Transcript Coulomb’s Law - University of Delaware

vC  VC sin t
Capacitor Load
qC  CvC
dqC
iC 
 CV cos t
dt
V
iC  C sin t  90  I C sin t  90
1
C
1
X

C
The capacitive reactance of a capacitor
C
VC
I

Generalized Ohm’s law: C
Notice Ic and VC are amplitudes
XC
ICE
vL  VL sin t
Inductive Load
vL  L
diL
dt
VL
 VL 
sin

dt



 cos t

L
 L 
V
iL  L sin t  90  I L sin t  90
L
iL 
The Inductive reactance of a inductive X L  L
Generalized Ohm’s law: I L  VL
XL
Notice IL and VL are amplitudes
ELI
The series RLC circuit
   m sin  d t
1. Same current through R, L, C
i  I sin  d t   
Same frequency as in the source
I
i
vR
VR
t-
VL
vC
VC
2. Consider VR, VC, VL
VR  iR  IR sin  d t   

I

vC  VC sin  d t    , VC  IX C 
2
d C



vL  VL sin   d t    , VL  IX L   d LI
2

  vR  vC  vL
The series RLC circuit: Continuous
  vR  vC  vL
Values at t.
This relation has to be maintained
as phosors are rotating
 m  VR  VL  VC
General rules:
1. KVL and KCL still hold, but
values at the same t have to
be used, i.e. vertical
components in phasor
diagram.
2. Vectors operation for
Amplitude
v
V
The series RLC circuit: Continuous
=IZ
 m  VR  VL  VC
m 
I
2
VL  VC 
 VR
2
X L  X C 
 R2
2
=IR
=I(XL-XC)
2
1 

2
 I  L 
 R
C 

I

m

R
C
2
L  1
tan  
L  1C
R

2
m
Z
Z is the impedance of the circuit
X L  XC

R
Examples
33-43P. A coil of inductance 88 mH
and unknown resistance and a 0.94 mF
are connected in series with an
alternating emf of frequency 930 Hz.
If the phase constant between the
applied voltage and current is 75,
what is the resistance of the coil.
f=930 Hz
d=2f
=IZ
=IR
=I(XL-XC)
=IZ
RLC Resonance
tan  
L  1C
R
=IR
X L  XC

R
XL>XC: inductive loading
=I(XL-XC)
XL=XC: Resonance
XC>XL: Capacitive loading
RLC Resonance: Cont
tan  
L  1C
R
X L  XC
 d L  1 C
d
d 
I max 
1
LC
m
R
X L  XC

R
I
m
L  1C   R
2
2
Conditions at Resonance
I

=IZ
m
 X L  X C 2  R 2
=IR
m
L  1C   R
2
2
• I is a maximum
=I(XL-XC)
• Z is at minimum; Z=R; Z is purely resistive
• XL=XC; inductive reactance cancels capacitive reactance;
net reactance is zero
• The phase angle is zero; the current is perfectly in phase
with applied emf; the tangent of the phase angle is zero.
• The driven frequency is identical to the natural frequency.
• The power factor is unity