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
