Transformer phasor diagrams

A transformer has no moving parts, yet everything inside it — flux, currents, EMFs — is a sinusoid chasing another sinusoid by some phase angle. Numbers alone (magnitudes of $V_1$ , $I_1$ , $E_2$ ...) cannot tell you whether the primary current leads or lags the applied voltage, or by how much, because that depends entirely on phase relationships that a bare list of magnitudes throws away. A phasor diagram is the tool that keeps magnitude and phase together in one picture. This node builds that picture in two stages: first on no-load, where the primary draws only a small magnetising/core-loss current $I_0$ with the secondary open, and then on full load, where a load current $I_2$ at some power factor is reflected back to the primary and combined with $I_0$ . The mutual flux $\Phi_m$ is the natural reference phasor throughout, since it links both windings. By the end you should be able to resolve $I_0$ into its two components, sketch the full-load diagram for lagging and leading loads, and add phasors by rectangular components to get the actual primary current and its power-factor angle — which, perhaps surprisingly, is never quite equal to the load's own power factor.

The core insight

Taking the mutual flux $\Phi_m$ as reference, the no-load current $I_0$ splits into a magnetising component $I_\mu = I_0\sin\phi_0$ (in phase with $\Phi_m$ ) and a core-loss component $I_w = I_0\cos\phi_0$ (in phase with $V_1$ ), and on full load the primary current is the phasor sum $I_1 = I_0 + I_2'$ of this no-load current with the reflected load current, which is why the primary power factor is always slightly different from the secondary's.

The core idea

Why phasors, not just numbers. All quantities in a transformer — $V_1$ , $E_1$ , $\Phi_m$ , $I_0$ , $E_2$ , $V_2$ , $I_2$ — oscillate at the same frequency but with different phase angles between them. A phasor diagram fixes one quantity as reference and draws every other quantity as an arrow at its correct angle relative to that reference, so magnitude and phase are both visible at a glance. The natural reference is the mutual (core) flux $\Phi_m$ , because it is the one quantity common to both the primary and secondary windings.

No-load current has two components. With the secondary open-circuited, the primary still draws a small current $I_0$ to set up the core flux and to supply the core (hysteresis + eddy-current) loss. Because $E_1$ leads $\Phi_m$ by exactly $90^\circ$ (from Faraday's law) and $V_1 \approx -E_1$ when the primary resistance/leakage drop is small, $I_0$ lags $V_1$ by an angle $\phi_0$ that is close to but less than $90^\circ$ . Resolving it along and perpendicular to $V_1$ gives:

I_w = I_0\cos\phi_0, \qquad I_\mu = I_0\sin\phi_0

where: $I_w$ is the core-loss (working) component, in phase with $V_1$ , supplying real power to core loss; $I_\mu$ is the magnetising component, in phase with $\Phi_m$ (i.e. $90^\circ$ behind $V_1$ ), producing the flux; $\phi_0$ is the no-load power-factor angle. The two are related back to $I_0$ by $I_0 = \sqrt{I_w^2 + I_\mu^2}$ .

Full-load: reflect and add. On load, the secondary supplies current $I_2$ to the load at terminal voltage $V_2$ and power-factor angle $\phi_2$ (lagging for inductive loads, leading for capacitive loads). $E_1$ and $E_2$ stay in phase with each other (both lead $\Phi_m$ by $90^\circ$ ), related by the turns ratio $E_1/E_2 = a$ . Referring the secondary current to the primary side, $I_2' = I_2/a$ , the primary current is approximately the phasor sum

I_1 = I_0 + I_2'

where: $I_0$ is the (unchanged) no-load current computed above, and $I_2'$ is the load current referred to the primary, at the same angle $\phi_2$ relative to $E_2$ (hence to $E_1$ ) that the actual load current makes relative to $V_2$ . Because $I_0$ is added vectorially, not scalar-added, $I_1$ 's magnitude is not simply $I_0 + I_2'$ , and its angle relative to $V_1$ is not exactly $\phi_2$ .

Lagging vs leading — the qualitative picture. For a lagging-pf load $I_2'$ lags $E_2$ ; since $I_0$ itself lags $V_1$ by nearly $90^\circ$ , adding it typically nudges the overall lag angle of $I_1$ slightly beyond $\phi_2$ . For a leading-pf load $I_2'$ leads $E_2$ , and the shift goes the other way. Either way, $I_1 \approx I_2'$ is a fair approximation only when $I_0 \ll I_2'$ , which is the normal case for a transformer operating well above no-load.

Figure

Phasor diagram of a transformer on full load showing flux as reference with E1 E2 in phase leading it by ninety degrees and I0 I2 prime and I1 as their phasor sum at the lagging power factor angle

Derivation AEstimating the primary current by phasor addition

Given the no-load current $I_0$ at angle $\phi_0$ behind the reference $E_1$ , and the primary-referred load current $I_2'$ at angle $\phi_2$ behind the same reference, find the magnitude and phase angle of the actual primary current $I_1$ by resolving both phasors into rectangular components and adding, since simply adding the two magnitudes arithmetically would ignore the fact that $I_0$ and $I_2'$ point in different directions and so cannot be summed like plain numbers.

Step 1. Write $E_1$ (or $E_2$ , since they are in phase) as the reference phasor at angle $0^\circ$ . Every other current is then expressed as a magnitude and an angle measured from this reference, lagging angles taken as negative.

Step 2. Resolve the reflected load current into rectangular form: $I_2' = |I_2'|\cos\phi_2 - j|I_2'|\sin\phi_2$ , where the real part is the component in phase with $E_1$ and the imaginary part is the component $90^\circ$ behind it (lagging, hence the minus sign for an inductive load).

Step 3. Resolve the no-load current the same way: $I_0 = |I_0|\cos\phi_0 - j|I_0|\sin\phi_0$ , using the no-load power-factor angle $\phi_0$ measured from the same reference.

Step 4. Add the real parts and the imaginary parts separately, since rectangular components along the same axis add like ordinary numbers even though the underlying phasors do not: $I_1 = \left(|I_2'|\cos\phi_2 + |I_0|\cos\phi_0\right) - j\left(|I_2'|\sin\phi_2 + |I_0|\sin\phi_0\right)$ .

Step 5. Recover the polar form from the summed rectangular components: $|I_1| = \sqrt{(\text{Re})^2+(\text{Im})^2}$ and $\phi_1 = \tan^{-1}\!\left(\dfrac{\text{Im}}{\text{Re}}\right)$ , which gives both the magnitude of the primary current and the angle by which it lags (or leads) $V_1$ — the actual primary power-factor angle, generally close to but not exactly equal to $\phi_2$ .

Worked example

A transformer's no-load current is $I_0 = 3\,\text{A}$ at $\phi_0 = 75^\circ$ lagging the applied voltage $V_1$ , and it supplies a load current $I_2' = 40\,\text{A}$ (referred to the primary) at a lagging power factor $\cos\phi_2 = 0.8$ , so $\phi_2 \approx 36.87^\circ$ . Find $I_1$ .

Step 1 — resolve $I_0$ into components, in phase with $V_1$ and in phase with $\Phi_m$ respectively:

I_w = I_0\cos\phi_0 = 3\times\cos75^\circ \approx 0.776\ \text{A}, \qquad I_\mu = I_0\sin\phi_0 = 3\times\sin75^\circ \approx 2.898\ \text{A}

so in rectangular form $I_0 = 0.776 - j2.898\ \text{A}$ (check: $\sqrt{0.776^2+2.898^2}\approx 3.0\,\text{A}$ ✓ — as expected, the magnetising part dominates the small no-load current).

Step 2 — resolve $I_2'$ into components, lagging $E_1$ by $36.87^\circ$ :

I_2' = 40\cos36.87^\circ - j40\sin36.87^\circ = 32.0 - j24.0\ \text{A}

Step 3 — add the rectangular parts:

I_1 = (32.0+0.776) - j(24.0+2.898) = 32.776 - j26.898\ \text{A}

Step 4 — convert back to polar form:

|I_1| = \sqrt{32.776^2+26.898^2} = \sqrt{1074.3+723.5} \approx 42.4\ \text{A}

\phi_1 = \tan^{-1}\!\left(\frac{26.898}{32.776}\right) \approx 39.4^\circ\ \text{lagging}

The primary current is $\boxed{I_1 \approx 42.4\ \text{A at } 39.4^\circ \text{ lagging}}$ — slightly larger in magnitude and lagging by slightly more than the load's own $36.87^\circ$ , because the no-load current's own lagging contribution adds a small extra lag.

Competitive stretch

If instead the same $40\,\text{A}$ load were at a leading power factor of $0.8$ ( $\phi_2 \approx 36.87^\circ$ leading), $I_2'$ becomes $32.0+j24.0\,\text{A}$ . Adding the same lagging $I_0 = 0.776 - j2.898\,\text{A}$ gives $I_1 = 32.776 + j21.102\,\text{A}$ , so $|I_1|\approx 39.0\,\text{A}$ at about $32.7^\circ$ leading — the primary angle shifts closer to unity power factor rather than further from it, showing the correction from $I_0$ works in opposite senses for lagging and leading loads.

Common traps

Treating $I_0$ as purely reactive (i.e. exactly $90^\circ$ behind $V_1$ ): it always has a real, in-phase-with- $V_1$ component $I_w$ that supplies core loss, so $\phi_0$ is close to but strictly less than $90^\circ$ .
Drawing $E_1$ in phase with $\Phi_m$ : they are $90^\circ$ apart by Faraday's law, not aligned.
Assuming the primary power factor equals the secondary (load) power factor exactly: the phasor addition of $I_0$ shifts $\phi_1$ away from $\phi_2$ , even if only slightly when $I_0 \ll I_2'$ .
Adding $I_0$ and $I_2'$ as plain magnitudes instead of rectangular components: this ignores their different phase angles and gives a wrong (too large) value for $I_1$ .

Ask the tutor: If the rectangular-component phasor addition of $I_0$ and $I_2'$ still feels shaky, ask the AI tutor to redo the worked example with a leading-power-factor load of your own choosing and check your answer step by step.

Why do we pick the mutual flux Phi_m as the reference phasor instead of V1?

Because $\Phi_m$ is the one quantity common to both windings — both $E_1$ and $E_2$ are induced by it and both lead it by exactly $90^\circ$ , so anchoring the diagram on $\Phi_m$ keeps the primary-side and secondary-side phasors tied together in one consistent picture. If you anchored on $V_1$ instead, you would still need to separately establish where $\Phi_m$ and hence $E_2$ sit relative to it, which is extra work for no benefit.

Is I1 = I0 + I2' an exact relationship or an approximation?

It is the standard approximate treatment used at this level, ignoring the (usually small) effect of primary leakage impedance drop on the phase relationship between $I_0$ and the flux that generates it, and treating $I_2'$ as referred by the simple turns ratio. In a fuller equivalent-circuit treatment (covered in the following topic) small corrections appear, but for phasor-diagram purposes this vector sum is accurate enough and is what you should use for problems and sketches at this stage.

Does the direction of the phase shift from I0 reverse between lagging and leading loads?

Yes. For a lagging load, $I_2'$ lags $E_2$ , and adding the lagging $I_0$ tends to push $I_1$ 's angle slightly beyond $\phi_2$ (more lag). For a leading load, $I_2'$ leads $E_2$ , and adding the same lagging $I_0$ pulls the resultant angle back toward — or even past — the in-phase axis, as the competitive-stretch example above shows, shifting $I_1$ closer to unity power factor. The direction of the shift always depends on whether $I_0$ is adding to or partly cancelling the reactive part of $I_2'$ .

Transformer phasor diagrams

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads