4. ELEMENTARY A.C. CIRCUITS
Main things to learn
e
~
R
•
•
•
•
Resistive, inductive and capacitive circuits
Reactance and impedance
Phase difference between current and voltage
Resonance
e
~
e = E m sin ω t
We assume that the frequency
C
e
~
i =?
f =ω / 2π is small, i.e.,
voltage and current change slowly
(for frequency below 20 MHz, it is still “slowly”)
Therefore, at every moment of time Kirchhoff’s rules can be applied.
L
RESISTIVE CIRCUIT
i
A
e
~
e = E m sin ω t
u R = u AB = e = U m sin ω t , U m = E m
R
iR = u R
B
uR U m
Um
=
sin ω t = I m sin ω t , I m =
i=
R
R
R
Resistance R acts in the same way as for direct current. Ohm’s law is valid
uR
t
u = U m sin ω t
i = I m sin ω t
iR
t
Phase difference is zero
Current and voltage are in-phase
CAPACITIVE CIRCUIT
CAPACITIVE REACTANCE
i
e
+q
−q
~
e = E m sin ω t
A
uC = u AB = e = U m sin ω t , U m = E m
C
dq
q = Cu AB ; i =
dt
q = CU m sin ω t
B
i = ω CU m cos ω t = I m sin(ω t + π ) , I m = ω CU m
2
In the form similar to the Ohm’s law this looks like
Im
Um
1
=
, where X C =
is equivalent to resistance
XC
ωC
XC is called capacitive reactance
1
XC is proportional to
ω
Units: Ohm [Ω]
XC is proportional to
1
C
u
PHASE DIFFERENCE IN A CAPACITIVE CIRCUIT
t
u = U m sin ω t
t
i = I m sin(ω t + π )
2
i
There is a phase difference between voltage and current
Current is ahead of voltage (current leads voltage)
Voltage is delayed as to current (voltage lags current)
by π
/2 - or 90o - or quarter-cycle
INDUCTIVE CIRCUIT
i
i
A
e
~
L
A
is equivalent to
e
B
~
~
ei
B
e = E m sin ω t
u AB = e = U m sin ω t , U m = E m
ei - EMF of self-induction
di
(current derivative)
e i = −L
dt
2nd Kirchhoff’s law
e + ei = 0
di
U m sin ω t − L = 0
dt
INDUCTIVE REACTANCE
di U m
=
sin ω t
dt
L
Um  1
Um
Um

sin ω t dt =
cos ω t =
i=∫
 − cos ω t  = −
ωL
L
L  ω

Um
Um
π
π


=
sin ω t −  = I m sin ω t −  , where I m =
2
2
ωL 
ωL

In the form similar to the Ohm’s law this looks like
Im
Um
=
, where X L = ω L is equivalent to resistance
XL
XL is called inductive reactance
XL is proportional to ω
Units: Ohm [Ω]
XL is proportional to L
u
PHASE DIFFERENCE IN AN INDUCTIVE CIRCUIT
t
u = U m sin ω t
t
i = I m sin(ω t − π )
2
i
There is a phase difference between voltage and current
Current is behind voltage (current lags voltage)
Voltage is ahead of current (voltage leads current)
by π
/2 - or 90o - or quarter-cycle
EXAMPLE: EFFECT OF THE PHASE DIFFERENCE
A
C
L
R
e
B
LRC are connected in series
~
How to analyse this circuit?
1st Kirchhoff’s law:
Current is the same
uC = I m X C sin(ω t − π 2)
i = I m sin ω t
where
XC = 1
where
XL = ω L
ωC
u R = I m R sin ω t
u L = I m X L sin(ω t + π 2)
2nd Kirchhoff’s law:
e = u AB
u AB
u AB = uC + u R + u L
= I m [X C sin(ω t − π 2) + R sin ω t + X L sin(ω t + π 2)]
IMPEDANCE
It is possible to show that uAB can be written as
u AB = I m Z sin(ω t + ϕ )
1 

Z = R 2 + ω L −

ωC 

2
1
X =ωL−
ωC
and
where
1

ω L −
ωC
ϕ = tan −1
R



- total reactance
ϕ - phase difference between voltage and current
Z - equivalent to resistance - impedance
Impedance = Resistance 2 + Reactance 2
Note that capacitive and inductive reactances are
not added but subtracted from each other






EXAMPLE: LC IN SERIES
C
L
e
~
R =0 ∴ Z = X
i = I m sin ω t
u = I m [X C sin(ω t − π 2) + X L sin(ω t + π 2)] =
= I m ( X L − X C ) sin(ω t + π 2)
• It is possible that both XL and XC are rather large
with Z
•
= X = XL - XC being much smaller
UL and UC may be much larger than Em
1
• Everything is frequency dependent: Z = ω L −
ωC
1
At ω =
Z = 0 !!!!
LC
At given voltage, current becomes very large - resonance