Alternating Current Circuits Chapter 33 continued Phasor Diagrams

Alternating Current Circuits Chapter 33 (continued)

Phasor Diagrams A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. Resistor Ip Vp Capacitor Inductor Vp Ip wt wt Vp

Reactance - Phasor Diagrams Resistor Ip Vp Capacitor Inductor Vp Ip wt wt Vp

“Impedance” of an AC Circuit R ~ L C The impedance, Z, of a circuit relates peak current to peak voltage: (Units: OHMS)

“Impedance” of an AC Circuit R ~ L C The impedance, Z, of a circuit relates peak current to peak voltage: (Units: OHMS) (This is the AC equivalent of Ohm’s law. )

Impedance of an RLC Circuit R E ~ L C As in DC circuits, we can use the loop method: E - VR - VC - VL = 0 I is same through all components.

Impedance of an RLC Circuit R E ~ L C As in DC circuits, we can use the loop method: E - VR - VC - VL = 0 I is same through all components. BUT: Voltages have different PHASES they add as PHASORS.

Phasors for a Series RLC Circuit Ip VLp VRp f (VCp- VLp) VP VCp

Phasors for a Series RLC Circuit Ip VLp VRp f (VCp- VLp) VP VCp By Pythagoras’ theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ]

Phasors for a Series RLC Circuit Ip VLp VRp f (VCp- VLp) VP VCp By Pythagoras’ theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ] = Ip 2 R 2 + (Ip XC - Ip XL) 2

Impedance of an RLC Circuit R Solve for the current: ~ L C

Impedance of an RLC Circuit R Solve for the current: Impedance: ~ L C

Impedance of an RLC Circuit The current’s magnitude depends on the driving frequency. When Z is a minimum, the current is a maximum. This happens at a resonance frequency: The circuit hits resonance when 1/w. C-w. L=0: w r=1/ When this happens the capacitor and inductor cancel each other and the circuit behaves purely resistively: IP=VP/R. IP R =10 W L=1 m. H C=10 m. F R = 1 0 0 W 0 1 0 wr 2 1 0 3 1 0 4 1 0 5 w The current dies away at both low and high frequencies.

Phase in an RLC Circuit VLp VRp (VCp- VLp) f Ip VP VCp We can also find the phase: or; or tan f = (VCp - VLp)/ VRp tan f = (XC-XL)/R. tan f = (1/w. C - w. L) / R

Phase in an RLC Circuit VLp VRp (VCp- VLp) f Ip VP VCp We can also find the phase: or; or tan f = (VCp - VLp)/ VRp tan f = (XC-XL)/R. tan f = (1/w. C - w. L) / R More generally, in terms of impedance: cos f = R/Z At resonance the phase goes to zero (when the circuit becomes purely resistive, the current and voltage are in phase).

Power in an AC Circuit V f= 0 p I 2 p wt V(t) = VP sin (wt) I(t) = IP sin (wt) (This is for a purely resistive circuit. ) P P(t) = IV = IP VP sin 2(wt) Note this oscillates twice as fast. p 2 p wt

Power in an AC Circuit The power is P=IV. Since both I and V vary in time, so does the power: P is a function of time. Use, V = VP sin (wt) and I = IP sin (w t+f ) : P(t) = Ip. Vpsin(wt) sin (w t+f ) This wiggles in time, usually very fast. What we usually care about is the time average of this: (T=1/f )

Power in an AC Circuit Now:

Power in an AC Circuit Now:

Power in an AC Circuit Now: Use: and: So

Power in an AC Circuit Now: Use: and: So which we usually write as

Power in an AC Circuit (f goes from -900 to 900, so the average power is positive) cos(f) is called the power factor. For a purely resistive circuit the power factor is 1. When R=0, cos(f)=0 (energy is traded but not dissipated). Usually the power factor depends on frequency.

Power in an AC Circuit What if f is not zero? I P V Here I and V are 900 out of phase. (f= 900) wt (It is purely reactive) The time average of P is zero.

Transformers use mutual inductance to change voltages: N 1 turns Iron Core V 1 Primary Power is conserved, though: (if 100% efficient. ) N 2 turns V 2 Secondary

Transformers & Power Transmission Transformers can be used to “step up” and “step down” voltages for power transmission. 110 turns Power =I 1 V 1=110 V 20, 000 turns V 2=20 k. V Power =I 2 V 2 We use high voltage (e. g. 365 k. V) to transmit electrical power over long distances. Why do we want to do this?

Transformers & Power Transmission Transformers can be used to “step up” and “step down” voltages, for power transmission and other applications. 110 turns Power =I 1 V 1=110 V 20, 000 turns V 2=20 k. V Power =I 2 V 2 We use high voltage (e. g. 365 k. V) to transmit electrical power over long distances. Why do we want to do this? P = I 2 R (P = power dissipation in the line - I is smaller at high voltages)