Chapter 21: RLC Circuits
[Pages:33]Chapter 21: RLC Circuits
PHY2054: Chapter 21
1
Voltage and Current in RLC Circuits
?AC emf source: "driving frequency" f
= m sint
= 2 f
?If circuit contains only R + emf source, current is simple
i
=
R
=
Im
sin (t )
Im
=
m
R
(current amplitude)
?If L and/or C present, current is not in phase with emf
i = Im sin (t - )
Im
=
m
Z
?Z, shown later
PHY2054: Chapter 21
2
AC Source and Resistor Only
?Driving voltage is = m sint
?Relation of current and voltage
i = /R
i = Im sint
Im
=
m
R
i
~
R
Current is in phase with voltage ( = 0)
PHY2054: Chapter 21
3
AC Source and Capacitor Only
?Voltage is
vC
=
q C
=
m
sin t
?Differentiate to find current
q = Cm sint
i
i = dq / dt = CVC cost
~
C
?Rewrite using phase (check this!)
i = CVC sin (t + 90?)
?Relation of current and voltage
i = Im sin (t + 90?)
Im
=
m
XC
( XC = 1/C)
?"Capacitive reactance": XC = 1/C
Current "leads" voltage by 90?
PHY2054: Chapter 21
4
AC Source and Inductor Only
?Voltage is vL = Ldi / dt = m sint
?Integrate di/dt to find current:
di / dt = (m / L)sint i = -(m /L)cost
?Rewrite using phase (check this!)
i = (m /L)sin (t - 90?)
i
~
L
?Relation of current and voltage
i = Im sin (t - 90?)
Im
=
m
XL
(XL =L)
?"Inductive reactance": X L = L
Current "lags" voltage by 90?
PHY2054: Chapter 21
5
General Solution for RLC Circuit
?We assume steady state solution of form i = Im sin (t - )
Im is current amplitude
is phase by which current "lags" the driving EMF
Must determine Im and
?Plug in solution: differentiate & integrate sin(t-)
i = Im sin (t - )
di dt
=
Im
cos (t
-)
Substitute
L
di dt
+
Ri
+
q C
=
m
sin t
q = - Im cos(t - )
I
m
L
cos
(t
-
)
+
Im
R
sin
(t
-
)
-
Im
C
cos
(t
-
)
=
m
sin
t
PHY2054: Chapter 21
6
General Solution for RLC Circuit (2)
Im
L
cos
(t
-
)
+
Im
R
sin
(t
-
)
-
Im
C
cos
(t
-
)
=
m
sin
t
?Expand sin & cos expressions
sin (t - ) = sint cos - cost sin cos(t - ) = cost cos + sint sin
High school trig!
?Collect sint & cost terms separately
(L -1/C )cos - R sin = 0 Im (L -1/C )sin + ImR cos = m
cost terms sint terms
?These equations can be solved for Im and (next slide)
PHY2054: Chapter 21
7
General Solution for RLC Circuit (3)
?Solve for and Im
tan = L -1/C X L - XC
R
R
Im
=
m
Z
?R, XL, XC and Z have dimensions of resistance
XL =L XC = 1/C
Inductive "reactance" Capacitive "reactance"
Z = R2 + ( X L - XC )2
Total "impedance"
?This is where , XL, XC and Z come from!
PHY2054: Chapter 21
8
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