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AC power through resistor 10.0 1.0 5.0 0.5 Current, A Voltage, V Example: i = IM cos(ωt); IM=1 A; R = 10 Ω v = VM cos(ωt); VM = IM*R = 10 V 0.0 -5.0 0.0 -0.5 -10.0 -1.0 0 2 4 6 8 10 12 0 2 4 ωt, rad 6 8 10 12 ωt, rad 10 The power dissipated in the resistor: At any point of time, p(t) = v(t)×i(t) = VM cos (ωt)× IM cos (ωt) = VMIM cos2 (ωt) Power, W 5 0 -5 p(t) is always positive -10 0 2 4 6 8 ωt, rad 10 12 Instantaneous and Average (“rms”) powers RMS = “root mean square” or the square root of the arithmetic mean (average) 10 10 Power, W 5 Power, W 5 0 -5 0 -5 -10 -10 0 2 4 6 8 10 12 0 2 4 6 8 10 12 ωt, rad ωt, rad Instantaneous power: p(t) = v(t)×i(t) = VMIM cos2 (ωt) In terms of Joule heat dissipation, instantaneous power is equivalent to a certain DC power (i.e. timeindependent). The equivalent DC power is called Average or “rms” power Average (“rms”) power dissipated in resistor 10 10 5 Power, W Power, W 5 0 -5 -10 0 2 4 6 8 10 12 ωt, rad 0 VM I M 2 -5 -10 0 2 4 6 8 10 12 ωt, rad p(t) = v(t)×i(t) = VMIM cos2 (ωt) 10 cos2 (ωt) = (1/2) [1+ cos(2 ωt)] VM I M VM I M + cos (2ωt ) 2 2 Prms = VMIM /2 Power, W p (t ) = 5 0 -5 VM I M cos ( 2ωt ) 2 -10 0 Veff =VM / 2 I eff = I M / 2 2 4 6 8 ωt, rad 10 12 Average (RMS) characteristics of sine waves The sine voltage source with the voltage amplitude VM produces the same power as the DC battery with the voltage VDC: VDC =VM / 2 = 0.707VM The equivalent DC voltage (producing the same Joule heat as the AC voltage) is called RMS voltage (Root Mean Square) or effective value Average (RMS) characteristics of sine waves VRMS = VM 2 = 0.707VM What is the amplitude of the AC source above? VM = VRMS /0.707 = 10V/0.707 = 14.14 V AC powers in R-L-C circuits There is a phase angle between the current and voltage: i = IM cos(ωt); v = VM cos(ωt + ϕ); The instantaneous power: p(t) = v(t)×i(t) = VM cos (ωt+ϕ) × IM cos (ωt) = VMIM cos (ωt+ϕ) cos (ωt) cos(a) cos(b) = (1/2) [cos(a+b) + cos(a-b)] a = ωt+ϕ; b = ωt; p (t ) = VM I M V I cos (ϕ ) + M M cos ( 2ωt + ϕ ) 2 2 cos(a+b) = cos(a) cos(b) - sin(a) sin(b) a = ωt+ϕ; b = ωt; cos(2ωt+ϕ) = cos(2ωt) cos(ϕ) - sin(2ωt) sin(ϕ) VM I M VM I M p (t ) = cos (ϕ ) ⋅ [1 + cos(2ωt ) ] − sin(ϕ )sin ( 2ωt ) 2 2 AC powers in R-L-C circuits In R-L-C circuits, the current and voltage may both have a phase angle with respect to the source signal: i = IM cos(ωt + ϕi); v = VM cos(ωt + ϕv); We can offset the origin of the current and voltage waveforms by ϕi: i = IM cos(ωt); v = VM cos(ωt + ϕv - ϕi) = VM cos(ωt + ϕ); where ϕ = ϕv – ϕi; we can now use the previous expressions derived for ϕi = 0 p (t ) = VM I M V I cos (ϕvi ) ⋅ [1 + cos(2ωt )] − M M sin(ϕvi ) sin ( 2ωt ) 2 2 where ϕvi = ϕv – ϕi is the phase angle between the voltage and the current Active (average) and reactive powers in R-L-C circuits p (t ) = VM I M V I cos (ϕvi ) ⋅ [1 + cos(2ωt )] − M M sin(ϕvi ) sin ( 2ωt ) 2 2 10 Power, arb. units Power, arb.units 10 5 0 VM I M cos(ϕvi ) 2 -5 -10 0 2 4 6 8 10 12 ωt, rad 5 0 -5 VM I M cos(ϕvi ) cos ( 2ωt ) 2 -10 0 2 4 Active (average) power: VM I M p A (t ) = P = cos (ϕvi ) = Veff I eff cos (ϕvi ) 2 6 8 ωt, rad 10 12 Active (average) and reactive powers in R-L-C circuits p (t ) = VM I M V I cos (ϕvi ) ⋅ [1 + cos(2ωt )] − M M sin(ϕvi ) sin ( 2ωt ) 2 2 Power, arb. units 10 5 0 -5 VM I M sin(ϕ )sin ( 2ωt ) 2 -10 0 2 4 6 8 10 12 ωt, rad Reactive power: pR (t ) = Q = Average reactive power = 0 VM I M sin(ϕvi ) = Veff I eff sin(ϕvi ) 2 Active (average) and reactive powers in R-L-C circuits p (t ) = P ⋅ [1 + cos(2ωt ) ] − Q sin(ϕvi ) sin ( 2ωt ) P= VM I M cos (ϕvi ) 2 Q= Examples: 1. Resistive circuit: ϕvi = 0; P = VMIM/2; Q = 0; 2. Inductance: ϕvi = -π/2; P = 0; Q = -VMIM/2; 2. Capacitance: ϕvi = π/2; P = 0; Q = +VMIM/2; VM I M sin(ϕvi ) 2 AC powers and complex amplitudes Given the complex current amplitude: and the complex voltage amplitude: IˆM = I M e jϕi VˆM =VM e jϕv Complex Power is defined as * S =Vˆeff ⋅ Iˆeff where Veff =VM / 2; I eff = I M / 2 Iˆ *eff = I eff e − jφi is a complex conjugate of the Ieff VM I M j (ϕv −ϕi ) VM I M VM I M * ˆ ˆ S = Veff ⋅ I eff = e = cos(ϕvi ) + j sin(ϕvi ) 2 2 2 Re(S) = P Im(S) = Q

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