The ideal transformer equations provide a lossless mathematical model linking primary and secondary voltages, currents, and turns ratios. In a perfectly ideal scenario, the ratio of primary to secondary voltage equals the turns ratio, while the current ratio is the exact inverse. This model assumes zero core losses, zero copper losses, and perfect magnetic coupling. Below is the complete mathematical framework, derivations, and practical worked examples to apply these equations on the bench or in the field.

The Core Ideal Transformer Equations

The master equation for an ideal transformer ties together voltage (V), number of turns (N), and current (I) into a single proportional relationship. The derivation stems from two fundamental physics principles: Faraday's Law of Induction and the Conservation of Energy.

According to Faraday's Law, the induced electromotive force (EMF) in a coil is proportional to its number of turns and the rate of change of magnetic flux. Assuming perfect coupling (all flux links both coils), the voltage per turn is identical on both sides: Vp/Np = Vs/Ns. Simultaneously, Conservation of Energy dictates that input power equals output power (Vp × Ip = Vs × Is). Combining these yields the master equation:

Vp / Vs = Np / Ns = Is / Ip = a

Where a is the dimensionless turns ratio. Below is the complete symbol definition table:

Symbol Parameter Standard Unit Description
Vp Primary Voltage Volts (V) RMS AC voltage applied to the primary winding.
Vs Secondary Voltage Volts (V) RMS AC voltage induced across the secondary winding.
Np Primary Turns Unitless (Count) Total number of wire loops on the primary side.
Ns Secondary Turns Unitless (Count) Total number of wire loops on the secondary side.
Ip Primary Current Amperes (A) RMS current drawn from the source by the primary.
Is Secondary Current Amperes (A) RMS current delivered to the load by the secondary.
a Turns Ratio Unitless The ratio of primary to secondary turns (Np/Ns).

For deeper reading on the magnetic principles underlying these ratios, HyperPhysics provides an excellent breakdown of mutual inductance and flux linkage.

Rearranged Forms: Solving for Every Variable

On the workbench, you rarely have all variables. Here is the master equation rearranged to solve for any single missing parameter, assuming you know the others:

  • Solve for Secondary Voltage: Vs = Vp × (Ns / Np) or Vs = Vp / a
  • Solve for Primary Voltage: Vp = Vs × (Np / Ns) or Vp = Vs × a
  • Solve for Secondary Current: Is = Ip × (Np / Ns) or Is = Ip × a
  • Solve for Primary Current: Ip = Is × (Ns / Np) or Ip = Is / a
  • Solve for Secondary Turns: Ns = Np × (Vs / Vp)
  • Solve for Primary Turns: Np = Ns × (Vp / Vs)
  • Solve for Turns Ratio (a): a = Vp / Vs = Np / Ns = Is / Ip

Assumptions, Realistic Magnitudes, and Limits

The ideal transformer equations are a mathematical abstraction. To use them correctly, you must understand the physical assumptions they make and where real-world physics diverges from the math.

When the Formula Applies (The 4 Assumptions)

  1. Zero Copper Losses: Assumes the windings have exactly 0 Ω resistance. No I²R heating occurs.
  2. Zero Core Losses: Assumes the magnetic core has no hysteresis loss and no eddy current losses.
  3. Perfect Magnetic Coupling (k=1): Assumes 100% of the magnetic flux generated by the primary links the secondary. Zero leakage inductance.
  4. Infinite Core Permeability: Assumes the core requires zero magnetizing current to establish the magnetic flux. The primary draws zero current when the secondary is open-circuited.

What a Realistic Answer Magnitude Looks Like

How close is ideal math to reality? It depends heavily on the transformer's physical size and VA rating. According to Electronics Tutorials, large utility distribution transformers (e.g., 50 kVA pole-top units) operate at 98% to 99% efficiency. For these, the ideal equations will predict currents and voltages within 1% to 2% of actual measured values under load.

However, for small PCB-mount transformers (e.g., a 2 VA Talema 70020), efficiency might only be 80% to 85%. If you use the ideal equations to predict primary current for a 2 VA unit, your calculated Ip will be 15% to 20% lower than what your multimeter actually reads, because the ideal math ignores the magnetizing current and copper losses required to sustain the magnetic field.

Worked Problems with Unit Tracking

Let's apply the equations to two common scenarios, tracking units at every step to prevent calculation errors.

Problem 1: Mains Step-Down Power Transformer

Given: A control transformer has a primary voltage (Vp) of 480 V RMS and a secondary voltage (Vs) of 120 V RMS. The secondary is connected to a resistive load drawing 15 A (Is). The primary winding has 800 turns (Np).
Find: The turns ratio (a), the primary current (Ip), and the number of secondary turns (Ns).

Step 1: Calculate the turns ratio (a)

  • Formula: a = Vp / Vs
  • Substitution: a = 480 V / 120 V
  • Unit tracking: Volts / Volts = Unitless
  • Result: a = 4 (This is a 4:1 step-down transformer)

Step 2: Calculate primary current (Ip)

  • Formula: Ip = Is / a
  • Substitution: Ip = 15 A / 4
  • Unit tracking: Amperes / Unitless = Amperes
  • Result: Ip = 3.75 A

Step 3: Calculate secondary turns (Ns)

  • Formula: Ns = Np / a
  • Substitution: Ns = 800 turns / 4
  • Unit tracking: Turns / Unitless = Turns
  • Result: Ns = 200 turns

Problem 2: Audio Impedance Matching Transformer

Given: An audio output transformer matches a vacuum tube amplifier with a primary impedance (Zp) of 5,000 Ω to an 8 Ω speaker (Zs).
Find: The required turns ratio (a).

Step 1: Identify the impedance relationship

In an ideal transformer, impedance transforms by the square of the turns ratio: Zp / Zs = a². (Derivation: Z = V/I. Since V scales by 'a' and I scales by '1/a', V/I scales by a / (1/a) = a²).

Step 2: Solve for a

  • Formula: a = √(Zp / Zs)
  • Substitution: a = √(5000 Ω / 8 Ω)
  • Unit tracking: √(Ω / Ω) = √(Unitless) = Unitless
  • Intermediate math: 5000 / 8 = 625
  • Result: a = √625 = 25

The transformer requires a 25:1 turns ratio to perfectly match the tube amplifier to the speaker.

Common Unit Mistakes That Break the Math

When your calculated values don't match your multimeter readings, you've likely fallen victim to one of these unit or convention traps:

  1. Mixing Peak and RMS Voltages: The ideal transformer equations strictly require RMS (Root Mean Square) values for AC waveforms. If your oscilloscope reads 170V peak on the primary, and you plug 170 into the formula alongside a 12V RMS secondary, your turns ratio will be wildly incorrect. Always convert peak to RMS first (Vrms = Vpeak / √2).
  2. Apparent Power (VA) vs. Real Power (W): The conservation of energy in the ideal model assumes VpIp = VsIs. This is true for Apparent Power (Volt-Amps). If your load is inductive (like a motor) with a power factor of 0.8, the Real Power (Watts) is lower. The transformer equations still hold for the RMS currents and voltages, but do not use Watts to calculate current unless the load is purely resistive (PF = 1.0).
  3. Current Scaling Direction: A frequent bench mistake is assuming current scales the same way as voltage. Remember: voltage and turns are directly proportional (more turns = more voltage), but current is inversely proportional (more turns = less current). High voltage side always carries the lower current.
  4. Turns Ratio Convention (a vs 1/a): In North American literature, 'a' is almost universally Np/Ns. In some older European texts, the ratio is sometimes expressed as secondary-to-primary. Always verify which convention a specific datasheet uses before winding a custom core. All About Circuits confirms the standard Np/Ns convention for modern engineering.

Frequently Asked Questions

Do ideal transformer equations work for DC voltage?

No. The ideal transformer equations rely on Faraday's Law of Induction, which requires a changing magnetic flux to induce a voltage. If you apply a steady DC voltage to the primary, the rate of change of flux (dΦ/dt) is zero once the magnetic field saturates. Consequently, no voltage is induced on the secondary, and the primary winding will simply act as a low-resistance short circuit, likely burning out the wire or tripping your power supply's overcurrent protection.

How do I calculate transformer efficiency using these equations?

You cannot calculate efficiency using only the ideal equations, because the ideal model assumes 100% efficiency by definition (Input Power = Output Power). To calculate real-world efficiency, you must measure the actual input power (Win) and actual output power (Wout) under load using a wattmeter. Efficiency (η) = (Wout / Win) × 100. The difference between your ideal calculated Ip and your measured Ip represents the losses (copper + core) that the ideal equations ignore.

Why is the secondary current inversely proportional to the turns ratio?

This is a direct consequence of the Conservation of Energy. A transformer cannot create power; it only changes the voltage/current ratio. If a transformer steps up the voltage by a factor of 10 (e.g., 12V to 120V), it must step down the current by a factor of 10 to ensure that Vp × Ip = Vs × Is. If it didn't, the secondary would be outputting more power than the primary is taking in, violating the first law of thermodynamics.

Can I use the ideal transformer equations for high-frequency switching power supplies?

You can use them for first-order approximations (like determining the baseline turns ratio for a flyback or forward converter), but they are highly inaccurate for detailed design. High-frequency switch-mode power supplies (SMPS) operate at 50 kHz to 1 MHz+. At these frequencies, leakage inductance, winding parasitic capacitance, and core switching losses become dominant factors. SMPS design requires modeling the transformer with an equivalent circuit that includes these parasitic elements, rather than relying on the lossless ideal model.