = Z/-90o
Z/-90 = Z*cos (-90o) + Z*sin (-90o)This article requires a knowledge of basic wholesale nfl jerseyselectrical theory, complex numbers, polar coordinates and trigonometry
APPARENT POWER, REACTIVE POWER AND TRUE POWER
Figure One shows the three types of power
Z/90o = 0 + jZ because XL = Z
I = V/0o divided by Z/90o = I/-90o Current lags world cup jerseysvoltage by 90 degrees Apparent Power = V*I/ao
Reactive Power = V*I*sine (ao)
True Power = V*I*cosine (ao)
Using Pythagorean Theorem
For this power calculation, the sign of the angle need not be considered. The reason is that there is no capacitive reactance in the circuit. However, for school work, go by what your instructor says.
Power = V/0o * I/-90o = (V * I)/-90o
(V * I)/-90o = V*I*cos (90o) - jV*I*sin (90o)Reactive Power = V*I*sin (90o)
In this case, the reactive power equals the apparent power because the true power is 0 watts.
The lower diagram shows XL with R = 0
Z/90 = Z*cos (90o) + Z*sin (90o)
Reactive Power = V*I*sin (90o)
The reactive power dissipated is the absolute value V*I*sine (90o)
In this case also, the reactive power equals the apparent power because the true power is 0 watts.
I
Z1 = 100/0o
Z2 consists of a resistor R3 and annba jerseysinductor L1
Z2 = 20/50o
Z3 consists of a resistor R2 and a capacitor C1f both inductive reactance and capacitive reactance were in the circuit, then the sign of the angle would be important because it would indicate whether the net power is inductive of capacitive in nature.
PROBLEM
The parallel circuit in figure two consists of three impedances. See figure four.
Z1 is a resistor R1
Z3 = 50/-80oFind the apparent power dissipated by the circuit
SOLUTION
The total power diss
The voltage source V output is 100/0o
ipated is the sum of the power dissipated by each branch of the circuit.
Finding the current and the apparent power for the branch containing Z1
The current I1 through Z1 equals Vs/0o divided by Z1/0o
I1 = (100/0o) volts divided by 100/0o ohms = 1/0o amperes
True Power = V*I = (100/0o) * 1/0o = 100/0o watts
In this case, the true power equals the apparent power.
Finding the current and the apparent power for the branch containing Z2
The current I2 through Z2 equals V/0o divided by Z2
I2 = 100/0o divided by 20/50o = 5/-50o
The power is equal to the voltage V multiplied by the current I2
Apparent power = 100/0o * 5/-50o = 500/-50o
Finding the current and the apparent power for the branch containing Z3
The current through Z3 equals the voltage V divided by the Z3
I3 = 100/0o divided by 50/-80o = 2/80o
Apparent Power = V * I = 100/0o * 2/80o = mlb jerseys200/80o
Finding the total apparent power
The total power is the sum of the power nhl jerseysthrough each branch
Now we add like terms
100 cos 0o + 500 cos 50o + 200 cos 80o = true power
100 + 500 * 0.643 0 + 200 * 0.174 =
492.75/22.20o
Hence the total apparent power is 492.75/22.20o watts.
