Ch 33. Alternating Current Circuits 33.1 AC Sources AC generator ...

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Alternating Current Circuits. 33.1 AC Sources. AC generator: Sinusoidal voltage. (emf): ε = Vmax sin ωt. Vmax Maximum voltage or valtage amplitude ω Angular ...
Ch 33. Alternating Current Circuits 33.2 Resistors in an AC Circuit 33.1 AC Sources AC generator:

Kirchhoff's loop rule: -vR = 0 Thus, the instantaneous voltage drop across the resistor: v R = Vmax sin t

Sinusoidal voltage (emf):

The current:

iR =

vR Vmax = sin t R R

iR = Imax sin t

 = Vmax sin t Vmax Maximum voltage or valtage amplitude  Angular frequency.

 =2f = 2/T (In US, f = 60 Hz and T = 1/60 seconds)

Vmax R • vR and iR are always in phase. Imax =

Phasor Diagram: A phasor is a vector whose length is proportional to the maximum value of the variable it represents (Vmax or Imax). The phasor rotates counterclockwise at the angular speed . The projection of the phasor onto the vertical axis represents the instantaneous value of V(t) or I(t). • Example: Phasors at t = T/8 and 9T/8

2 I 2 RImax = R max   2 2

But,

2 Pav = RImax sin 2 t =

Thus,

I rms =

Imax = 0.707Imax 2

For voltage:

V rms =

Vmax = 0.707Vmax 2

Example: 120 V ac voltage: Vrms = 120 V, Vmax = 2Vrms = (1.414)(120V ) = 169.7V

• The power dissipated is time dependent: 2 P(t) = V (t)I(t) = RI(t)2 = RI max sin2 t

P is always  0, but is not a constant.

33.3 Inductors in an AC Circu Kirchhoff's loop rule:   L

L

di = Vmax sin t dt

The solution: • The average power dissipated:

( )

Pav = RI 2 = R

I2

2

2 = RIrms

2 Pav = RIrms Irms is called "the root mean square of current".

di =0 dt

 Vmax V  cos t = max sin  t   L L 2   iL = Imax sin t   2

iL = 

The current and voltage are out of phase by /2 or 90°. For a sinusoidal applied voltage, The current in an inductor always lags behind the voltage across the inductor by 90°.

Imax =

Vmax Vmax = L XL

X L = L

"Inductive reactance". Unit  has a frequency dependence.

dQ = CVmax cos t = CVmax sin (t + 2 ) dt iC = Imax sin (t + 2 ) Ic and Vc are out of phase. iC =

For a sinusoidally applied emf, the current always leads the voltage across a capacitor by 90°.

Imax = CVmax = 33.4 Capacitors in an AC Circuit Kirchhoff's loop rule: Q   vC =   = 0 C

Q = CVmax sin t The current:

XC =

Vmax XC

1 , "Capacitive reactance", Unit: . C

Summary R circuit

L circuit

v = Vmax sin t

v = Vmax sin t

iR = Imax sin t Imax =

Vmax R

2 Pav = RIrms

Imax =

Vmax XL

v = Vmax sin t iC = Imax sin (t + 2 ) Imax =

Vmax XC

1 C

X L = L

XC =

Pav = 0

Pav = 0

Vmax = 0.707Vmax 2 I = max = 0.707Imax 2

Vrms = Irms

iL = Imax sin(t  2 )

C circuit

Example: In an R circuit, v = 200 sin t and R = 100 . • Find Vrms: Vmax = 200 V, V 200V Vrms = max = = 141V 2 2 • Find Imax: V 200V Imax = max = = 2.0A R 100 • Find Irms: V 141V Irms = rms = = 1.41A 100 R Example: In an L circuit, L = 25 mH, Vrms = 150 V, and f = 60Hz. (a) Find the inductive reactance XL:  = 2f = 2 (60) = 377 /s XL = L = (377 /s)(25 10-3 H) = 9.43  (b) Find rms current: V 150V Irms = rms = = 15.9 A X L 9.43 Example: In a C circuit, C = 8 μF, f = 60 Hz, and Vrms = 150 V. • Find the capacitive reactance XC:  = 2f = 2 (60) = 377 /s XC = 1/C = 1/(377 /s)(8 10-6 F) = 332  • Find Irms and Imax: V 150V Irms = rms = = 0.451A ; XC 332 Imax = Irms 2 = 0.451 2 = 0.639A

(

)

(

• Find I(t): IC (t) = Imax sin t + 2 = 0.639sin 377t + 2

)