Transformers

Step off a train near any electric substation and you will see them: humming grey boxes wound with heavy copper coils, quietly converting hundreds of kilovolts down to something a house can use. A transformer does this with no moving parts and (ideally) no wasted energy — it simply exploits the fact that a changing magnetic flux induces an EMF, and if two coils share the same flux, their voltages come out in the exact ratio of their turns. That single idea, mutual induction between two windings on a common iron core, is why India's power grid can send electricity hundreds of kilometres at very high voltage and still deliver a safe 230 V to a wall socket. This article builds the turns-ratio and power relations from Faraday's law, works through both an ideal-transformer and a real-transformer numerical, and lists the loss mechanisms an exam expects you to name.

The core insight

A transformer links a primary and secondary coil through a shared, changing magnetic flux in a common core, so their voltages scale with turns ( $V_s/V_p = N_s/N_p$ ) and, for an ideal (lossless) device, their currents scale inversely, letting AC power be stepped up for efficient transmission and stepped down for safe use.

The core idea

A transformer has two windings on a common soft-iron core: the primary ( $N_p$ turns, fed the AC input) and the secondary ( $N_s$ turns, delivering the output). The core is a good conductor of magnetic flux, so almost all the flux made by the primary current threads the secondary too — that shared flux is the entire mechanism.

Idealisations. An ideal transformer assumes zero winding resistance, zero flux leakage (all flux from the primary reaches the secondary), and zero core loss. Real transformers depart from this only slightly (90–99% efficient), so the ideal case is the right starting model for the syllabus. Because the same flux $\phi(t)$ threads every turn of both coils, Faraday's law gives each winding an EMF proportional to its own turns:

\varepsilon_p = -N_p\frac{d\phi}{dt}, \qquad \varepsilon_s = -N_s\frac{d\phi}{dt}

where: $\varepsilon_p,\varepsilon_s$ are the induced EMFs in the primary and secondary, $N_p,N_s$ are the turns, and $d\phi/dt$ is the common rate of change of flux per turn. Dividing one equation by the other cancels $d\phi/dt$ , giving the transformer's central relation:

\frac{V_s}{V_p} = \frac{N_s}{N_p}

where: $V_p,V_s$ are the primary and secondary AC voltages (equal to $\varepsilon_p,\varepsilon_s$ for an ideal transformer). If $N_s > N_p$ the device is step-up; if $N_s < N_p$ it is step-down. For an ideal transformer no power is lost in the transformer itself, so input power equals output power:

V_pI_p = V_sI_s \quad\Rightarrow\quad \frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{V_p}{V_s}

where: $I_p,I_s$ are primary and secondary currents. A step-up transformer raises voltage but lowers current in the same proportion — this inverse pairing is the single most tested idea in this topic.

Why the grid needs this. A transmission line of resistance $R_{line}$ carrying current $I$ wastes power as $I^2R_{line}$ , regardless of the voltage riding on it. For a fixed power $P = VI$ delivered, raising $V$ lowers $I$ proportionally, so line loss falls as the square of that reduction. This is why generating stations step voltage up to hundreds of kV for transmission, and local substations step it back down near consumers.

Figure

A step down transformer with a taller primary coil of more turns and a shorter secondary coil of fewer turns joined by a shared iron core loop, with a larger input voltage arrow on the primary side and a smaller output voltage arrow on the secondary side

Derivation ATurns-ratio and current relations for an ideal transformer

The goal is to derive, purely from Faraday's law applied twice to a shared flux, both the voltage turns-ratio $V_s/V_p = N_s/N_p$ and the inverse current relation $I_s/I_p = N_p/N_s$ that follows from ideal (lossless) power conservation, so that either voltage or current in one winding can be found from the other winding's value and the turns count alone.

Step 1. Model the shared flux. In an ideal transformer, the soft-iron core carries essentially all the magnetic flux made by the primary current, so the same flux $\phi(t)$ links every turn of the primary and every turn of the secondary — there is no leakage flux to treat separately.

Step 2. Apply Faraday's law to each winding. Each turn of a coil contributes an EMF equal to $-d\phi/dt$ , so a coil of $N$ turns contributes $N$ times that: $\varepsilon_p = -N_p\,d\phi/dt$ for the primary and $\varepsilon_s = -N_s\,d\phi/dt$ for the secondary. Since $\phi(t)$ is identical in both expressions, $d\phi/dt$ is the same number in each.

Step 3. Divide to eliminate the shared, generally unknown, $d\phi/dt$ . Taking magnitudes and identifying the winding EMFs with the terminal voltages $V_p,V_s$ for an ideal transformer, $\varepsilon_s/\varepsilon_p = N_s/N_p$ becomes $V_s/V_p = N_s/N_p$ — the turns-ratio law, independent of frequency or the size of the flux.

Step 4. Bring in power conservation to get the current relation. An ideal transformer has no resistive, hysteresis, eddy-current, or leakage losses, so every watt entering the primary leaves through the secondary: $V_pI_p = V_sI_s$ . Substituting $V_s = V_p(N_s/N_p)$ gives $V_pI_p = V_p(N_s/N_p)I_s$ , and dividing both sides by $V_p$ leaves $I_p = (N_s/N_p)I_s$ , i.e. $I_s/I_p = N_p/N_s$ — current scales inversely with the same turns ratio that scales voltage forward. Result: $V_s/V_p = N_s/N_p$ and $I_s/I_p = N_p/N_s = V_p/V_s$ , both traceable to one shared flux and the conservation of power.

Worked example

An ideal transformer has 1500 turns on the primary and 250 turns on the secondary, with the primary connected to a 230 V AC supply and the secondary delivering a current of 6 A. Find the type of transformer, the secondary voltage, and the primary current.

Step 1. Compare turns to classify the transformer. $N_s = 250$ is less than $N_p = 1500$ , so this is a step-down transformer.

Step 2. Find the secondary voltage from the turns-ratio law. $V_s = V_p \times \dfrac{N_s}{N_p} = 230 \times \dfrac{250}{1500} = 38.3\ \text{V}$ .

Step 3. Find the primary current from ideal power conservation, $V_pI_p = V_sI_s$ . $I_p = \dfrac{V_sI_s}{V_p} = \dfrac{38.3\times6}{230} = 1.0\ \text{A}$ .

Final answer: step-down transformer, $V_s = 38.3\ \text{V}$ , $I_p = 1.0\ \text{A}$ .

Competitive stretch

The same transformer's specification sheet also reports a real-world efficiency test: a transformer draws 4 A at 220 V on its primary and delivers 40 A at 20 V on its secondary. Input power is $P_{in} = 220\times4 = 880\ \text{W}$ and output power is $P_{out} = 20\times40 = 800\ \text{W}$ , so $\eta = (800/880)\times100\% = 90.9\%$ , and the power lost as heat is $880 - 800 = 80\ \text{W}$ , dissipated as copper, hysteresis, and eddy-current losses. This shows why every real transformer's efficiency is quoted below 100%, never above.

Common traps

Assuming current scales the same way as voltage — it scales inversely: a step-up transformer's secondary current is smaller than its primary current, not larger.
Thinking a transformer works on steady DC — it needs a changing flux, so it functions only with AC (or switched/pulsed DC); constant DC induces no EMF at all.
Mixing up eddy-current loss (reduced by laminating the core into thin insulated sheets) with hysteresis loss (reduced by choosing a soft magnetic core material with a narrow loop) — they are separate mechanisms with separate fixes.
Believing an ideal transformer changes the amount of power delivered — it changes voltage and current in inverse proportion but, ideally, conserves power exactly.

Ask the tutor: ask for a walkthrough connecting transformer losses to why India's power grid runs long-distance lines at very high voltage.

Why can a transformer step up voltage without breaking energy conservation?

It is not creating energy — it is trading current for voltage. An ideal transformer keeps input power equal to output power, $V_pI_p = V_sI_s$ , so whenever voltage goes up by some factor, current must come down by exactly the same factor to keep the product unchanged. Real transformers lose a small fraction as heat (typically 1 to 10 percent), which is why efficiency is always quoted below 100 percent, but the basic step-up mechanism itself conserves power by construction, not by violating it.

Why does high-voltage transmission actually save energy on power lines?

Transmission-line loss is $I^2R_{line}$ , set by the line's resistance and the current flowing through it, not by the voltage. For a fixed power $P = VI$ delivered to a city, raising the transmission voltage lowers the current in the same proportion, and since loss depends on the square of current, even a modest voltage increase cuts losses sharply. That is why step-up transformers push transmission voltage to hundreds of kilovolts right at the generating station, with step-down transformers bringing it back to safe household levels only near the consumer.

Does a transformer work if I connect its primary to a battery?

No, not in steady state. A transformer only induces an EMF in the secondary when the flux through the core is changing, and Faraday's law gives zero EMF for a constant flux. A battery connected directly to the primary produces a constant current and hence a constant flux once any initial surge settles, so the secondary reads zero volts after that brief transient. This is exactly why transformers are AC devices: the continuously reversing current keeps the flux changing at every instant.

Transformers

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads