A transformer calculation starts with two labelled sides: the primary receives electrical energy from the supply, and the secondary supplies a load. Keep those labels attached to every voltage, current and number of turns. A step-up transformer increases voltage; it does not create extra power. A step-down transformer decreases voltage; that does not mean its secondary current must also decrease.
These original transformer calculations use supplied classroom data. Voltages and currents are effective, or rms, values. The loads are treated as resistive and the elementary model takes real power as P = VI. Where losses matter, an efficiency is explicitly supplied. These are paper calculations, not instructions for opening, wiring, testing or building electrical equipment.
Choose the relationship that answers the question
| Primary quantity | Secondary quantity | Meaning |
|---|---|---|
| Np | Ns | Number of turns in each winding |
| Vp | Vs | Voltage across each winding, in volts |
| Ip | Is | Current in each circuit, in amperes |
| Pin | Pout | Input and useful output power, in watts |
For an ideal transformer, Vs/Vp = Ns/Np. Voltage and turns have the same ratio. If the secondary has one quarter as many turns, its voltage is one quarter of the primary voltage. The ideal model assumes full flux linkage and no energy losses.
For that lossless model, Pin = Pout, so VpIp = VsIs. Therefore Is/Ip = Np/Ns. The current ratio is reversed. Write the power equality first if remembering two different ratios causes confusion.
When efficiency η is given, use η = Pout/Pin, with a percentage converted to a decimal. Thus Pout = ηVpIp and Is = Pout/Vs. Do not combine a non-ideal efficiency with the lossless current ratio. In a real loaded transformer the voltage ratio may also depart from the ideal approximation; use the model or measured voltages stated in the question.
What the transformer diagram means
The schematic shows two electrically separate windings on a common magnetic core. Alternating current in the primary produces changing magnetic flux; the changing flux linked with the secondary induces an emf there. The drawn loop counts are symbolic and do not replace the stated numerical turns in an exercise. There is no direct wire connection between the two windings in this schematic.
A steady direct current does not maintain the changing flux needed for continuous transformer action. A brief change when a circuit is switched is different from sustained operation. Never interpret this explanation as permission to connect a transformer to an unsuitable supply.
Worked case 1: step down voltage and check current
Data: An ideal transformer has 400 primary turns and 100 secondary turns. Vp = 80 V, and the load draws Is = 1.5 A. Find the secondary voltage and primary current.
First label the ratio: Ns/Np = 100/400 = 1/4. Hence Vs = 80 × 1/4 = 20 V. The output power is 20 × 1.5 = 30 W. With no losses, the input power is also 30 W, so Ip = 30/80 = 0.375 A.
The voltage falls by a factor of four while the current rises by a factor of four from primary to secondary. Both circuits carry the same power. A proposed primary current of 6 A would imply 480 W entering and only 30 W leaving, contradicting the explicitly ideal model.
Worked case 2: step up without increasing power
Data: An ideal transformer has Np = 150, Ns = 900 and Vp = 30 V. Its secondary current is 0.20 A. Find Vs and Ip.
The turns ratio, secondary to primary, is 900/150 = 6. Therefore Vs = 30 × 6 = 180 V. Output power is 180 × 0.20 = 36 W, giving Ip = 36/30 = 1.2 A. The primary current is six times the secondary current.
If asked for the ratio of primary turns to secondary turns, answer 150:900 = 1:6. If asked for secondary to primary, answer 6:1. Those are different orders, even though they describe the same transformer. Read the requested order instead of reporting an unlabelled “six”.
Worked case 3: account for stated losses
Data: Vp = 100 V, Ip = 0.40 A, Vs = 16 V and efficiency is 80%. Find the useful output power, secondary current and power lost.
Input power = 100 × 0.40 = 40 W. Convert 80% to 0.80: Pout = 0.80 × 40 = 32 W. Then Is = 32/16 = 2.0 A. Power lost = 40 − 32 = 8 W.
The output is less than the input, as expected. Using Is = 40/16 would give 2.5 A and silently assume 100% efficiency. Dividing 40 by 0.80 would give an output greater than the input. Neither answer follows the stated direction of energy transfer.
Worked case 4: separate a turns estimate from input power
Data: A question permits the ideal turns/voltage approximation for a transformer with Np = 600, Vp = 90 V and Vs = 18 V. At this operating point, useful output power is 9.0 W and efficiency is 90%. Find the estimated secondary turns and primary current.
Ns = 600 × 18/90 = 120 turns. For current, first find Pin = 9.0/0.90 = 10 W. Therefore Ip = 10/90 ≈ 0.111 A. Also Is = 9.0/18 = 0.50 A.
The actual current ratio in this model is 0.50/0.111… = 4.5, not five. The distinction is intentional: the turns estimate uses the permitted approximation, while the power calculation includes losses. State both assumptions rather than pretending the transformer is simultaneously perfectly lossless and 90% efficient.
Three independent tasks with solutions
Practice 1: a fresh step-down calculation
An ideal transformer has 720 primary turns, 180 secondary turns and a 96 V primary supply. Its secondary current is 0.80 A. Calculate both the secondary voltage and primary current. The voltage is 96 × 180/720 = 24 V. Output power is 24 × 0.80 = 19.2 W, so primary current is 19.2/96 = 0.20 A. Check that 96 × 0.20 = 24 × 0.80.
Practice 2: diagnose a reversed current ratio
An ideal transformer has 60 primary turns and 300 secondary turns. Vp = 24 V and Ip = 2.5 A. A learner predicts a secondary current of 12.5 A. Explain and correct the error. The secondary voltage is 24 × 5 = 120 V. Input power is 60 W, so Is = 60/120 = 0.50 A. Multiplying both voltage and current by five would multiply power by twenty-five, contradicting conservation of energy.
Practice 3: calculate useful and wasted power
A transformer takes 0.50 A at 60 V and supplies a resistive load at 12 V with 84% efficiency. Find secondary current and power lost. Pin = 30 W; Pout = 0.84 × 30 = 25.2 W; Is = 25.2/12 = 2.1 A. Loss is 30 − 25.2 = 4.8 W. The useful and lost powers sum to the input.
Explain losses with a cause and a response
Winding resistance causes heating; using suitably low-resistance windings reduces this loss. Changing magnetic flux can induce eddy currents in the core; insulated laminations restrict their paths. Repeated magnetisation also dissipates energy through hysteresis, so suitable core material matters. These are distinct mechanisms, not three names for one effect.
For a fixed transmitted power in the simplified model, increasing transmission voltage reduces current. Line heating is I²R, so halving current reduces that heating to one quarter at unchanged resistance. A transformer changes the voltage/current balance; it does not make line resistance disappear.
Frequently asked questions
Does more secondary voltage mean more output power?
No. In the ideal model the higher voltage is accompanied by a lower current. With losses, useful output power is less than input power.
Which side is the primary?
The primary is the winding receiving energy from the supply in the stated arrangement. It is not automatically the side with more turns.
Must turns be written with a unit?
Turns are a count. Write “120 turns” for clarity; a ratio of turns has no physical unit.
Can I use the ideal current ratio at 80% efficiency?
No. Use the power relation with efficiency and the supplied voltages. The lossless current ratio would discard the stated losses.
Why is 80% written as 0.80?
Efficiency in Pout = ηPin is a fraction. Eighty per cent means eighty hundredths.
Why does a transformer need changing flux?
Induced emf depends on a change in magnetic flux linkage. A steady flux does not maintain an induced secondary emf.
What is the quickest numerical check?
Compare input and output power, then check whether voltage changes in the same direction as turns. An impossible power gain often exposes a reversed ratio.
Further revision and reference
The Physics topical revision collection provides existing resource navigation; its contents may require membership. No particular download is promised. The basic transformer model is also explained in OpenStax Physics: motors, generators and transformers. All numerical cases and the schematic here are original.
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