This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Physics 212 LCR Circuits
10/28/2009 Fall 2009 Summary
AC Voltage RMS (RootMeanSquare) VRMS, IRMS V = V0 sin(ωt ) ω = 2πf € f= 1
T ω= € IRMS = 2π
T IPVP = ISVS Magnetic ﬂux change is the same
due to iron core N
VS = S V p
NP IS = IP NS>NP  step up; NS<NP  step down € € I0
2 keep power constant € € 12
PAVG = I0 R
2 Average Power VP
N
= IP P
VS
NS Step up the voltage
and
step down the current Transformer The ac adapter for a laptop computer contains a transformer. The input of the adapter is
the 120 volts from the ac wall outlet. The output from the transformer is 20 volts.
What is the turns ratio of the transformer?
a) 0.17
b) 6
c) 100
d) This cannot be determined without knowing how many turns one of
the coils in the transformer has. What is a Phasor?
Vector that rotates about the origin  represents oscillating quantities
Angular speed  rotation speed is equal to oscillation angular frequency ω
Length  size of the vector represents the amplitude of the alternating quantity
Projection  projection onto vertical axis is the instantaneous value at time (t)
Rotation Angle  Phase of the quantity at time (t) Resistive Load
Peak Current and Reactance
Voltage across R
I0 I0 = v R = V0 sin(ωt ) V0 V0
R ⇒ V0 = I0 R Current
€ €
I0 v
V
iR = R = 0 sin(ωt )
RR € Time varying vR or iR can be
depicted by projection onto yaxis of rotating phasor V0
Current and voltage are “in phase” Peak current is not frequency dependent Capacitative Reactance
Voltage across C
vC = V0 sin(ωt )
I0
V0 Peak Current and Reactance
I0 = CV0ω ⇒ V0 = Charge and Current φ = 90° = π 2 rad € € XC = q = CvC = CV0 sin(ωt ) iC = CV0ω sin(ωt + 90°) 1
ωC V0 = I0 X C dq
d sin(ωt )
i=
= CV0
= CV0ω cos(ωt )
dt
dt € I0
ωC € € I0
V0 € Time varying vC or iC can be
depicted by projection onto yaxis of rotating phasor
Current “leads” the voltage Peak current is frequency dependent
Circuit “reacts” differently for low and high f Example Consider a circuit with a purely capacitative load where C=15.0 μF and the sinusoidal emf device
operates at amplitude of V0=36.0 V and f=60.0 Hz, a) what is the potential difference across C and
what is the current in the circuit?
Reactance
Angular Frequency
1
1
XC =
=
ω = 2πf = 2π (60 s−1 ) = 120π
ωC 2πfC
Voltage
vC ( t ) = ( 36.0 V ) sin(120πt ) XC = Units € I0 = Current V0
36 V
=
= 0.203 A
X C 177 Ω iC ( t ) = (0.203 A) sin(120πt + π 2)
€ € € 1
= 177 Ω
2π (60 s )(15 E − 6 F ) XC = −1 1
1
V
V
=
=
= =Ω
1C C
( s )( F )
A
s
sV
−1 Inductive Reactance
Voltage across L Peak Current and Reactance v L = V0 sin(ωt ) Current φ = −90° = −π 2 rad V0
I0 diL
= V0 sin(ωt )
dt
1
V
iL = ∫ V0 sin(ωt ) dt = − 0 cos(ωt )
L
ωL
V
iL = 0 sin(ωt − 90°)
ωL € € V0 €
I0 I0 = Current “lags” the voltage ⇒ V0 = I0ωL
X L = ωL L Time varying vL or iL can be
depicted by projection onto yaxis of rotating phasor V0
ωL V0 = I0 X L € Peak current is frequency dependent
Circuit “reacts” differently for low and high f Example Consider a circuit with a purely inductive load where L=230 mH and the sinusoidal emf device
operates at amplitude of V0=36.0 V and f=60.0 Hz, a) what is the potential difference across L and
what is the current in the circuit?
Reactance
Angular Frequency
X L = ωL = 2πfL ω = 2πf = 2π (60 s−1 ) = 120π Voltage
v L ( t ) = ( 36.0 V ) sin(120πt ) X L = 2π (60 s−1 )(230 E − 3 H ) = 86.7 Ω
Units 1 Tm 2 1 N m 2 Nm J 1 V
X L = ( € )( H ) =
s
=
=
=
= =Ω
sA
s C m A CA C A A
s
−1 € Current I0 = V0
36. V
=
= 0.415 A
X L 86.7 Ω €
iL ( t ) = (0.415 A) sin(120πt − π 2)
€ € RCL Series Circuit
V = εm sin(ωt ) Applied emf i = I0 sin(ωt + φ ) Solve for amplitude I0 and phase ϕ Use phasors to simplify the analysis
€ AC source creates
oscillating current Individual voltage
drops across
components Use vector addition
to ﬁnd magnitude of
applied EMF Inductive Reactance
Four circuits are set up so that each contains an ac source and an inductor. Circuit A has a
50Hz source and a 15H inductor. Circuit B has a 60Hz source and a 10H inductor. Circuit
C has a 100Hz source and a 2H inductor. Circuit D has a 50Hz source and a 4H inductor.
Rank the inductive reactance for these circuits in order, from largest to smallest.
a) A > B > C > D
b) A > B > C = D
c) D > C > A > B
d) B > A > C > D Next Time LCR Oscillations ...
View
Full
Document
This note was uploaded on 10/03/2011 for the course PHYS 212 taught by Professor Tedeschi during the Spring '08 term at South Carolina.
 Spring '08
 TEDESCHI
 Physics

Click to edit the document details