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Physics · Alternating Current

Transformers

A transformer changes an alternating voltage from one value to a greater or smaller one using the principle of mutual induction. NCERT §7.8 builds the device from the turns-ratio relation $\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$ and the ideal power-conservation rule $V_pI_p=V_sI_s$. For NEET, transformers reward a single clear idea — voltage and current trade off so power stays fixed — and a tidy list of where real devices lose energy.

What a transformer does

For many purposes it is necessary to change, or transform, an alternating voltage from one value to a greater or smaller one. NCERT §7.8 introduces the transformer as the device that does this, working entirely on the principle of mutual induction. It consists of two sets of coils, insulated from each other and wound on a soft-iron core — either one coil on top of the other, or on separate limbs of the core.

One coil, the primary, has $N_p$ turns and is usually the input coil. The other, the secondary, has $N_s$ turns and is usually the output coil. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux. This flux links the secondary and induces an emf in it; the value of that emf depends on how many turns the secondary has.

Figure 1 — Construction Laminated soft-iron core Primary $N_p$ turns Secondary $N_s$ turns AC input $\phi$

The analysis assumes an ideal transformer: the primary has negligible resistance, and all the flux in the core links both windings. These idealisations let us derive clean relations; real devices depart from them only slightly, since a well-designed transformer can exceed 95% efficiency.

Turns ratio and the ideal relations

Let $\phi$ be the flux in each turn of the core at time $t$. By Faraday's law the induced emf in the secondary, with $N_s$ turns, is $\varepsilon_s=-N_s\,\dfrac{d\phi}{dt}$, and the back emf in the primary, with $N_p$ turns, is $\varepsilon_p=-N_p\,\dfrac{d\phi}{dt}$. Because the primary resistance is negligible, $\varepsilon_p=v_p$ (otherwise the primary current would be infinite). If the secondary current is small, $\varepsilon_s=v_s$. Dividing one relation by the other eliminates $d\phi/dt$:

$$\frac{v_s}{v_p}=\frac{N_s}{N_p}$$

This is the turns-ratio relation. It rests on three assumptions stated in NCERT: the primary resistance and current are small; the same flux links both windings because very little escapes the core; and the secondary current is small. Since $v$ and $i$ oscillate at the source frequency, the same ratio holds for rms values and for amplitudes.

Quantity ratioIdeal relationMeaning
VoltageVs/Vp = Ns/NpOutput voltage scales with the turns ratio
CurrentIs/Ip = Np/NsOutput current scales inversely with turns ratio
CombinedVs/Vp = Ns/Np = Ip/IsVoltage rises exactly as current falls
PowerVp Ip = Vs IsInput power equals output power (ideal)

Combining the turns-ratio relation with power conservation (next section) gives the master statement of the topic:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$$

Step-up versus step-down

The turns ratio decides whether the device raises or lowers the voltage. If the secondary has more turns than the primary, $N_s>N_p$, the voltage is stepped up, $V_s>V_p$, and the device is a step-up transformer. Because power is fixed, the secondary then carries less current than the primary, $I_sstep-down transformer.

Figure 2 — Step-up vs step-down STEP-UP $N_p$ small $N_s$ large $V_s\!\uparrow\;,\;I_s\!\downarrow$ STEP-DOWN $N_p$ large $N_s$ small $V_s\!\downarrow\;,\;I_s\!\uparrow$

NCERT gives a concrete case: a primary of 100 turns and a secondary of 200 turns has $N_s/N_p=2$, so a 220 V input at 10 A steps up to a 440 V output at 5.0 A. The voltage doubled and the current halved; the product $V\!I$ — the power — stayed at 2200 W.

NEET Trap

"Step-up means more energy" and "transformers run on DC"

A step-up transformer raises the voltage but lowers the current by exactly the same factor, so the power is unchanged — it does not create energy. A second common error is to assume a transformer can step a battery's dc voltage. It cannot: a steady current gives a constant flux, $d\phi/dt=0$, so no emf is induced in the secondary, and the low-resistance primary would draw a damaging current.

Transformers conserve power ($V_pI_p=V_sI_s$) and require AC, because induction needs a changing flux.

Power conservation in an ideal transformer

If the transformer is assumed to be 100% efficient, the input power equals the output power. Since instantaneous power is $p=iv$, this is

$$i_p v_p = i_s v_s$$

Combining this with the turns-ratio relation gives the inverse current relation. Where the voltage is multiplied by $N_s/N_p$, the current is divided by the same factor:

$$V_s=\left(\frac{N_s}{N_p}\right)V_p \qquad\text{and}\qquad I_s=\left(\frac{N_p}{N_s}\right)I_p$$

∿
Build the foundation

Transformer ratings are quoted in rms values. Revisit AC applied to a resistor to see why rms is the working measure for ac voltage and current.

NCERT notes in its "Points to Ponder" that a step-up transformer changing a low voltage into a high voltage does not violate conservation of energy precisely because the current is reduced in the same proportion. This single sentence is the conceptual core of every transformer MCQ.

Energy losses in a real transformer

The ideal relations assume no losses. Real transformers lose a small amount of energy through four mechanisms identified in NCERT, each with a standard countermeasure.

LossCauseReduced by
Flux leakageNot all primary flux passes through the secondary (poor core design or air gaps)Winding the primary and secondary one over the other
Copper loss ($I^2R$)Resistance of the winding wire produces Joule heatingUsing thick wire in high-current, low-voltage windings
Eddy currentsAlternating flux induces circulating currents in the iron core, heating itUsing a laminated core
HysteresisEnergy spent repeatedly reversing the core's magnetisation appears as heatUsing a material with low hysteresis loss

The copper loss is named for the winding metal and scales as $I^2R$, so it matters most in the high-current secondary of a step-down transformer; thick wire keeps $R$ small. Eddy-current loss is cut by lamination — building the core from thin insulated sheets so induced currents cannot circulate freely. Hysteresis loss depends on the magnetic material, which is why soft iron with a narrow hysteresis loop is chosen. Even with all four present, a good transformer exceeds 95% efficiency.

NEET Trap

Matching each loss to its fix

Examiners often pair a loss with the wrong remedy. Lamination addresses eddy currents, not hysteresis. Thick wire addresses copper (I²R) loss, not flux leakage. A low-hysteresis material addresses hysteresis. Winding the coils together addresses flux leakage.

Lamination → eddy currents; thick wire → copper loss; soft magnetic material → hysteresis; coils wound together → flux leakage.

Transformers in power transmission

The large-scale transmission and distribution of electrical energy over long distances relies on transformers. The reason traces directly to the transmission-line loss, which is $I^2R$ where $R$ is the resistance of the line. For a fixed amount of transmitted power, raising the voltage lowers the current, and because the loss depends on the square of the current, a modest cut in current produces a large cut in wasted heat.

Figure 3 — Transmission chain G Generator ↑ Step-up High V, low I small $I^2R$ loss ↓ Sub-station ↓ Utility pole ~240 V home

The chain runs as NCERT describes it: the generator output is stepped up so the current — and therefore the $I^2R$ loss — is reduced for the long-distance leg. The high-voltage line carries the power to an area sub-station near the consumers, where the voltage is stepped down. It is stepped down again at distributing sub-stations and at utility poles before a supply of about 240 V reaches homes. This ability to convert ac voltages cheaply and efficiently is precisely why ac, not dc, is used for transmission.

Worked example

NCERT-based · §7.8

The primary coil of a transformer has 100 turns and the secondary has 200 turns. A 220 V input is supplied at 10 A. Find the secondary voltage and current, and confirm the power is conserved.

Turns ratio. $\dfrac{N_s}{N_p}=\dfrac{200}{100}=2$, so this is a step-up transformer.

Secondary voltage. $V_s=\dfrac{N_s}{N_p}\,V_p = 2\times 220 = 440\ \text{V}.$

Secondary current. $I_s=\dfrac{N_p}{N_s}\,I_p = \dfrac{1}{2}\times 10 = 5.0\ \text{A}.$

Power check. Input $V_pI_p = 220\times 10 = 2200\ \text{W}$; output $V_sI_s = 440\times 5.0 = 2200\ \text{W}.$ The power is unchanged, as required for an ideal transformer.

Quick Recap

Transformers in one screen

  • A transformer changes an ac voltage by mutual induction; it does not work on dc.
  • Turns-ratio relation: $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$.
  • Ideal power conservation: $V_pI_p=V_sI_s$; step-up raises $V$ but lowers $I$, so power is fixed.
  • Step-up: $N_s>N_p$, $V_s>V_p$, $I_s
  • Four losses: flux leakage, copper ($I^2R$), eddy currents (cut by lamination), hysteresis (cut by soft material).
  • Power is transmitted at high voltage, low current to slash the line $I^2R$ loss, then stepped down near consumers.

NEET PYQ Snapshot — Transformers

Real transformer questions from recent NEET papers — turns ratio and ideal power conservation.

NEET 2024 · Q.9

In an ideal transformer, the turns ratio is $N_P/N_S = 1/2$. The ratio $V_S : V_P$ is equal to (the symbols carry their usual meaning):

  1. 1 : 2
  2. 2 : 1
  3. 1 : 1
  4. 1 : 4
Answer: (2) 2 : 1

For an ideal transformer $\dfrac{V_S}{V_P}=\dfrac{N_S}{N_P}$. Given $\dfrac{N_P}{N_S}=\dfrac12$, so $\dfrac{N_S}{N_P}=2$, giving $V_S:V_P = 2:1$.

NEET 2021 · Q.45

A step-down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

  1. 4 A
  2. 0.2 A
  3. 0.4 A
  4. 2 A
Answer: (2) 0.2 A

For an ideal transformer input power equals output power: $V_pI_p = V_sI_s = 44$ W. So $220\times I_p = 44$, giving $I_p = 0.2$ A.

NEET 2023 · Q.29

A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?

  1. 3.7 A
  2. 0.27 A
  3. 2.7 A
  4. 0.37 A
Answer: (2) 0.27 A

Ideal transformer: $P_{\text{input}}=P_{\text{output}}$, so $220\times I_p = 60$ W, giving $I_p = 0.27$ A.

FAQs — Transformers

Common doubts on turns ratio, power conservation, and losses.

Why do transformers work only with alternating current and not with direct current?
A transformer operates by mutual induction, which requires a continuously changing magnetic flux through the core. An alternating primary current produces an alternating flux, which induces an emf in the secondary. A steady direct current produces a constant flux, so dφ/dt is zero and no emf is induced in the secondary. A transformer therefore cannot step direct current up or down. Connecting a transformer primary to a steady dc source also draws a large current because the primary has very low resistance, which can damage the winding.
Does a step-up transformer increase electrical power?
No. A step-up transformer raises the secondary voltage but lowers the secondary current in the same proportion, so the power is unchanged. For an ideal transformer the input power equals the output power, VpIp = VsIs. A step-up transformer changing low voltage into high voltage does not violate conservation of energy because the current is reduced by the same factor that the voltage is increased.
What is the turns-ratio relation for an ideal transformer?
For an ideal transformer the secondary-to-primary voltage ratio equals the turns ratio: Vs/Vp = Ns/Np. Because input power equals output power, the current ratio is the inverse: Ip/Is = Ns/Np, or equivalently Is/Ip = Np/Ns. Combining these gives Vs/Vp = Ns/Np = Ip/Is.
What are the four main energy losses in a real transformer?
Real transformers lose energy through four mechanisms. Flux leakage occurs when not all the primary flux links the secondary; it is reduced by winding the coils one over the other. Copper loss is the I²R heating in the resistance of the windings, reduced by using thick wire in high-current windings. Eddy-current loss arises from currents induced in the iron core, reduced by using a laminated core. Hysteresis loss is energy spent in repeatedly reversing the magnetisation of the core, reduced by using a magnetic material with low hysteresis loss.
Why is electrical power transmitted at high voltage over long distances?
The power lost as heat in the transmission line is I²R, where R is the line resistance. Stepping the generator voltage up reduces the current for the same transmitted power, which cuts the I²R loss sharply. The power is transmitted at high voltage and low current to a sub-station near the consumers, where it is stepped down, and stepped down further at distributing sub-stations before a supply of about 240 V reaches homes.
Why is the core of a transformer made of laminated soft iron?
Soft iron is used because it has high permeability, so it concentrates the magnetic flux and links it efficiently between the coils, and it has low hysteresis loss. The core is laminated, that is built from thin insulated sheets, to interrupt the paths of induced eddy currents in the core. Thinner laminations mean smaller eddy currents and less heating, which improves efficiency.
Related Topics
Alternating Current Back to the full chapter overview and all subtopics. AC Applied to a Resistor rms values and Joule heating — the working measures for ac. Phasors Rotating-vector representation of ac voltage and current. AC Applied to an Inductor Inductive reactance and the lagging current. AC Applied to a Capacitor Capacitive reactance and the leading current. Series LCR Circuit & Resonance Impedance, phase angle and the resonant frequency. Power in AC Circuit & Power Factor Average power, the power factor and wattless current.