If an {\it isolation transformer} (a transformer with the same number of ``turns’’ in the primary and secondary coils) is connected between an AC source and an AC load, we will measure the same voltage and the same current at both source and load terminals:
If we calculate power output by the source and power dissipated by the load, the value is the same: 420 Watts ($P = IV$).
Now suppose we analyze a circuit containing a {\it step-up} transformer (one with more turns of wire in the secondary coil than in the primary coil). With a step-up transformer, the load voltage will be greater than the supply voltage. In this example, I show a step-up transformer with a 1:2 step ratio:
Assuming the load resistance is completely different from the first (isolation transformer) circuit, what can you deduce about the load current and the power (both source and load) in this circuit? Is the load current less than the source current? Is the load current greater than the source current? Is the load power greater than the source power? Explain your answers.
The basic physical law known as {\it The Law of Conservation of Energy} tells us that power cannot come from nowhere, or disappear into nowhere. If the power source is sending 420 watts into the transformer, then the load must be receiving 420 watts (neglecting any inefficiencies internal to the transformer). The transformer’s step ratio is completely irrelevant as far as {\it power} is concerned!
For a real transformer (with less than 100
The only reason I hesitate to tell students they can calculate load current {\it precisely} is because it was not stated whether or not the transformer is ``lossy’’ at all. No real transformer is 100
I have found that the Conservation of Energy approach not only makes sense to students as they learn to calculate transformer behavior, but it is an excellent reinforcement of a basic physical law, a good understanding of which will serve them well throughout their careers.
Industrial {\it control power transformers} are used to step down 480 or 240 volts to a level more acceptable for relay control circuitry: usually 120 volts. Some control power transformers are built with multiple primary windings, to facilitate connection to either a 480 volt or 240 volt AC power source:
Such transformers are usually advertised as having ``240 $\times$ 480’’ primary windings, the ``$\times$’’ symbol representing two independent windings with four connection points (H1 through H4).
Show the connections on the four ``H’’ terminals necessary for 240 volt operation, and also for 480 volt operation, on the following illustrations:
This type of transformer is {\it very} common in industrial control systems. Discuss with your students why the primary winding terminals are arranged as they are (H1-H3-H2-H4), to facilitate near-terminal jumpering with metal clips.
Calculate all listed values for this transformer circuit:
{\bullet} $V_{primary} = $
{\bullet} $V_{secondary} = $
{\bullet} $I_{primary} = $
{\bullet} $I_{secondary} = $
Explain whether this is a {\it step-up}, {\it step-down}, or {\it isolation} transformer, and also explain what distinguishes the ``primary’’ winding from the ``secondary’’ winding in any transformer.
{\bullet} $V_{primary} = 48 \hbox{ volts}$
{\bullet} $V_{secondary} = 14.77 \hbox{ volts}$
{\bullet} $I_{primary} = 30.3 \hbox{ mA}$
{\bullet} $I_{secondary} = 98.5 \hbox{ mA}$
This is a {\it step-down} transformer.
Most transformer problems are nothing more than ratios, but some students find ratios difficult to handle. Questions such as this are great for having students come up to the board in the front of the classroom and demonstrating how they obtained the results.
Calculate the load current and load voltage in this transformer circuit:
$I_{load}$ = {hskip 80pt}$V_{load}$ =
$I_{load}$ = 23.77 mA {hskip 80pt}$V_{load}$ = 8.318 V
Most transformer problems are nothing more than ratios, but some students find ratios difficult to handle. Questions such as this are great for having students come up to the board in the front of the classroom and demonstrating how they obtained the results.
Calculate the source current and load current in this transformer circuit:
$I_{source}$ = {hskip 80pt}$I_{load}$ =
$I_{source}$ = 187.5 mA {hskip 80pt}$I_{load}$ = 72.73 mA
Most transformer problems are nothing more than ratios, but some students find ratios difficult to handle. Questions such as this are great for having students come up to the board in the front of the classroom and demonstrating how they obtained the results.
Calculate all voltages and all currents in this circuit, given the component values and the number of turns in each of the transformer’s windings:
$V_R =$ 750 V
$I_R =$ 340.9 mA
$V_{source} =$ 50 V
$I_{source} =$ 5.114 A
This question checks students’ ability to relate the winding ratio to voltage and current ratios in a transformer circuit. The symbolism here is common in Europe, but not so common in the United States.
Calculate all voltages and all currents in this transformer circuit, assuming the 170 ohm resistor carries a current of 5.8 mA:
{\bullet} $V_{primary}$ =
{\bullet} $V_{secondary}$ =
{\bullet} $I_{primary}$ =
{\bullet} $I_{secondary}$ =
{\bullet} $V_{primary}$ = 21.04 volts
{\bullet} $V_{secondary}$ = 3.045 volts
{\bullet} $I_{primary}$ = 839.5 microamps
{\bullet} $I_{secondary}$ = 5.8 milliamps
Most transformer problems are nothing more than ratios, but some students find ratios difficult to handle. Questions such as this are great for having students come up to the board in the front of the classroom and demonstrating how they obtained the results.
Calculate all voltages and all currents in this transformer circuit, assuming the 5 ohm resistor carries a current of 10 amps:
{\bullet} $V_{primary}$ =
{\bullet} $V_{secondary}$ =
{\bullet} $I_{primary}$ =
{\bullet} $I_{secondary}$ =
{\bullet} $V_{primary}$ = 80.56 volts
{\bullet} $V_{secondary}$ = 50 volts
{\bullet} $I_{primary}$ = 13.97 amps
{\bullet} $I_{secondary}$ = 22.5 amps
Most transformer problems are nothing more than ratios, but some students find ratios difficult to handle. Questions such as this are great for having students come up to the board in the front of the classroom and demonstrating how they obtained the results.
Calculate all voltages and all currents in this transformer circuit, assuming the 3.3 k$\Omega$ resistor drops 40 volts:
{\bullet} $V_{source}$ =
{\bullet} $V_{primary}$ =
{\bullet} $V_{secondary}$ =
{\bullet} $I_{source}$ =
{\bullet} $I_{primary}$ =
{\bullet} $I_{secondary}$ =
{\bullet} $V_{source}$ = 18.84 V
{\bullet} $V_{primary}$ = 18.84 V
{\bullet} $V_{secondary}$ = 46.06 V
{\bullet} $I_{source}$ = 76.74 mA
{\bullet} $I_{primary}$ = 29.63 mA
{\bullet} $I_{secondary}$ = 12.12 mA
Most transformer problems are nothing more than ratios, but some students find ratios difficult to handle. Questions such as this are great for having students come up to the board in the front of the classroom and demonstrating how they obtained the results.
Calculate all voltages and all currents in this transformer circuit, assuming the 3.3 k$\Omega$ resistor drops 13 volts:
{\bullet} $V_{source}$ =
{\bullet} $V_{primary}$ =
{\bullet} $V_{secondary}$ =
{\bullet} $I_{source}$ =
{\bullet} $I_{primary}$ =
{\bullet} $I_{secondary}$ =
{\bullet} $V_{source}$ = 5.32 V
{\bullet} $V_{primary}$ = 4.763 V
{\bullet} $V_{secondary}$ = 14.97 V
{\bullet} $I_{source}$ = 12.38 mA
{\bullet} $I_{primary}$ = 12.38 mA
{\bullet} $I_{secondary}$ = 3.939 mA
Suppose a power system were delivering AC power to a resistive load drawing 150 amps:
Calculate the load voltage, load power dissipation, the power dissipated by the wire resistance ($R_{wire}$), and the overall power efficiency ($\eta = {P_{load} \over P_{source}}$).
$E_{load} = $
$P_{load} = $
$P_{lines} = $
$\eta = $
Now, suppose we were to use a pair of perfectly efficient 10:1 transformers to step the voltage up for transmission, and back down again for use at the load. Re-calculate the load voltage, load power, wasted power, and overall efficiency of this system:
$E_{load} = $
$P_{load} = $
$P_{lines} = $
$\eta = $
Simple system (no transformers):
$E_{load} = 210$ volts
$P_{load} = 31.5$ kW
$P_{lines} = 4.5$ kW
$\eta = 87.5
Complex system (with transformers):
$E_{load} = 239.7$ volts
$P_{load} = 35.96$ kW
$P_{lines} = 45$ W
$\eta = 99.88
An example like this usually clarifies the benefits of using AC instead of DC for transmission of large amounts of electrical power over substantial distances, better than simply telling students why transformers are used in power systems. Even with modest power losses in the transformers (say, 3
In discussing the follow-up question, be sure to bring up safety as a consideration if none of your students do.
How much current will be output by a current transformer if the load current is 350 amps and the CT ratio is 600:5?
Secondary current = 2.917 amps
This question is an exercise in mathematical ratios.
Calculate the primary winding current (magnitude and phase angle) for this resistively loaded isolation transformer, with primary and secondary inductances of 18 Henrys each:
Also, draw an equivalent schematic diagram (with no transformer in it) illustrating the impedance ``seen’’ by the AC power source. Assume no winding resistance in either transformer winding, and a magnetic coupling coefficient between the two windings of exactly 1.
$I_{primary} = 1.2001 \hbox{ A} \> \angle -0.84^o$
This question illustrates how reflected load impedance is ``seen’’ by the source, and how it interacts with the transformer’s intrinsic winding impedance.
Published under the terms and conditions of the Creative Commons Attribution License
by Jeff Kerns