Transformer

PHYInductance & Transformers

A transformer changes AC voltage by mutual induction between a primary and a secondary coil wound on a common laminated soft-iron core; it cannot change DC. Students compute the secondary voltage/current via the turns ratio, find efficiency from input/output power, and back out the core flux from the EMF equation $V_{rms} = 4.44\,fN\phi_{max}$ .

It is a guaranteed high-yield numerical and theory topic.

Turns / voltage / current ratio (ideal)

\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}

N_p, N_s

= primary/secondary turns;

V

= voltages;

I

= currents. Voltage scales WITH turns, current scales INVERSELY (ideal, 100% efficient).

Efficiency

\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{V_s I_s}{V_p I_p}

P_{in} = V_p I_p

(primary),

P_{out} = V_s I_s

(delivered to load);

\eta < 1

for a real transformer (multiply by 100 to express as a percentage).

EMF (flux) equation

V_{rms} = 4.44\, f N \phi_{max}

f

= supply frequency (Hz),

N

= turns,

\phi_{max}

= peak core flux (Wb); the factor

4.44 = 2\pi/\sqrt{2}

converts the peak rate-of-change of flux to an RMS voltage.

Step-up: $N_s > N_p$ raises voltage but lowers current; step-down: $N_s < N_p$ lowers voltage and raises current — power (ideal) is conserved.
The ideal turns relation assumes 100% flux linkage and no losses; use it only when the problem says 'ideal' or 'efficiency 100%'.
For a real transformer with efficiency $\eta$ , find the output power as $P_{out} = \eta\, P_{in}$ , then get $I_s = P_{out}/V_s$ — do NOT use $I_p/I_s = N_s/N_p$ for the current when $\eta \ne 1$ .
Secondary voltage still follows $V_s = V_p (N_s/N_p)$ even for a real transformer (it is fixed by induction); it is the current relation that breaks when there are losses.
Main losses: copper ( $I^2R$ in the windings), eddy currents (reduced by laminating the core), hysteresis (reduced by a soft-iron / silicon-steel core), and flux leakage.
Transformers work only on AC because a steady (DC) current gives $d\phi/dt = 0$ and hence no induced EMF in the secondary.
In the EMF equation $V_{rms}$ is the RMS secondary voltage while $\phi_{max}$ is the peak core flux; the 4.44 factor already accounts for this, so substitute $V_{rms}$ on the left and solve for $\phi_{max}$ on the right.

Where the marks go

Applying the ideal current ratio $I_p/I_s = N_s/N_p$ when efficiency is below 100% — compute $I_s$ from $P_{out}/V_s$ instead.
Inverting the turns ratio (using $N_p/N_s$ where $N_s/N_p$ is needed) and getting a step-down instead of a step-up.
Forgetting the 4.44 factor or writing it as $4.4$ or $\pi$ in the EMF equation.
Thinking a transformer steps up power as well as voltage — output power can never exceed input power.

How the board asks it

Numericalturns ratio with efficiency; secondary current from output power
A step-up transformer has $100$ turns in its primary and $1500$ turns in its secondary. The primary is connected to a $220\,V$ AC supply and the secondary delivers a current of $2\,A$ at an efficiency of $90\%$ . Calculate (i) the secondary voltage and (ii) the current drawn from the primary.
Numericalemf equation $V_{rms} = 4.44\,fN\phi_{max}$
The secondary coil of a transformer has $500$ turns and develops an RMS voltage of $220\,V$ when operated on a $50\,Hz$ supply. Calculate the maximum value of the magnetic flux in the core.
Give reasonsno induced emf for steady current since $d\phi/dt = 0$
Give a reason: A transformer cannot be used to step up a DC voltage.
Define / statecopper, eddy current, hysteresis and flux leakage losses
State two main sources of energy loss in a transformer and explain how each is reduced in its construction.
Applicationconservation of power; step-up lowers current
Explain why a step-up transformer, which raises the voltage, does not increase the power output, and state how its output current compares with its input current.
Assertion–Reasonlaminated soft-iron core reduces eddy-current loss
Assertion: The core of a transformer is made of laminated sheets of soft iron. Reason: Lamination reduces energy loss due to eddy currents in the core. Choose the correct option regarding these two statements.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.