Exp 9

Electric lab

: Objective In this experiment we explore the many : ramification of power in AC circuit . we examine instantaneous power .we examine average power we examine complex power . we examine power factor correction : Theory : Instantaneous power (V(t)=Vp cos(wt+θv (I(t)=Ip cos (wt+θi P=V(t)*I(t) P=(Vp*Ip)/2* {cos (θv{(θi)+cos(2wt+θv+ θi

Pavg=(1/T (Pavg=1/2*Vp*Ip*cos(θv-θi Pure resistance circuit Pavg=(1/2) *Vp*Ip=(1/2)*(Vp2/R)=(1/2)*Ip2*R At reactive P=0 (Pavg=Vrms*Irms *cos(θv-θi 1

Electric lab

(Power factor (pf)=cos(θv-θi P=R+jX =P (Θ=tan-1(X/R S=Vrms*Irms Power factor =Pavg/S

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

Source voltage (amplitude(v

2

Source voltage angle

0.0

Table #1

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Electric lab

The source voltage is ↔ V = 2cos (6280t) W=2π*1000=6280 rad.s

Load voltage Load voltage amplitude( (v 1Vp

Load voltage angle

θ =22.3

Table #2

The load voltage is ↔ V = 1cos )6280t(22.3+ w=2π*1000=6280 rad.s The Figure # 2 is the sinusoidal wave form before use the capacitor (use power factor correction )

Figure # 2

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Electric lab

Load current Load current amplitude( (A

Load current angle

1Vp

a=0.5 , b=1.1

θ =27 Table #3

→ θ = sin-1(a/b) →Pf = cos(θv-θi) = cos(27) = 0.89 →Pave = Vrms Irms cos(θv-θi) =0.7*0.7cos 27 = 0.43 w →Apparent power = Vrms Irms = 0.7*0.7 = 0.49 W

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

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Electric lab

The Figure # 3 is the sinusoidal wave form after use ( the capacitor (use power factor correction

Figure # 3

load load current current Amplitude Angle 1Vp

θ =17

Table #4 P f = c os (θv -θ i) = co s( 17) = 0 .95

Conclusions:  the parallel capacitor increasing the power factor and the inductor decrease it,so that we can decrease the loss of power by adding a capacitor parallel to the load in order to consume most of the power dissipated

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Electric lab



the power factor in the capacitive and inductive load is zero because cosθZL = cos90= cos90 =0

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