Note 11: Alternating Current (AC) Circuits

Transcription

1 Note 11: Alternating Current (AC) Circuits V R No phase difference between the voltage difference and the current and max For alternating voltage Vmax sin t, the resistor current is ir sin t. the instantaneous power is ( Vmax ) sin (W) P t irv t R 1 The time average of sin t is. Therefore, the average power is 1 ( Vmax ) 1 Pave = R( Imax ) (W) R

2 RMS Current and Voltage Root-mean-square (RMS) value of sinusoidal voltage and current are 1 1 Vrms Vmax, Irms I rms The time averaged power is Pave R( Irms ). max ( V ) R Example 1. Voltage source 00sin t (V) is connected to a 100 resistor. The rms voltage is 00 / V, the rms current is A, and the average power delivered to the reistor is Pave = VrmsIrms =00.0 W.

3 max Inductors If voltage source V sin t is connected to an inductance L, the inductor current i t can be found from di 1 t L Vmax Vmax L Vmax sin t il Vmax sin tdt cos t sin t 90 dt L L L The current lags the voltage by phase angle = 90. Since the current and voltage are out of phase by 90, the average of iv only store magnetic energy without dissipation. L L L vanishes. This is expected since inductors

4 Example. 60 Hz ac voltage source with RMS value of 150 V is connected to a 5 mh inductor. Find the RMS current. Sol. If the voltage is V sint, the current is described by rms Vrms 150 il t cost L If the current is i sin t, the volatge is 3 cost 15.9 cos t (A) di vl t L Lirms cost dt The quantity L has the same dimensions as resistance ( ) but called as inductive reactan ce. Proof of Then L rms Ohms. 1 Wb V sec, L sec A A V L Ohms A

5 RL AC Circuit To find the current in series RL circuit, we apply Kirchhoff's loop rule, Vmax sint Ri t L dt Solution can be found by assuming i t Asin t B cost di Noting Acost B sin t, we find dt V sint R Asint B cost L Acost B sint max di t This must hold at any time. Then V RA LB, RB LA 0 max V RV L V LV These yield A, B R L R R L R L R R L max max max max / R / V R L R t R L R L Vmax Vmax cos sin t sin cost sin t R L R L max i t sint cost 1 where tan. L R

6 Example 3. In the LR circuit shown, the power source is 170sin t (V), the resistance is and the inductance is mh. The frequency is 60Hz. Find the current waveform, I average power in the resistor. 3 Sol. The reactance is X L = The current is Vmax 170 V I t = sin t sin t 79.55sin t A R X.137 X R Ir ms = = 56.5 A. The average power dissipation is 1 1 where tan =tan The RMS current is P ave R I ( rms ) 56.5 W 6.33 kw. The voltage in the resisor is 56.5=11.5 V. The volatge in the inductor di t LI t 60cos t is 4.4 V. VL, rms LI rms VL L dt max cos V RI t 79.55sin t 159 sin t. R rms and the

7 EMF voltage 170sint (black) Resistor voltage 159.1sin t (red) Inductor voltage 59.96cos t (green) =0.66

8 Complex Method The equation di Ri L Vmax sin t dt can be solved using complex calculus as follows. jt jt de jt Since sin( t)= Im e, and je, the current can be found from dt jt jt V R maxe jl e Vmax i t Im Vmax Im R sint L cost R jl R L R L

9 Current in capacitor q 1 dv v t i t dt i t C CV t CV t C C dt The current leads the voltage by 90 t C 0cos 0sin 90

10 Example Hz voltatge source of 150 (V, RMS) is connected to a cpacitor of 8 F. Find the RMS current. Sol. If the voltage source is 150sin t (V), the current through the capacitor is 6 ic t CVr ms cost cost 0.45 cos t (A) 1 The quantity C has dimensions of 1/Ohms. Capacitive reactance is (Ohms) C Example 5. In the CR circuit shown, R 50 but the capacitance is unknown. When connected to 110 V (RMS), 60 Hz line, average power dissipation of 00 W is observed. Detrmine the capacitance. Sol. The RMS current can be found from R( I rms ) 00, I rms.0 A. The impedance is Z R 1 V 110 V 55 C F. rms 4 C Irms A 1 1 where =tan tan t I t t If the voltage wave is 110sin V, the current is = sin 1 RC

11 EMF voltage (black) 155.6sint Capacitor voltage (green) 64.81cos Resistor voltage (red) 141.5sin 4.6 t t

12 max LCR in Series For current I sin t, the voltages across each element are: 1 VR = RImaxsin t, VL = LI maxcos t, VC = I maxcost C The sum of all voltages is 1 R sint L cost cost I C max Using sin cos sin A t B t A B t where tan B A, we find 1 1 sin cos cos max max sin R t L t t I I R L t C C 1 L tan C, and the total voltage is R 1 Vtotal t Imax R L t C sin

13 Example 6. LCR series circuit is connected to voltage source of 150 V and f 60 Hz. When R 45, L 1.5 H, and C 3.5 F, determine the impedance and the current. 1 1 Sol. The reactance is X L = = 87 6 C (capacitive). The impedance is Z R X 51.8 Vmax The amplitude of the current is imax 0.93 A Z The phase difference between the voltage and current is =tan X 87 tan 34.0 R Power in the resistor is Ri ( rms ) W

14 Use of complex calculus AC voltage V cos t is the real part of V e V cost j sin t. jt max max max The current in an LCR circuit V R X V i t t t t R X R X R X R X can be found as max max cos sin cos jt Vm axe Vmax it Re Re cos t j sin t R jx R jx R X Vmax Re R cost X sint jr sint jx cost R X Vmax R cost X sint R X For sin voltage source, the imaginary part should be chosen.

15 Example 7. Redo Example 6 using complex analysis. R 45, X 87. If the voltage source is V t =150cos t, the current is it 150e 150 i t t j t j 45 j Re Re cos sin cost sin t 0.9 cos t 34.1 The impedance can be written as Z 45 j e 513e j34.1 j34.1 jt 150e Then the current is i t Re Re 0.9e 0.9 cos t 34.1 j e jt j34.1

16 Example 8. In the circuit, find the voltage across the capacitor. Sol. The impedance of last branch is Z 0 j10 which is parallel with the 10 resistor. The impedance is Z 1 The impedance seen by the source is Z 7 j j10 7 j9 The current is The current through the capacitor branch is C 10 0 j10 0 j10 0 j10 3 j 70 j10 7 j 30 j10 3 j j9 i j ( 5.1 ) 7 j9 7 9 i 10 3 j i j10 10 The capacitor voltage is v i j C C 1 mh 0 Ohms 10 V rad/s 10 Ohms 10 F

17 Resonance in LCR Series Circuit 1 The impedance Z R L becomes minimum when C 1 1 the reactance vanishes, X L 0, 0 C LC The current becomes maximum at the resonance frequency. Example. When L 5H, C nf in a series LCR circuit, the resonance frequency is at LC f MHz 7 10 rad/sec

18 Transformer Step-up or step-down transformer d d N v N, v N v v dt dt N1 provided the magnetic flux is confined along the iron core without leakage. The current ratio is I N I 1 1 N and power conservation holds, I v I v 1 1