Ch 33. Alternating Current Circuits 33.1 AC Sources AC generator: Sinusoidal voltage (emf): = V max sin t V max Maximum voltage or valtage amplitude Angular frequency. =2 f = 2 /T (In US, f = 60 Hz and T = 1/60 seconds) 33.2 Resistors in an AC Circuit Kirchhoff's loop rule: -v R = 0 Thus, the instantaneous voltage drop across the resistor: v R = V max sin t The current: i R = v R R = V max R sin t i R = I max sin t I max = V max R • v R and i R are always in phase.
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Ch 33. Alternating Current Circuits 33.1 AC Sources AC generator: Sinusoidal voltage (emf): = Vmax sin t Vmax Maximum voltage or valtage amplitude Angular frequency.
=2 f = 2 /T
(In US, f = 60 Hz and T = 1/60 seconds)
33.2 Resistors in an AC Circuit Kirchhoff's loop rule: -vR = 0 Thus, the instantaneous voltage drop across the resistor: vR = Vmax sin t
The current: iR =vRR=VmaxRsin t
iR = Imax sin t
Imax =VmaxR
• vR and iR are always in phase.
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 • The power dissipated is time dependent:
P(t) =V (t)I(t) = RI(t)2 = RImax2 sin2 t
P is always 0, but is not a constant.
• The average power dissipated:
Pav = RI2 = R I 2( )
2= RIrms
2
Pav = RIrms2
Irms is called "the root mean square of current".
But, Pav = RImax2 sin2 t =
RImax2
2= R
Imax2
2
Thus, I rms=Imax2= 0.707Imax
For voltage: Vrms=Vmax2= 0.707Vmax
Example: 120 V ac voltage: Vrms = 120 V,
Vmax = 2Vrms = (1.414)(120V ) =169.7V
33.3 Inductors in an AC Circu
Kirchhoff's loop rule: Ldi
dt= 0
Ldi
dt= Vmax sin t
The solution: iL =VmaxLcos t =
VmaxLsin t
2
iL = Imax sin t2
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 =VmaxL=VmaxXL
XL = L "Inductive reactance". Unit has a frequency dependence.
33.4 Capacitors in an AC Circuit Kirchhoff's loop rule:
vC =Q
C= 0
Q = CVmax sin t
The current:
iC =dQ
dt= CVmax cos t = CVmax sin t + 2( )
iC = Imax sin t + 2( ) Ic and Vc are out of phase. For a sinusoidally applied emf, the current always leads the voltage across a capacitor by 90°.
Imax = CVmax =VmaxXC
XC =1
C, "Capacitive reactance", Unit: .
Summary
R circuit L circuit C circuit
v = Vmax sin t v = Vmax sin t v = Vmax sin t iR = Imax sin t iL = Imax sin t 2( ) iC = Imax sin t + 2( )
Imax =VmaxR
Imax =VmaxXL
Imax =VmaxXC
XL = L XC =1
C
Pav = RIrms
2 Pav = 0 Pav = 0
Vrms =Vmax2= 0.707Vmax
Irms =Imax2= 0.707Imax
Example: In an R circuit, v = 200 sin t and R = 100 . • Find Vrms:
Vmax = 200 V,
Vrms =Vmax2=200V
2= 141V
• Find Imax:
Imax =VmaxR
=200V
100= 2.0A
• Find Irms:
Irms =VrmsR
=141V
100= 1.41A
Example: In an L circuit, L = 25 mH, Vrms = 150 V, and f = 60Hz. (a) Find the inductive reactance XL: = 2 f = 2 (60) = 377 /s XL = L = (377 /s)(25 10-3 H) = 9.43 (b) Find rms current:
Irms =VrmsXL
=150V
9.43= 15.9A
Example: In a C circuit, C = 8 μF, f = 60 Hz, and Vrms = 150 V. • Find the capacitive reactance XC: = 2 f = 2 (60) = 377 /s XC = 1/ C = 1/(377 /s)(8 10-6 F) = 332 • Find Irms and Imax:
Irms =VrmsXC
=150V332
= 0.451A;
Imax = Irms 2 = 0.451 2 = 0.639A
• Find I(t): IC (t) = Imax sin t + 2( ) = 0.639sin 377t + 2( )
Related Documents
