Alternating Current Circuits Chapter 33 (continued)

Phasor Diagrams V Rp IpIp  t V Cp IpIp  t V Lp IpIp  t ResistorCapacitor Inductor 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. The “y component” is the actual current or voltage. 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. The “y component” is the actual current or voltage.

Impedance in AC Circuits The impedance Z of a circuit or circuit element relates peak current to peak voltage: (Units: Ohms) V R ~ C L (This is the AC equivalent of Ohm’s law.)

Phasor Diagrams Circuit elementImpedanceAmplitudePhase ResistorRV R = I P RI, V in phase Capacitor X c =1/  C V C =I P X c I leads V by 90° Inductor XL=LXL=L V L =I P X c I lags V by 90° V Rp IpIp  t V Cp IpIp  t V Lp IpIp  t ResistorCapacitor Inductor

RLC Circuit Use the loop method: V - V R - V C - V L = 0 I is same through all components. V R ~ C L BUT: Voltages have different PHASES  they add as PHASORS.

RLC Circuit IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

RLC Circuit By Pythagoras’s theorem: (V P ) 2 = [ (V Rp ) 2 + (V Cp - V Lp ) 2 ] = I p 2 R 2 + (I p X C - I p X L ) 2 IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

RLC Circuit Solve for the current: V R ~ C L

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

The circuit hits resonance when 1/  C-  L=0:  r =1/ When this happens the capacitor and inductor cancel each other and the circuit behaves purely resistively: I P =V P /R. 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 current dies away at both low and high frequencies. rr L=1mH C=10  F

Phase in an RLC Circuit IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp We can also find the phase: tan  = (V Cp - V Lp )/ V Rp = (X C -X L )/R = (1/  C -  L) / R

Phase in an RLC Circuit At resonance the phase goes to zero (when the circuit becomes purely resistive, the current and voltage are in phase). IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp We can also find the phase: tan  = (V Cp - V Lp )/ V Rp = (X C -X L )/R = (1/  C -  L) / R More generally, in terms of impedance: cos  R/Z

The power dissipated in an AC circuit is P=IV. Since both I and V vary in time, so does the power: P is a function of time. Power in an AC Circuit Use V = V P sin (  t) and I = I P sin (  t+  ) : P(t) = I p V p sin(  t) sin (  t+  ) 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 Use: and: So Now:

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

Power in an AC Circuit  goes from to 90 0, so the average power is positive) cos(  is called the power factor. For a purely resistive circuit the power factor is 1. When R=0, cos(  )=0 (energy is traded but not dissipated). Usually the power factor depends on frequency, and usually 0<cos(  )<1.

Power in a purely resistive circuit V(t) = V P sin (  t) I(t) = I P sin (  t) P(t) = IV = I P V P sin 2 (  t) Note this oscillates twice as fast. V tt   I tt  P  = 0 (This is for a purely resistive circuit.)

Power in a purely reactive circuit The opposite limit is a purely reactive circuit, with R=0. I V P This happens with an LC circuit. Then  90 0 The time average of P is zero. tt

Transformers Transformers use mutual inductance to change voltages: Primary (applied voltage) Secondary (produced voltage) N s turns VpVp VsVs N p turns Iron Core Faraday’s law on the left: If the flux per turn is  then V p =N p (d  /dt). Faraday’s law on the right: The flux per turn is also , so V s =N s (d  /dt). 

Transformers Transformers use mutual inductance to change voltages: Primary (applied voltage) Secondary (produced voltage) N s turns VpVp VsVs N p turns Iron Core In the ideal case, no power is dissipated in the transformer itself. Then I p V p =I s V s 

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

Transformers & Power Transmission 20,000 turns V 1 =110V V 2 =20kV 110 turns Transformers can be used to “step up” and “step down” voltages, for power transmission and other applications. Power =I 1 V 1 Power =I 2 V 2 We use high voltage (e.g. 365 kV) 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)