# I t - Purdue University ```PHYS 219
Spring semester 2016
Lecture 16:
ac Voltages, ac currents and
Transformers
Ron Reifenberger
Birck Nanotechnology Center
Purdue University
Exam III
When: Wednesday, May 4, 2016
7:00-9:00 PM
Where: PHYS 112
Lecture 16
V(t)
AC Generators
T1
f1 =
1
1
T1
time, t(s)
T2
if T2= &frac12;T1,
then f2 = 2f1
T3
Sinusoidal
waveforms
Notation
t

V  t  = Vmax sin 2πft  = Vmax sin  2π  = Vmax sin  ωt 
T

OR
t

V  t  = Vo sin 2πft  = Vo sin  2π  = Vo sin  ωt 
T

if T1 = 0.0167 s
1
= 60 s-1 = 60Hz
0.0167
ω = 2π radians &times; 60 s-1 = 377 rad2 / s
f1 =
1
Resistors in AC Circuits
• Assume a circuit consisting of an
AC generator and a resistor
• The voltage across the output of
the AC source varies with time
according to
V(t) = Vmax sin (2 π ƒ t)
– Vmax (or Vo) is the amplitude
of the AC voltage
– V(t) is the instantaneous
potential difference
V(t1) = Vmax sin (2 π ƒ t1)
t1
3
What is the Current?
• Applying Ohm’s Law:
I =V
Voltage applied
R
• Since the voltage varies
sinusoidally, so does the current
I=
or
Vmax
sin 2πƒt 
R
Response
I = Imax sin 2πƒt 
where
Imax =
Vmax
R
I(t) is in-phase with V(t)
4
2
to the time-domain representation of the waveform.
voltage phasor
Θ=ωt
Projection of
vector onto y-axis
• The voltage in an AC circuit can be represented by a voltage phasor
• The current in an AC circuit can also be represented by a current phasor
• The voltage and current phasors always make the same angle with the horizontal
axis as time passes
• For a circuit with only resistors. the current and voltage are in phase
5
Phasor Simulation
Θ=2 π ƒ t
Instantaneous voltage
Θ
http://www.jhu.edu/signals/phasorapplet2/phasorappletindex.htm
6
3
Specifying the rms value (effective average) of a sine wave
V  t  = Vmax sin 2πft 
• What is the average voltage of an
AC source?
• Is there a better way?
• To specify voltage values varying
sinusoidally with time, the rms value
• “rms” stands for Root Mean Square:
Find average (mean) value of V2 and
take square root
2
V 2  t  = Vmax
sin 2πft 
2
• For a sinusoidally varying voltage
Vrms =
=
V 
2
ave

V   V   ....  V 
2
1
2
2
2
N
N
 1 2  Vmax
= 0.707 Vmax
 Vmax  =
2
2

t1 t2 t3
.
.
. tN
7
Power lost in a resistor
• The instantaneous power is the
product of the instantaneous
voltage and instantaneous current
P = I V
• Since both I and V vary with time,
the power also varies with time:
P = Vmax Imax sin2 (2πƒt)
• The instantaneous power varies
between 0 and Vmax Imax
• The average power is &frac12; the
maximum power
Pave = &frac12; (VmaxImax )
 V  I
=  max   max
 2  2

 = VrmsIrms

• Ohm’s Law can be used
to express the average
power in different ways
Pave =
2
Vrms
2
= Irms
R
R
8
4
Example: ac circuit analysis
Vmax = 170 V
f=60 Hz
R=20 Ω
What are Vrms, Irms, and average power dissipated in load?
What are Vmax, Imax, and maximum power dissipated in load?
Average values
Vrms =
Irms =
Vmax
2
=
170V
= 120V
1.414
Maximum values
Vmax = 170V
Vrms 120V
=
=6A
R
20 Ω
Imax =
Pave = IrmsVrms =  6 A120V  = 720W
Vmax 170V
=
= 8.5 A
R
20 Ω
Pmax = Imax Vmax =  8.5 A170V  = 1445W
9
An ac isolator

B (t )
I p (t )
I s (t )
Primary
Secondary
The Secondary is electrically isolated from the Primary
Input: ac +dc signal
I p (t )
Output: ac signal ONLY
Ioffset
I s (t )
Air coupling not very
efficient
10
5
A Practical Transformer: Trapping
Magnetic Flux with an Iron Core
iron
core
Coil 2
Coil 1
11
An Iron-Core Transformer
Np=N1
Primary
(Coil 1)
Ns=N2
Secondary
(Coil 2)
12
6
Electric transformer: Primary and Secondary Coils
Resistive
Input voltage must
vary with time
Important
equations for
an ideal
transformer
ΔV1 N 1
=
ΔV2 N 2
Pin = Pout
Both input voltage
and input current can
be “transformed”
&quot; output &quot;
I1 ΔV1 = I2 ΔV2
I2 ΔV1 N 1
=
=
I1 ΔV2 N 2
I2 N 1
=
I1 N 2
I2 =
&quot;input&quot;
N1
N2
ΔV2 =
I1
N2
ΔV1
N1
13
schematic diagrams
Primary
Vmax
8 turns
Secondary
&frac12;Vmax
1 V ac
4 turns
14
7
EXAMPLE
A transformer has 10 turns on the primary and 100 turns on the
secondary. If 110V rms is applied to the primary, what is the
peak-to-peak voltage across the secondary?
ΔV1 N 1
=
ΔV2 N 2
110V rms
ΔV2 rms
=
V2
10
100


ΔV2 rms = 10 110V rms = 1100V rms
Vrms =
Vp to p
Vmax
Vp to p
Vmax
 Vmax = 1555V
2
= 2Vmax = 3111 V
15
If a 100 Ω load is connected to the secondary, what is the rms
current in the primary?
ΔV1 N 1 I2
=
=
ΔV2 N 2 I1
110V rms
ΔV2 rms
110 V
100 Ω
10
=
100


ΔV2 rms = 10 110V rms = 1100V rms
I2 rms =
ΔV2 rms
N 1 I2
=
N 2 I1

=
1100V rms
100 Ω
= 11A rms
10 11A
=
100 I1
 100 
I1 rms = 
 11A = 110A
 10 
16
8
Up Next – Optics and Optical Phenomena
17
APPENDIX: Transformers in the US Power Grid
Why is power transmitted at such high voltages?
Vrms = 345, 000V
Typical values
Irms = 1000A
PTransmitted = Vrms Irms = 345, 000V 1000A  = 345MW
SUPPOSE 10% of power is lost
PLost = 10% PTransmitted = 34.5MW
PLost = I2rms Rline
Rline = 34.5 Ω
Now suppose Vrms is decreased by 9/10 and
Irms is increased by 10/9, so the average power
transmitted remains the same:
9
&times; 345, 000V  = 310,500 V
10
10
Irms = &times; 1000A  =1111A
9
PTransmitted = VrmsIrms = 310,500 V 1111A 
Vrms =
= 345MW (same as before)
How much power is now lost?
Rline = 34.5 Ω (same as before)
PLost = I2rmsRline = 1111A 34.5 Ω  = 42.6MW
2
PLost
42.6MW
=
= power lost increases by 12.3%
PTransmitted 345MW
18