# What is no-load voltage in a transformer?

From what "law" of transformers is the following equation for a practical transformer derived? I presume it is not V1/N1 = V2/N2, which only pertains to ideal transformers.

\begin{align} V_{2,NL}&=V_1/a \end{align}

V2,NL is the no-load voltage across the secondary, V1 is the voltage across the primary, and a is the turns ratio.

• Where is V1 in the diagram? What does the 192 ohms represent? Is it eddy current losses? What about magnetization inductance? That doesn't appear to be shown in the diagram? Feb 15 at 18:14
• V1 is the 1200V applied voltage on the primary winding. The 192 ohms represent the load on the secondary winding, when referred to the primary circuit. The magnetizing current is ignored in this example. Thanks for clarifying Andy Feb 15 at 18:46
• If 192 ohms is the referred load and magnetizing current is negligible, then with no load a * V2 = V1 = 1200 V since the referred load is a²Zl with Zl = infinity.
– vir
Feb 15 at 18:59
• If "magnetizing current is ignored" then no load means no current flow whatsoever. This contradicts the "practical transformer" however. It seems you will have to refer to the text to see why that has been asked for (perhaps it's accounted for in a later step?). Feb 16 at 1:04
• If magnetizing current and associated losses in primary winding resistance then you still have hysteresis losses in the core. Feb 20 at 17:59

There is an old saying that "All models are wrong, some models are useful.". Engineering is often about determining what effects need to be considered and what can be ignored.

The concept of "an ideal x" wouldn't exist if it wasn't useful. An "ideal x" is the most basic approximation of a "real x". In the case of a transformer, the ideal model comes when we build an electromagnetic model of a transformer using zero resistance wire, wound round a core of infinite permeability and we ignore capacitive effects.

If an "ideal x" is not a good enough model we move to better models. A common model is shown below. This represents a real transformer in terms of an ideal transformer, three resistors and three inductors.

$$\R_P\$$ represents the primary resistance, $$\X_P\$$ represents the primary leakage inductance, $$\R_S\$$ represents the secondary resistance and $$\X_S\$$ represents the secondary leakage inductance. $$\R_C\$$ represents core losses and $$\X_M\$$ represents the magnetising inductance. The resistors representing the secondary resistance and leakage have been transferred across the ideal transformer to make the circuit easier to analyse.

In an ideal transformer the resistances and inductances representing leakage inductance and winding resistance would be zero while those representing core loss and magnetisation current would be infinite.

So how does this relate to the "no load voltage"?

Since there is no load, $$\R_S\$$ and $$\X_S\$$ do not affect the output voltage. $$\R_P\$$, $$\X_P\$$, $$\R_C\$$ and $$\X_M\$$ however can have an effect. Current flowing in $$\R_C\$$ and $$\X_M\$$ will cause voltage drop in $$\R_P\$$ and $$\X_P\$$.

Therefore, the underlying assumption of your equation is that $$\R_P\$$ and $$\X_P\$$ are much smaller than $$\R_C\$$ and $$\X_M\$$ and hence the voltage drop in $$\R_P\$$ and $$\X_P\$$ is neligable.

• Are you saying that V1/N1 = V2/N2 holds for the illustrated transformer if we assume that Rs, Xs, Rp, Xp are negligible? But why would the equation still hold when we have non-zero Rc and Xm? Feb 20 at 21:25
• If Rc and Xm are infinite or Rp and Xp are zero then it holds exactly, if Rp and Xp are much smaller than Rc and Xm then it holds approximately. Feb 20 at 21:28
• Do Rc and Xm not affect output voltage because they are in parallel with the primary winding? Feb 20 at 21:37
• Yes, a resistor or inductor with an impedance of zero drops no voltage, also a component carrying no current drops no voltage. Feb 20 at 21:48
• It was in response to a comment that has now disappeard about voltages across the top of the circuit. Rc and Xm do not directly affect the output voltage, but they can cause current flow in Rp and Xp which in turn will affect the output voltage. Feb 20 at 22:00

V1/N1 = V2/N2

$$\\dfrac{N_1}{N_2} ~=~a\$$

This makes your equations equivalent for an ideal transformer.

The practical transformer design has many variables for voltage tolerance , temp rise at rated load and harmonic content at max voltage from near saturation. The voltage must excite the core to mutually couple the output. From primary impedance, this is about 10% of the rated load current and it’s effective primary impedance.

Due to the cost of copper + core vs. the cost of lost power, the tradeoff for practical transformers, the ratios depend on cost.

For a step-down 10:1 or a=10 or any ratio step-up, the losses must be compensated (k) with a higher output voltage such that real ratio = k*a for k>1

• Small < 100W, k = 1.1
• Large > 5MVA k = 1.01

This is not a “law” but a “Rule of Thumb” and there will be exceptions.

From what "law" of transformers is the following equation for a practical transformer derived?

The "law of transformers" is actually Faraday's Law of induction

A current through a primary inductor will produce a magnetic flux ($$\\phi\$$) within a region according to Ampere's Law and the self-inductance associated with the inductor.

The voltage ($$\V_{L_P}\$$) across the primary inductance ($$\L_P\$$) forms according to Faraday's Law of Induction. The voltage ($$\V_P)\$$ across the inductor will be the sum of $$\V_{L_P}\$$ and the voltage across the RC parasitics.

$$V_{L_P}=-N_P\frac{d\phi_P}{dt}\tag{1}$$

A second inductor is brought close to the first so that it encounters a portion (k) of the flux from the primary inductor. The voltage across the secondary inductance will be:$$V_{L_S}=-N_S\frac{d\phi_S}{dt}$$

The secondary flux will be $$\\phi_S=k\phi_P\$$ so: $$V_{L_S}=-kN_S\frac{d\phi_P}{dt}\tag{2}$$

The ratio of equation 2 to equation 1 yields the familiar transformer equation.

$$\frac{V_{L_S}}{V_{L_P}}=\frac{-kN_S\frac{d\phi_P}{dt}}{-N_P\frac{d\phi_P}{dt}}=\frac{kN_S}{N_P}$$

Reforming and letting $$\a=\frac{N_P}{N_S}\$$ reproduces the OP's equation if $$\k=1\$$ which is almost true for power transformers.

$$V_{L_S}=\frac{kV_{L_P}}{a}$$

Notice that this result is true for ideal and real transformers. The coupling constant, k, allows leakage flux to be included in Faraday's description.

This is not a model but derived directly from a "Law" as was requested.

The model in Peter Green's answer (+1) can be obtained from this result as a starting point and electronic analysis.