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PHYS 219 Spring semester 2016 Lecture 16: ac Voltages, ac currents and Transformers Ron Reifenberger Birck Nanotechnology Center Purdue University Exam III When: Wednesday, May 4, 2016 7:00-9:00 PM Where: PHYS 112 Lecture 16 V(t) AC Generators T1 f1 = 1 1 T1 time, t(s) T2 if T2= ½T1, then f2 = 2f1 T3 Sinusoidal waveforms Notation t V t = Vmax sin 2πft = Vmax sin 2π = Vmax sin ωt T OR t V t = Vo sin 2πft = Vo sin 2π = Vo sin ωt T if T1 = 0.0167 s 1 = 60 s-1 = 60Hz 0.0167 ω = 2π radians × 60 s-1 = 377 rad2 / s f1 = 1 Resistors in AC Circuits • Assume a circuit consisting of an AC generator and a resistor • The voltage across the output of the AC source varies with time according to V(t) = Vmax sin (2 π ƒ t) – Vmax (or Vo) is the amplitude of the AC voltage – V(t) is the instantaneous potential difference V(t1) = Vmax sin (2 π ƒ t1) t1 3 What is the Current? • Applying Ohm’s Law: I =V Voltage applied R • Since the voltage varies sinusoidally, so does the current I= or Vmax sin 2πƒt R Response I = Imax sin 2πƒt where Imax = Vmax R I(t) is in-phase with V(t) 4 2 Phasors: linking a rotating vector to the time-domain representation of the waveform. voltage phasor Θ=ωt Projection of vector onto y-axis • The voltage in an AC circuit can be represented by a voltage phasor • The current in an AC circuit can also be represented by a current phasor • The voltage and current phasors always make the same angle with the horizontal axis as time passes • For a circuit with only resistors. the current and voltage are in phase 5 Phasor Simulation Θ=2 π ƒ t Instantaneous voltage Θ http://www.jhu.edu/signals/phasorapplet2/phasorappletindex.htm http://www.animations.physics.unsw.edu.au/jw/phasor-addition.html 6 3 Specifying the rms value (effective average) of a sine wave V t = Vmax sin 2πft • What is the average voltage of an AC source? • Is there a better way? • To specify voltage values varying sinusoidally with time, the rms value has been universally adopted • “rms” stands for Root Mean Square: Find average (mean) value of V2 and take square root 2 V 2 t = Vmax sin 2πft 2 • For a sinusoidally varying voltage Vrms = = V 2 ave V V .... V 2 1 2 2 2 N N 1 2 Vmax = 0.707 Vmax Vmax = 2 2 t1 t2 t3 . . . tN 7 Power lost in a resistor • The instantaneous power is the product of the instantaneous voltage and instantaneous current P = I V • Since both I and V vary with time, the power also varies with time: P = Vmax Imax sin2 (2πƒt) • The instantaneous power varies between 0 and Vmax Imax • The average power is ½ the maximum power Pave = ½ (VmaxImax ) V I = max max 2 2 = VrmsIrms • Ohm’s Law can be used to express the average power in different ways Pave = 2 Vrms 2 = Irms R R 8 4 Example: ac circuit analysis Vmax = 170 V f=60 Hz R=20 Ω What are Vrms, Irms, and average power dissipated in load? What are Vmax, Imax, and maximum power dissipated in load? Average values Vrms = Irms = Vmax 2 = 170V = 120V 1.414 Maximum values Vmax = 170V Vrms 120V = =6A R 20 Ω Imax = Pave = IrmsVrms = 6 A120V = 720W Vmax 170V = = 8.5 A R 20 Ω Pmax = Imax Vmax = 8.5 A170V = 1445W 9 An ac isolator B (t ) I p (t ) I s (t ) Primary Secondary The Secondary is electrically isolated from the Primary Input: ac +dc signal I p (t ) Output: ac signal ONLY Ioffset I s (t ) Air coupling not very efficient 10 5 A Practical Transformer: Trapping Magnetic Flux with an Iron Core iron core Coil 2 Coil 1 11 An Iron-Core Transformer Np=N1 Primary (Coil 1) Ns=N2 Secondary (Coil 2) 12 6 Electric transformer: Primary and Secondary Coils Resistive Load Input voltage must vary with time Important equations for an ideal transformer ΔV1 N 1 = ΔV2 N 2 Pin = Pout Both input voltage and input current can be “transformed” " output " I1 ΔV1 = I2 ΔV2 I2 ΔV1 N 1 = = I1 ΔV2 N 2 I2 N 1 = I1 N 2 I2 = "input" N1 N2 ΔV2 = I1 N2 ΔV1 N1 13 Adjusting the Output Voltage schematic diagrams Primary Vmax 8 turns Secondary ½Vmax 1 V ac 4 turns 14 7 EXAMPLE A transformer has 10 turns on the primary and 100 turns on the secondary. If 110V rms is applied to the primary, what is the peak-to-peak voltage across the secondary? ΔV1 N 1 = ΔV2 N 2 110V rms ΔV2 rms = V2 10 100 ΔV2 rms = 10 110V rms = 1100V rms Vrms = Vp to p Vmax Vp to p Vmax Vmax = 1555V 2 = 2Vmax = 3111 V 15 If a 100 Ω load is connected to the secondary, what is the rms current in the primary? ΔV1 N 1 I2 = = ΔV2 N 2 I1 110V rms ΔV2 rms 110 V 100 Ω 10 = 100 ΔV2 rms = 10 110V rms = 1100V rms I2 rms = ΔV2 rms N 1 I2 = N 2 I1 Rload = 1100V rms 100 Ω = 11A rms 10 11A = 100 I1 100 I1 rms = 11A = 110A 10 16 8 Up Next – Optics and Optical Phenomena 17 APPENDIX: Transformers in the US Power Grid Why is power transmitted at such high voltages? Vrms = 345, 000V Typical values Irms = 1000A PTransmitted = Vrms Irms = 345, 000V 1000A = 345MW SUPPOSE 10% of power is lost PLost = 10% PTransmitted = 34.5MW PLost = I2rms Rline Rline = 34.5 Ω Now suppose Vrms is decreased by 9/10 and Irms is increased by 10/9, so the average power transmitted remains the same: 9 × 345, 000V = 310,500 V 10 10 Irms = × 1000A =1111A 9 PTransmitted = VrmsIrms = 310,500 V 1111A Vrms = = 345MW (same as before) How much power is now lost? Rline = 34.5 Ω (same as before) PLost = I2rmsRline = 1111A 34.5 Ω = 42.6MW 2 PLost 42.6MW = = power lost increases by 12.3% PTransmitted 345MW 18 http://adogreen.com/blog/power-generation/overview-from-power-plant-to-consumer/ 9