2. Objectives:
Objectives: After completing this
After completing this
module, you should be able to:
module, you should be able to:
•
• Write and apply equations for calculating
Write and apply equations for calculating
the
the inductive and capacitive reactances
inductive and capacitive reactances for
for
inductors and capacitors in an ac circuit.
inductors and capacitors in an ac circuit.
•
• Describe, with diagrams and equations, the
Describe, with diagrams and equations, the
phase relationships
phase relationships for circuits containing
for circuits containing
resistance,
resistance, capacitance,
capacitance, and
and inductance
inductance.
.
•
• Describe the sinusoidal variation in
Describe the sinusoidal variation in ac
ac
current and voltage
current and voltage, and calculate their
, and calculate their
effective
effective values.
values.
3. Objectives (Cont.)
Objectives (Cont.)
•
• Write and apply equations for calculating the
Write and apply equations for calculating the
impedance
impedance, the
, the phase angle
phase angle, the
, the effective
effective
current
current, the
, the average power
average power, and the
, and the resonant
resonant
frequency
frequency for a series ac circuit.
for a series ac circuit.
•
• Describe the basic operation of a
Describe the basic operation of a step
step-
-
up
up and a
and a step
step-
-down transformer
down transformer.
.
•
• Write and apply the
Write and apply the transformer equation
transformer equation
and determine the
and determine the efficiency
efficiency of a
of a
transformer.
transformer.
4. Alternating Currents
Alternating Currents
An alternating current such as that produced
by a generator has no direction in the sense
that direct current has. The magnitudes vary
sinusoidally with time as given by:
An
An alternating current
alternating current such as that produced
such as that produced
by a generator has no direction in the sense
by a generator has no direction in the sense
that direct current has. The magnitudes vary
that direct current has. The magnitudes vary
sinusoidally
sinusoidally with time as given by:
with time as given by:
Emax
i
imax
max
time, t
E = Emax sin
i = imax sin
AC-voltage
and current
5.
450 900 1350
1800 2700 3600
E
R = Emax
E = Emax sin
Rotating Vector Description
Rotating Vector Description
The coordinate of the emf at any instant is the
value of Emax sin Observe for incremental
angles in steps of 450. Same is true for i.
The coordinate of the emf at any instant is the
value of Emax sin Observe for incremental
angles in steps of 450. Same is true for i.
450 900 1350
1800 2700 3600
E
Radius = Emax
E = Emax sin
6. Effective AC Current
Effective AC Current
i
imax
max
The average current
The average current
in a cycle is zero
in a cycle is zero—
—
half + and half
half + and half -
-.
.
But energy is expended,
But energy is expended,
regardless of direction.
regardless of direction.
So the
So the “
“root
root-
-mean
mean-
-
square
square”
” value is useful.
value is useful.
2
2 0.707
rms
I I
I
I = imax
The
The rms
rms value
value I
Irms
rms is
is
sometimes called the
sometimes called the
effective
effective current
current I
Ieff
eff:
:
The effective ac current:
ieff = 0.707 imax
7. AC Definitions
AC Definitions
One
One effective ampere
effective ampere is that ac current for
is that ac current for
which the power is the same as for one
which the power is the same as for one
ampere of dc current.
ampere of dc current.
One
One effective volt
effective volt is that ac voltage that
is that ac voltage that
gives an effective ampere through a
gives an effective ampere through a
resistance of one ohm.
resistance of one ohm.
Effective current: ieff = 0.707 imax
Effective current: ieff = 0.707 imax
Effective voltage: Veff = 0.707 Vmax
Effective voltage: Veff = 0.707 Vmax
8. Example 1:
Example 1: For a particular device, the house
For a particular device, the house
ac voltage is
ac voltage is 120
120-
-V
V and the ac current is
and the ac current is 10 A
10 A.
.
What are their
What are their maximum
maximum values?
values?
ieff = 0.707 imax
ieff = 0.707 imax Veff = 0.707 Vmax
Veff = 0.707 Vmax
max
10 A
0.707 0.707
eff
i
i max
120V
0.707 0.707
eff
V
V
imax = 14.14 A
imax = 14.14 A Vmax = 170 V
Vmax = 170 V
The ac voltage actually varies from +170 V to
-170 V and the current from 14.1 A to –14.1 A.
The ac voltage actually varies from
The ac voltage actually varies from +170 V to
+170 V to
-
-170 V
170 V and the current from
and the current from 14.1 A to
14.1 A to –
–14.1 A
14.1 A.
.
9. Pure Resistance in AC Circuits
Pure Resistance in AC Circuits
A
a.c. Source
R
V
Voltage and current are in phase, and Ohm’s
law applies for effective currents and voltages.
Voltage and current are in phase, and Ohm
Voltage and current are in phase, and Ohm’
’s
s
law applies for effective currents and voltages.
law applies for effective currents and voltages.
Ohm’s law: Veff = ieffR
Vmax
i
imax
max
Voltage
Current
10. AC and Inductors
AC and Inductors
Time, t
I
i
Current
Current
Rise
Rise
0.63I
Inductor
The voltage
The voltage V
V peaks first, causing rapid rise in
peaks first, causing rapid rise in i
i
current which then peaks as the emf goes to zero.
current which then peaks as the emf goes to zero.
Voltage
Voltage leads
leads (
(peaks before
peaks before) the current by 90
) the current by 900
0.
.
Voltage and current are out of phase
Voltage and current are out of phase.
.
Time, t
I i
Current
Current
Decay
Decay
0.37I
Inductor
11. A Pure Inductor in AC Circuit
A Pure Inductor in AC Circuit
A
L
V
a.c.
Vmax
i
imax
max
Voltage
Current
The voltage peaks 900 before the current peaks.
One builds as the other falls and vice versa.
The voltage peaks 90
The voltage peaks 900
0 before the current peaks.
before the current peaks.
One builds as the other falls and vice versa.
One builds as the other falls and vice versa.
The
The reactance
reactance may be defined as the
may be defined as the nonresistive
nonresistive
opposition
opposition to the flow of ac current.
to the flow of ac current.
12. Inductive Reactance
Inductive Reactance
A
L
V
a.c.
The
The back
back emf
emf induced
induced
by a changing current
by a changing current
provides opposition to
provides opposition to
current, called
current, called inductive
inductive
reactance X
reactance XL
L.
.
Such losses are
Such losses are temporary
temporary, however, since the
, however, since the
current
current changes direction
changes direction, periodically re
, periodically re-
-supplying
supplying
energy so that no net power is lost in one cycle.
energy so that no net power is lost in one cycle.
Inductive reactance
Inductive reactance X
XL
L is a function of both the
is a function of both the
inductance
inductance and the
and the frequency
frequency of the ac current.
of the ac current.
13. Calculating Inductive Reactance
Calculating Inductive Reactance
A
L
V
a.c.
Inductive Reactance:
2 Unitisthe
L
X fL
Ohm's law: L L
V iX
The
The voltage
voltage reading
reading V
V in the above circuit at
in the above circuit at
the instant the
the instant the ac
ac current is
current is i
i can be found from
can be found from
the
the inductance
inductance in
in H
H and the
and the frequency
frequency in
in Hz
Hz.
.
(2 )
L
V i fL
Ohm’s law: VL = ieffXL
14. Example 2:
Example 2: A coil having an inductance of
A coil having an inductance of
0.6 H
0.6 H is connected to a
is connected to a 120
120-
-V
V,
, 60 Hz
60 Hz ac
ac
source. Neglecting resistance, what is the
source. Neglecting resistance, what is the
effective current through the coil?
effective current through the coil?
A
L = 0.6 H
V
120 V, 60 Hz
Reactance: X
Reactance: XL
L =
= 2
2
fL
fL
X
XL
L =
= 2
2
(60 Hz)(0.6 H)
(60 Hz)(0.6 H)
X
XL
L = 226
= 226
120V
226
eff
eff
L
V
i
X
ieff = 0.531 A
ieff = 0.531 A
Show that the peak current is
Show that the peak current is I
Imax
max =
= 0.750
0.750 A
A
15. AC and Capacitance
AC and Capacitance
Time, t
Qmax
q
Rise in
Rise in
Charge
Charge
Capacitor
0.63 I
Time, t
I
i
Current
Current
Decay
Decay
Capacitor
0.37 I
The voltage
The voltage V
V peaks
peaks ¼
¼ of a cycle after the
of a cycle after the
current
current i
i reaches its maximum. The voltage
reaches its maximum. The voltage lags
lags
the current.
the current. Current
Current i
i and V out of phase
and V out of phase.
.
16. A Pure Capacitor in AC Circuit
A Pure Capacitor in AC Circuit
Vmax
i
imax
max
Voltage
Current
A V
a.c.
C
The voltage peaks 900 after the current peaks.
One builds as the other falls and vice versa.
The voltage peaks 90
The voltage peaks 900
0 after
after the current peaks.
the current peaks.
One builds as the other falls and vice versa.
One builds as the other falls and vice versa.
The diminishing current i builds charge on C
which increases the back emf of VC.
The diminishing current
The diminishing current i
i builds charge on
builds charge on C
C
which increases the
which increases the back emf
back emf of
of V
VC
C.
.
17. Capacitive Reactance
Capacitive Reactance
No
No net power
net power is lost in a complete cycle, even
is lost in a complete cycle, even
though the capacitor does provide nonresistive
though the capacitor does provide nonresistive
opposition (
opposition (reactance
reactance) to the flow of ac current.
) to the flow of ac current.
Capacitive reactance
Capacitive reactance X
XC
C is affected by both the
is affected by both the
capacitance
capacitance and the
and the frequency
frequency of the ac current.
of the ac current.
A V
a.c.
C
Energy
Energy gains and losses
gains and losses
are also
are also temporary
temporary for
for
capacitors due to the
capacitors due to the
constantly changing ac
constantly changing ac
current.
current.
18. Calculating Inductive Reactance
Calculating Inductive Reactance
Capacitive Reactance:
1
Unitisthe
2
C
X
fC
Ohm's law: VC C
iX
The
The voltage
voltage reading
reading V
V in the above circuit at
in the above circuit at
the instant the
the instant the ac
ac current is
current is i
i can be found from
can be found from
the
the inductance
inductance in
in F
F and the
and the frequency
frequency in
in Hz
Hz.
.
2
L
i
V
fL
A V
a.c.
C
Ohm’s law: VC = ieffXC
19. Example 3:
Example 3: A 2
A 2-
-
F capacitor is connected to
F capacitor is connected to
a 120
a 120-
-V, 60 Hz ac source. Neglecting
V, 60 Hz ac source. Neglecting
resistance, what is the effective current
resistance, what is the effective current
through the coil?
through the coil?
Reactance:
Reactance:
X
XC
C = 1330
= 1330
120V
1330
eff
eff
C
V
i
X
ieff = 90.5 mA
ieff = 90.5 mA
Show that the peak current is
Show that the peak current is i
imax
max =
= 128 mA
128 mA
A V
C = 2 F
120 V, 60 Hz
1
2
C
X
fC
-6
1
2 (60Hz)(2 x 10 F)
C
X
20. Memory Aid for AC Elements
Memory Aid for AC Elements
An
An old
old, but very
, but very
effective, way to
effective, way to
remember the
remember the phase
phase
differences
differences for
for inductors
inductors
and
and capacitors
capacitors is :
is :
“
“E
E L I
L I”
” the
the “
“i
i C
C E
E”
” Man
Man
Emf E is before current i in inductors L;
Emf E is after current i in capacitors C.
Emf
Emf E
E is
is before
before current
current i
i in inductors
in inductors L
L;
;
Emf
Emf E
E is
is after
after current
current i
i in capacitors
in capacitors C.
C.
“E
E L
L i”
“I C E
E”
man
the
21. Frequency and AC Circuits
Frequency and AC Circuits
f
f
R, X
R, X
1
2
C
X
fC
2
L
X fL
Resistance
Resistance R
R is constant and not affected by
is constant and not affected by f.
f.
Inductive reactance
Inductive reactance X
XL
L
varies directly with
varies directly with
frequency as expected
frequency as expected
since
since E
E
i/
i/
t
t.
.
Capacitive reactance
Capacitive reactance X
XC
C varies
varies
inversely
inversely with
with f
f since rapid ac
since rapid ac
allows little time for charge to
allows little time for charge to
build up on capacitors.
build up on capacitors.
R
R
X
XL
L
X
XC
C
22. Series LRC Circuits
Series LRC Circuits
L
VR VC
C
R
a.c.
VL
VT
A
Series ac circuit
Consider an inductor L, a capacitor C, and
a resistor R all connected in series with an
ac source. The instantaneous current and
voltages can be measured with meters.
Consider an
Consider an inductor
inductor L
L,
, a
a capacitor
capacitor C
C,
, and
and
a
a resistor
resistor R
R all connected in
all connected in series
series with
with an
an
ac source
ac source. The instantaneous current and
. The instantaneous current and
voltages can be measured with meters.
voltages can be measured with meters.
23. Phase in a Series AC Circuit
Phase in a Series AC Circuit
The voltage
The voltage leads
leads current in an inductor and
current in an inductor and lags
lags
current in a capacitor.
current in a capacitor. In phase
In phase for resistance
for resistance R
R.
.
450 900 1350
1800 2700 3600
V V = Vmax sin
VR
VC
VL
Rotating
Rotating phasor
phasor diagram
diagram generates voltage waves
generates voltage waves
for each element
for each element R
R,
, L
L, and
, and C
C showing phase
showing phase
relations. Current
relations. Current i
i is always
is always in phase
in phase with
with V
VR.
R.
24. Phasors
Phasors and Voltage
and Voltage
At time t = 0, suppose we read
At time t = 0, suppose we read V
VL
L,
, V
VR
R and
and V
VC
C for an
for an
ac series circuit. What is the source voltage
ac series circuit. What is the source voltage V
VT
T?
?
We handle phase differences by finding the
We handle phase differences by finding the
vector sum
vector sum of these readings.
of these readings. V
VT
T =
=
V
Vi
i.
. The
The
angle
angle
is the
is the phase angle
phase angle for the ac circuit.
for the ac circuit.
VR
VL - VC
V
VT
T
Source voltage
Source voltage
VR
VC
VL
Phasor
Phasor
Diagram
Diagram
25. Calculating Total Source Voltage
Calculating Total Source Voltage
VR
VL - VC
V
VT
T
Source voltage
Source voltage Treating as vectors, we find:
Treating as vectors, we find:
2 2
( )
T R L C
V V V V
tan L C
R
V V
V
Now recall that:
Now recall that: V
VR
R =
= iR
iR;
; V
VL
L =
= iX
iXL
L;
; and
and V
VC
C =
= iV
iVC
C
Substitution into the above voltage equation gives:
Substitution into the above voltage equation gives:
2 2
( )
T L C
V i R X X
26. Impedance in an AC Circuit
Impedance in an AC Circuit
R
XL - XC
Z
Z
Impedance
Impedance 2 2
( )
T L C
V i R X X
Impedance
Impedance Z
Z is defined:
is defined:
2 2
( )
L C
Z R X X
Ohm
Ohm’
’s law for ac current
s law for ac current
and impedance:
and impedance:
or T
T
V
V iZ i
Z
The impedance is the combined opposition to ac
current consisting of both resistance and reactance.
The impedance is the combined opposition to ac
current consisting of both resistance and reactance.
27. Example 3:
Example 3: A
A 60
60-
-
resistor, a
resistor, a 0.5 H
0.5 H inductor, and
inductor, and
an
an 8
8-
-
F
F capacitor are connected in series with a
capacitor are connected in series with a
120
120-
-V, 60 Hz
V, 60 Hz ac source. Calculate the impedance
ac source. Calculate the impedance
for this circuit.
for this circuit.
A
60 Hz
0.5 H
60
120 V
8 F
1
2 and
2
L C
X fL X
fC
2 (60Hz)(0.6 H) = 226
L
X
-6
1
332
2 (60Hz)(8 x 10 F)
C
X
2 2 2 2
( ) (60 ) (226 332 )
L C
Z R X X
Thus, the impedance is:
Thus, the impedance is: Z = 122
28. Example 4:
Example 4: Find the effective current and the
Find the effective current and the
phase angle for the previous example.
phase angle for the previous example.
A
60 Hz
0.5 H
60
120 V
8 F
X
XL
L =
= 226
226
X
XC
C =
= 332
332
R =
R = 60
60
Z =
Z = 122
122
120 V
122
T
eff
V
i
Z
ieff = 0.985 A
ieff = 0.985 A
Next we find the
Next we find the phase angle
phase angle:
:
R
XL - XC
Z
Z
Impedance
Impedance
X
XL
L –
– X
XC
C = 226
= 226 –
– 332 =
332 = -
-106
106
R = 60
R = 60
tan L C
X X
R
Continued . . .
Continued . . .
29. Example 4 (Cont.):
Example 4 (Cont.): Find the
Find the phase angle
phase angle
for
for
the previous example.
the previous example.
-106
60
Z
Z
X
XL
L –
– X
XC
C = 226
= 226 –
– 332 =
332 = -
-106
106
R = 60
R = 60
tan L C
X X
R
106
tan
60
= -60.50
= -60.50
The negative phase angle means that the ac
voltage lags the current by 60.50. This is
known as a capacitive circuit.
The
The negative
negative phase angle means that the ac
phase angle means that the ac
voltage
voltage lags
lags the current by 60.5
the current by 60.50
0. This is
. This is
known as a
known as a capacitive
capacitive circuit.
circuit.
30. Resonant Frequency
Resonant Frequency
Because inductance causes the voltage to lead
the current and capacitance causes it to lag the
current, they tend to cancel each other out.
Because
Because inductance
inductance causes the voltage to
causes the voltage to lead
lead
the current and
the current and capacitance
capacitance causes it to
causes it to lag
lag the
the
current, they tend to
current, they tend to cancel
cancel each other out.
each other out.
Resonance
Resonance (Maximum Power)
(Maximum Power)
occurs when X
occurs when XL
L = X
= XC
C
R
XC
XL X
XL
L =
= X
XC
C
2 2
( )
L C
Z R X X R
1
2
2
fL
fC
1
2
r
f
LC
Resonant
Resonant f
fr
r
X
XL
L = X
= XC
C
31. Example 5:
Example 5: Find the resonant frequency for the
Find the resonant frequency for the
previous circuit example: L = .5 H, C = 8
previous circuit example: L = .5 H, C = 8
F
F
1
2
r
f
LC
-6
1
2 (0.5H)(8x10 F
f
Resonant fr = 79.6 Hz
Resonant fr = 79.6 Hz
At resonant frequency, there is zero reactance (only
resistance) and the circuit has a phase angle of zero.
At resonant frequency, there is zero reactance (
At resonant frequency, there is zero reactance (only
only
resistance
resistance) and the circuit has a phase angle of zero.
) and the circuit has a phase angle of zero.
A
? Hz
0.5 H
60
120 V
8 F
Resonance XL = XC
32. Power in an AC Circuit
Power in an AC Circuit
No power is consumed by inductance or
capacitance. Thus power is a function of the
component of the impedance along resistance:
No power is consumed by inductance or
No power is consumed by inductance or
capacitance. Thus power is a function of the
capacitance. Thus power is a function of the
component of the impedance along resistance:
component of the impedance along resistance:
In terms of ac voltage:
In terms of ac voltage:
P = iV cos
P = iV cos
In terms of the resistance R:
In terms of the resistance R:
P = i2R
P = i2R
R
XL - XC
Z
Z
Impedance
Impedance
P
P lost in
lost in R
R only
only
The fraction
The fraction Cos
Cos
is known as the
is known as the power factor.
power factor.
33. Example 6:
Example 6: What is the average power loss for
What is the average power loss for
the previous example:
the previous example: V
V = 120 V,
= 120 V,
=
= -
-60.5
60.50
0,
,
i
i = 90.5 A, and R = 60
= 90.5 A, and R = 60
.
.
The higher the power factor, the more
efficient is the circuit in its use of ac power.
The
The higher
higher the power factor, the more
the power factor, the more
efficient
efficient is the circuit in its use of ac power.
is the circuit in its use of ac power.
A
? Hz
0.5 H
60
120 V
8 F
Resonance XL = XC
P = i
P = i2
2R
R = (0.0905 A)
= (0.0905 A)2
2(60
(60
Average P = 0.491 W
Average P = 0.491 W
The power factor is:
The power factor is: Cos 60.5
Cos 60.50
0
Cos = 0.492 or 49.2%
Cos = 0.492 or 49.2%
34. The Transformer
The Transformer
A
A transformer
transformer is a device that uses induction
is a device that uses induction
and ac current to step voltages up or down.
and ac current to step voltages up or down.
R
a.c.
Np Ns
Transformer
P P
N
t
E S S
N
t
E
Induced
emf’s are:
Induced
emf’s are:
An ac source of emf
Ep is connected to
primary coil with Np
turns. Secondary has
Ns turns and emf of
Es.
An ac source of emf
An ac source of emf
E
Ep
p is connected to
is connected to
primary coil with
primary coil with N
Np
p
turns. Secondary has
turns. Secondary has
N
Ns
s turns and emf of
turns and emf of
E
Es
s.
.
35. Transformers (Continued):
Transformers (Continued):
R
a.c.
Np Ns
Transformer
P P
N
t
E
S S
N
t
E
Recognizing that
Recognizing that
/
/
t
t is the same in each coil,
is the same in each coil,
we divide first relation by second and obtain:
we divide first relation by second and obtain:
The transformer
equation:
The transformer
equation:
P P
S S
N
N
E
E
36. Example 7:
Example 7: A generator produces 10 A at
A generator produces 10 A at
600 V. The primary coil in a transformer has
600 V. The primary coil in a transformer has
20 turns. How many secondary turns are
20 turns. How many secondary turns are
needed to step up the voltage to 2400 V?
needed to step up the voltage to 2400 V?
R
a.c.
Np Ns
I = 10 A; Vp = 600 V
20
turns
P P
S S
V N
V N
Applying the
Applying the
transformer equation:
transformer equation:
(20)(2400V)
600V
P S
S
P
N V
N
V
NS = 80 turns
NS = 80 turns
This is a step-up transformer; reversing coils
will make it a step-down transformer.
This is a
This is a step
step-
-up transformer
up transformer; reversing coils
; reversing coils
will make it a step
will make it a step-
-down transformer.
down transformer.
37. Transformer Efficiency
Transformer Efficiency
There is no power gain in stepping up the voltage
There is no power gain in stepping up the voltage
since voltage is increased by reducing current. In
since voltage is increased by reducing current. In
an ideal transformer with no internal losses:
an ideal transformer with no internal losses:
or S
P
P P S S
s P
i
i i
i
E
E E
E
An ideal
An ideal
transformer:
transformer:
R
a.c.
Np Ns
Ideal Transformer
The above equation assumes no internal energy
losses due to heat or flux changes. Actual
efficiencies are usually between 90 and 100%.
The above equation assumes no internal energy
The above equation assumes no internal energy
losses due to heat or flux changes.
losses due to heat or flux changes. Actual
Actual
efficiencies
efficiencies are usually between
are usually between 90 and 100%.
90 and 100%.
38. Example 7:
Example 7: The transformer in
The transformer in Ex. 6
Ex. 6 is
is
connected to a power line whose resistance
connected to a power line whose resistance
is
is 12
12
. How much of the power is lost in
. How much of the power is lost in
the transmission line?
the transmission line?
V
VS
S = 2400
= 2400 V
V
R
a.c.
Np Ns
I = 10 A; Vp = 600 V
20
turns
12
P P
P P S S S
S
i
i i i
E
E E
E
(600V)(10A)
2.50 A
2400V
S
i
P
Plost
lost = i
= i2
2R
R = (2.50 A)
= (2.50 A)2
2(12
(12
)
) P
Plost
lost = 75.0 W
= 75.0 W
P
Pin
in = (600 V)(10 A) = 6000 W
= (600 V)(10 A) = 6000 W
%Power Lost = (75 W/6000 W)(100%) = 1.25%
%Power Lost = (75 W/6000 W)(100%) = 1.25%
%Power Lost = (75 W/6000 W)(100%) = 1.25%
39. Summary
Summary
Effective current: ieff = 0.707 imax
Effective current: ieff = 0.707 imax
Effective voltage: Veff = 0.707 Vmax
Effective voltage: Veff = 0.707 Vmax
Inductive Reactance:
2 Unitisthe
L
X fL
Ohm's law: L L
V iX
Capacitive Reactance:
1
Unitisthe
2
C
X
fC
Ohm's law: VC C
iX
40. Summary (Cont.)
Summary (Cont.)
2 2
( )
T R L C
V V V V
tan L C
R
V V
V
2 2
( )
L C
Z R X X
or T
T
V
V iZ i
Z
tan L C
X X
R
1
2
r
f
LC
41. Summary (Cont.)
Summary (Cont.)
In terms of ac voltage:
In terms of ac voltage:
P = iV cos
P = iV cos
In terms of the resistance R:
In terms of the resistance R:
P = i2R
P = i2R
Power in AC Circuits:
Power in AC Circuits:
P P
S S
N
N
E
E P P S S
i i
E E
Transformers:
Transformers:
