Download Lab 40 ELECTRICAL TRANSFORMERS

LAB 40 ELECTRICAL TRANSFORMERS
Friday, April 18, 2014
OVERVIEW
An electrical transformer is used to increase or decrease AC (alternating current) voltage.
It consists of two coils of wire sharing an iron core; the iron core improves the magnetic link
between the coils. AC voltage is applied to the input (or primary) coil. The back and forth
sloshing of the AC current produces an alternating magnetic field, which causes the electrons in
the output (or secondary coil) to slosh back and forth. This process, called electromagnetic
induction, allows the primary AC power to be transmitted to the secondary coil without any wire
connections. Under ideal conditions, all of the electrical power in the primary coil would be
transmitted to the secondary coil.
Power is proportional to ΔV/N, where N is the number of turns in the coil. If the secondary coil
has more turns than the primary coil, the secondary voltage has to be higher too. This
configuration (𝑁𝑜𝑢𝑡 > 𝑁𝑖𝑛 ) is called a step-up transformer. Since 𝑃 = 𝛥𝑉 • 𝐼, the increase in
output voltage means a decrease in output current (as ΔV↑, I↓). If the primary coil has more turns
than the secondary (𝑁𝑜𝑢𝑡 > 𝑁𝑖𝑛 ), the transformer is called a step-down transformer. Since the
secondary has fewer turns, it will have a lower voltage and therefore a higher current (as ΔV↑,
I↓).The transformer "steps down" in voltage.
The ideal mechanical advantage (IMA) of a transformer is the ratio of the number of turns in the
output coil to the number of turns in the input coil (𝑁𝑜𝑢𝑡 > 𝑁𝑖𝑛 ). Like all IMA's, it is the
mechanical advantage that could be obtained if the efficiency (η) was 100% and it can be
determined without actually operating the system:
𝑁𝑜𝑢𝑡
𝐼𝑀𝐴 =
𝑁𝑖𝑛
The actual mechanical advantage (AMA) is the mechanical advantage you actually get from the
transformer. This is the ratio of the output voltage to the input voltage:
𝛥𝑉𝑜𝑢𝑡
𝐴𝑀𝐴 =
𝛥𝑉𝑖𝑛
One measure of the efficiency of the electrical transformer is the ratio of the actual mechanical
advantage to the ideal mechanical advantage:
𝐴𝑀𝐴
𝜂=
• 100%
𝐼𝑀𝐴
Another measure of the efficiency (the one you'll calculate) is the ratio of the output power to the
input power:
𝑃𝑜𝑢𝑡
𝜂=
𝑃𝑖𝑛
A segment of the core of the transformer you will use is removable. When it is removed, it will
reduce the magnetic field transmitted from the primary coil to the secondary. This means less
power will be transmitted to the secondary coil and the efficiency will drop.
In the experiment you will find the power in a transformer when it is operated in both directions,
with and without the core segment attached. You will also find the IMA and efficiency (η) of each
construction.
1
LAB 40 ELECTRICAL TRANSFORMERS
Friday, April 18, 2014
OBJECTIVES
A. Use a transformer to step-down and step-up voltage.
B. Find the IMA and the AMA of a transformer.
C. Find the input power, the output power and the efficiency of a simple transformer.
EQUIPMENT REQUIRED
Transformer Assembly
Resistors (lΏ, 10Ώ, 100Ώ)
AC/DC Power Supply
Universal Lead Set
Circuit Panel and Easel
Digital Multimeter
PROCEDURE
A. Step-Down Transformer
1. Mount the transformer assembly in the center of the circuit panel. Orient the transformer so
the coils are as shown in Figure 3 (coil B on the left, coil A on the right). Check to make sure the
panel pinholes are not damaged to ensure proper connections.
2. Using you lead set, connect the circuit in Figure 3 using a 1 Ώ resistor on the primary side and
a 10 Ώ resistor on the secondary side. Use the 0-24 VAC setting on the power supply. Connect
your leads to the power supply's AC inputs, the two yellow jacks on the right side of the panel.
Be sure than coil B is the primary coil and coil A is the secondary coil. This arrangement will
give you a step-down transformer. Place the core segments on the transformer as in Figure 2
(don't worry that the Figure has the coils on the wrong sides).
3. Turn on the power supply and adjust the output to 5 VAC. Using the digital multimeter
(DMM), measure the voltages across the transformer primary and secondary coils. To measure
voltage (jumpers installed):
- Turn DMM OFF. Select the 20V (~) scale or set the AC/DC to AC.
- Place meter probe jacks in "common" and "V-Ώ".
- Place meter probe tips at "V" points in the figure below on the primary or secondary
sides. Turn the DMM 𝑂𝑁, lower the voltage scale to obtain the optimum reading
accuracy. Record the AC voltage.
- Record the primary and secondary voltages in Data Table 1.
- Turn the DMM OFF before removing the meter probes.
4. Using the DMM, measure the current in the primary and secondary circuits. Current must be
measured in series with the leads in the DMM current inputs. To measure the AC current
(jumper removed):
- With the DMM OFF, select 2A (~) or set the AC/DC to AC.
Place the meter probe jacks in "common" and "A".
Replace the jumper with the meter probes with the wire extenders (the "A" points) as shown in
the Figure below. This puts the DMM in series with the circuit.
Turn the DMM on and lower the scale to obtain the most accurate reading. Record the primary
and secondary currents in Data Table 1.Turn the DMM OFF before removing the meter leads.
2
LAB 40 ELECTRICAL TRANSFORMERS
Friday, April 18, 2014
5. Remove the core and repeat the procedures described in steps 3 and 4.
6. Turn the voltage control on the power supply all the way down and turn off the power supply.
Replace the core segment on the transformer.
B. Step-Up Transformer
1. Wire the circuit as shown in Figure 5. Now coil A is the primary coil and coil B is the
secondary coil. This is a step-up transformer.
2. Replace the 10 Ώ resistor on the output side with the 100 Ώ resistor.
3. Make sure the voltage control is turned completely counterclockwise before turning on the
power supply. Turn on the power supply and adjust it for 5 volts.
4. Measure the voltage across the transformer primary and secondary coils using the DMM.
Record your measurements in Data Table 1.
5. Turn off the power supply. See if you can get your calculations completed before you take
apart your lab set-up, just in case you made some measurement mistakes.
C. Calculations
1. Calculate the ratio of the number of output to input turns for the step-up transformer and the
step-down transformer. Coil A has either 300 turns or 200 turns (check the coil!) and coil B has
600 turns. Enter your answers in Data Table
2. The ideal mechanical advantage (IMA) is given by
𝑁𝑜𝑢𝑡
𝐼𝑀𝐴 =
𝑁𝑖𝑛
3. Calculate the power in the primary and secondary coils using the equations
𝑃𝑖𝑛 = 𝛥𝑉𝑖𝑛 • 𝐼𝑖𝑛
𝑃𝑜𝑢𝑡 = 𝛥𝑉𝑜𝑢𝑡 • 𝐼𝑜𝑢𝑡
Record the values in Data Table 2. Remember to convert your values for current from milliamps
(mA) to amps (A) before solving for power (1 mA = 1 x 10−3 A).
4. Calculate the efficiencies (η) using the relation
𝑃𝑜𝑢𝑡
𝜂=
𝑃𝑖𝑛
5. Put away your lab set-up and complete the analysis portion of the lab report.
3
LAB 40 ELECTRICAL TRANSFORMERS
Friday, April 18, 2014
OBJECTIVES:
SKETCH:
DATA TABLE 1
Transformer Condition
Primary
Coil
Secondary
Coil
Core
Segment
B
A
In Place
B
A
Removed
A
B
In Place
A
B
Removed
Voltage
V (V)
Primary
A-B
Secondary
C-D
Current
I (mA)
Primary
A-B
Secondary
C-D
DATA TABLE 2
Transformer Condition
Primary
Coil
Secondary
Coil
Core
Segment
B
A
In Place
B
A
Removed
A
B
In Place
A
B
Removed
Power
Primary
Secondary
P (Watts)
IMA
Efficiency
4
LAB 40 ELECTRICAL TRANSFORMERS
Friday, April 18, 2014
LAB 40 ANALYSIS
1. Which transformer arrangement had the greatest efficiency?
2. What effect does the removable core segment have on the efficiency of the transformer?
3. What happens to the energy that is lost between the primary and secondary sides of the
transformer?
4. Suppose a transformer has 150 turns in the primary and 90 turns in the secondary. What is
the IMA for this transformer? (Show your calculation)
5. Is the transformer in question 4 a step-up or step-down transformer? Explain your reason.
6. Suppose the power input of the transformer in question 4 is 300 watts and it is 75% efficient.
What is the power output? Show your calculation.
7. A transformer has 4 turns in the primary and 8 turns in the secondary. The input voltage is 3
V. What would be the maximum voltage in the secondary coil?
_____________________________
5