Experiement 9

AC Steady State Power

Objectives: 1. We 2. We 3. We 4. We

examine examine examine examine

instantaneous power. average power. complex power. power factor correction.

Equipment: 1. Breadboard. 2. Function generator. 3. PSpice software. 4. Connecting wires. 5. Oscilloscope. 6. Resistors, Capacitor, Inductor.

Contents: 1. Ramifications of power. 2. Power factor correction.

Theory: Done by: Shady1Arafeh & Ahmad Mujahed

AC Steady State Power We have many types of power instantaneous, average, apparent, active, reactive, and complex power. the instantaneous power is p(t) = v(t)i(t) & it can be written as p(t) = 0.5(VpIp)[cos(θv-θi) + cos(2ωt + θv + θi )] ,And we have Another power is the average power Pave = 0.5(VpIp) cos(θv-θi) and it can be written as Pave = Vrms Irms cos(θv-θi). the power factor is Pf = cos(θv-θi) = cosθZL and the load in RC circuit has a leading power factor & a lagging power factor in RL circuits. And the complex power is defined as S = Vrms I*rms , S = P + jQ where Q is the reactive power and the magnitude of the complex power is the apparent power and the phase is the power factor angle.

Calculation and data analysis: Done by: Shady2Arafeh & Ahmad Mujahed

AC Steady State Power



Ramifications of Power:

We connect the circuit shown below and then take some readings and fill the following table :

Source voltage Amplitude 2v

Source voltage Angle 0

the source voltage is V = 2cos (6280t) w=2π*1000=6280 rad.s load voltage Amplitude

load voltage Angle

2v

0

the load voltage is V = 2cos (6280t) w=2π*1000=6280 rad.s

load current Amplitude

load current Angle

Done by: Shady3Arafeh & Ahmad Mujahed

AC Steady State Power 11mA

A=0.8 , B=1.4 so the angle =34.8

Pf = cos(θv-θi) = cos(34.8) = 0.82 Pave = Vrms Irms cos(θv-θi) =0.5*2*0.011cos34.8 = 0.00902 w Apparent power = Vrms Irms = 0.5*2 *0.011 = 0.011w



power factor correction:

We connect a parallel capacitor with the circuit and measure the power factor then we fill the following table:

load current Amplitude 1.7mA

load current Angle A=0.2 , B=0.4 So the angle is 30o

Pf = cos(θv-θi) = cos(30) = 0.866

Conclusions:  the parallel capacitor increasing the power

factor and the inductor decrease it,so that we

Done by: Shady4Arafeh & Ahmad Mujahed

AC Steady State Power can decrease the loss of power by adding a capacitor parallel to the load in order to consume most of the power dissipated  the power factor in the capacitive and

inductive load is zero because cosθZL = cos-90= cos90 =0

Done by: Shady5Arafeh & Ahmad Mujahed