Transformer Circuit Calculations

The basic physical law known as 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 power is concerned!

If the coils are to be considered a resistors, since the same series/parallel equations apply to both, then we could consider how high and low voltage might affect resistors.

With a 240 volt supply, this (relatively) low voltage can be dropped across both inductors, and placing them in parallel provides more energy and less loss, while inducing a voltage on the secondary coil.

In the case of (relatively) high 480 volts, the coils placed in series will allow both coils to drop 240 volts, exactly the same as before. Therefore, in both cases, two coils are dropping 240 volts, and the resulting voltage on the secondary coil will be the same.

To begin, we immediately know the voltage applied to the primary coil from the basic information provided.

{\bullet} $V_{primary} = 48~volts$

To identify the secondary voltage, the turns ratio of the transformer is 13000:4000, so the secondary will be lower voltage by a 13:4 ratio.

$${V_{primary} \over V_{secondary}} = {13 \over 4}$$
{\bullet} $V_{secondary} = 14.77~volts$

The secondary current is the next simple calculation, based on Ohm’s Law:

$$I_{secondary}={V_{secondary} \over R_{secondary}}$$
{\bullet} $I_{secondary} = 98.5~mA$

Finally, the current in the primary is found be either letting the power (V x I) of the secondary equal the power of the primary, or use the same turns ratio equations for voltage, but reverse the ratio, as current and turns re inversely proportional.
{\bullet} $I_{primary} = 30.3~mA$

This is a step-down transformer.

The primary winding is connected to the source. There is no limitation on which side should have more or less windings, so it is unwise to assume that the higher turns indicate the primary side.

The first order of business is to determine the secondary (load) voltage by dividing the source voltage by the turns ratio.

$$Turns~Ratio={2390 \over 710}=3.366$$

$$V_{load}={V_{source} \over 3.366$$

$V_{load} = 8.318~V$

The load current is a simple application of Ohm’s Law:

$$I_{load}={V_{load} \over R_{load}}$$

$I_{load} = 23.77~mA$

First, find the load current using the exact same method as the previous problem.

One important detail to note is the turns ratio, which is:

$$Turns~Ratio={1400 \over 3610}=0.388$$

This means when you divide the source voltage by the turns ratio, the load voltage will be higher. This is a step-up transformer.

$I_{load} = 72.73~mA$

Since this is a step up transformer, and the change in voltage and current are inversely related, a higher load voltage means we will actually have a smaller load current.

Calculate the power consumed by the load, and that will equal the power produced by the source.

$$P= V_{load} \cdot I_{load}=V_{source} \cdot I_{source}$$

And then solve for $I_{source}$

$I_{source} = 187.5~mA$

One obvious known property is the source voltage.

$V_{source} = 50~V$

This is a step-up transformer, with a turns ratio of 300:4500 (notice the primary is on the right side, along with the source).

Divide source voltage by turns ratio to determine the resistor’s voltage

$V_R = 750~V$

Ohm’s Law to determine the resistor’s current:

$I_R = 340.9~mA$

Source current can be found by assuming the $P_{source}$ equals the $P_{load}$, but there is another trick. The turns ratio was useful in determining the load voltage, but going back the other direction, it can be used the same way to determine the source current.

$$V_{load}={V_{source} \over Turns~Ratio}$$

$$I_{source}={I_{load} \over Turns~Ratio}$$

Note how we sitch the place of source and load as we toggle between voltage and current. This is because in the power equation, I and V will vary by equal, but opposite magnitudes to preserve power.

$I_{source} = 5.114~A$

First, we know the current in the secondary coil because it is a series circuit and the current of one resistor is given.

{\bullet} $I_{secondary} = 5.8~mA$

The secondary voltage is found by adding all of the resistor voltages in series; these can be found by using Ohm’s Law $V=I \cdot R$.

{\bullet} $V_{secondary} = 3.045~V$

Since this is a step-down transformer (more turns on the primary), we can find the primary voltage by multiplying the secondary voltage by the turns ratio: 3800:550.

{\bullet} $V_{primary} = 21.04~V$

In a likewise fashion, the primary current is found by dividing the secondary current by the turns ration 3800:550.

{\bullet} $I_{primary} = 839.5~\mu A$

The secondary voltage is the first objective. Since we know the current and resistance of one parallel load, this will directly determine the voltage of the secondary coil: $V=I \cdot R$.

{\bullet} $V_{secondary} = 50~V$

Knowing the voltage of both load resistors, we can calculate the current through the 4 ohm resistor as well.

The total secondary current is the sum of both resistor currents.

{\bullet} $I_{secondary} = 22.5~A$

As before, the secondary voltage multiplied by the turns ratio will yield the primary voltage.

As a logic check, which coil should have higher voltage? The one with more coils: the primary coil in this case.

1450:900 is the ratio.

{\bullet} $V_{primary} = 80.56~V$

The primary current can be found by equating the primary and secondary power and solving for current.

Alternatively, the secondary current can be divided by the turns ratio to yield the primary current.

As a logic check, which side should have higher current? The one with fewer coils: the secondary side in this case.

{\bullet} $I_{primary} = 13.97~A$

In this example, we will solve the voltages and currents in the most reasonable order, based on what we know.

First, we can calculate the secondary current by applying Ohm’s Law to 3.3 kΩ resistor with 40 volts.

{\bullet} $I_{secondary} = 12.12~mA$

Use this same current to solve for the voltage of the other resistor, and the sum of those voltages is the secondary voltage.

{\bullet} $V_{secondary} = 46.06~V$

Secondary voltage multiplied by the turns ratio (900:2200) is the primary voltage. Remember, which one should have more voltage? The one with more turns: the secondary, in this case.

{\bullet} $V_{primary} = 18.84~V$

Since the primary side of the circuit is a parallel circuit, the primary voltage is also the voltage across the 400 Ω resistor, and it is also the source voltage.

{\bullet} $V_{source} = 18.84~V$

The primary current is determined by dividing the secondary current by the turns ratio (explanation in previous problems).

Again as a logical guide, the side with fewer turns should have more current.

{\bullet} $I_{primary} = 29.63~mA$

Finally, the source current is the sum of the primary current + the current through the 400 Ω resistor, the latter of which is found through use of Ohm’s Law, since we know the resistor’s voltage.

{\bullet} $I_{source} = 76.74~mA$

In this example, we will solve the voltages and currents in the most reasonable order, based on what we know.

First, we can calculate the secondary current by applying Ohm’s Law to 3.3 kΩ resistor with 13 volts.

{\bullet} $I_{secondary} = 3.939~mA$

Use this same current to solve for the voltage of the other resistor, and the sum of those voltages is the secondary voltage.

{\bullet} $V_{secondary} = 14.97~V$

Secondary voltage multiplied by the turns ratio (700:2200) is the primary voltage. Remember, which one should have more voltage? The one with more turns: the secondary, in this case.

{\bullet} $V_{primary} = 4.763~V$

The primary current is determined by dividing the secondary current by the turns ratio (explanation in previous problems).

Again as a logical guide, the side with fewer turns should have more current.

{\bullet} $I_{primary} = 12.38~mA$

Since the primary side features a series circuit, the 45 Ω resistor as well as the source, both share the primary current.

{\bullet} $I_{source} = 12.38~mA$

Finally, the source voltage is the sum of the primary voltage + the voltage of the 45 Ω resistor, which is found by applying Ohm’s Law to the resistor: $V= I \cdot R$.

{\bullet} $V_{source} = 5.32~V$

Simple system (no transformers):

The losses in the wire result from 150 amps passing through 2x 0.1 Ω resistors. Although not much, this causes a loss of 30 V total, leaving only 210 V for the load.

Multiplying this 210 V by the 150 A current for power, and comparing this to the 240 V source supplying 150 A, the resulting efficiency is quite poor.

$E_{load} = 210~V$
$P_{load} = 31.5~kW$
$P_{lines} = 4.5~kW$
$\eta = 87.5$

Complex system (with transformers):

In this case, the transformers provide a high-voltage and low-current situation in the wires. Now with only 15 A in the wire, the total voltage loss from the wire is only 3 V total. Now, stepping this voltage loss back down the load, the loss of voltage at the load is only 0.3 V, leaving 239.7 V for the load.

Comparing the voltage at the load and source (since current is identical) the efficiency is FAR better.

$E_{load} = 239.7~V$
$P_{load} = 35.96~kW$
$P_{lines} = 45~W$
$\eta = 99.88$

The CT ratio is a measure of the current at the primary to the current at the secondary. Given the ratio of 600:5, which reduces to 120:1, the current in the primary is 120x higher than the current in the secondary.

$${I_{primary} \over I_{secondary}}={600 \over 5}$$

Secondary current = 2.917 amps