A charger for a phone contains a small transformer. It is connected to the 230 V a.c. mains supply and must output a secondary voltage of Vs = 5 V to charge the phone's battery. The primary coil of the transformer has Np = 2300 turns. (a) Explain, in terms of the number of turns…

Answer: ((a)) A step-down transformer has fewer turns on the secondary coil than on the primary coil (Ns < Np), so by Vp/Vs = Np/Ns the output (secondary) voltage is smaller than the input (primary) voltage. ((b)) Ns = 50 turns ((c)) A transformer needs a continuously changing magnetic field to induce an e.m.f. in the secondary coil; a d.c. supply produces a constant, unchanging field (after the initial instant), so almost no e.m.f. would be induced in the secondary coil, and no voltage transformation would…. Part (a): Step-down vs step-up For an ideal transformer, Vp/Vs = Np/Ns. A step-down transformer reduces the voltage, so Vs < Vp. From the equation above, this happens when Ns < Np. That is, when the secondary coil has fewer turns than the…

IGCSE Physics 0625: Step-Down Transformer Turns Ratio Calculation

A charger for a phone contains a small transformer. It is connected to the $230\text{ V}$ a.c. mains supply and must output a secondary voltage of $V_s = 5\text{ V}$ to charge the phone's battery. The primary coil of the transformer has $N_p = 2300$ turns.

(a) Explain, in terms of the number of turns on the primary and secondary coils, what makes a transformer a step-down transformer rather than a step-up transformer. [1]

(b) Calculate the number of turns $N_s$ required on the secondary coil, so that the transformer outputs $5\text{ V}$ from the $230\text{ V}$ primary supply. [3]

(c) State two separate reasons why this transformer would not work correctly if it were connected directly to a d.c. supply instead of an a.c. supply. [2]

Part (a): Step-down vs step-up

For an ideal transformer, $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$ .

A step-down transformer reduces the voltage, so $V_s < V_p$ . From the equation above, this happens when $N_s < N_p$ . That is, when the secondary coil has fewer turns than the primary coil. (A step-up transformer has the opposite: more turns on the secondary than the primary.)

Part (b): Finding the number of secondary turns

Start from the transformer equation:

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

Rearrange to make $N_s$ the subject:

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

Substitute $N_p = 2300$ , $V_s = 5\text{ V}$ and $V_p = 230\text{ V}$ :

$N_s = 2300 \times \frac{5}{230} = 2300 \times \frac{1}{46}$

$N_s = \boxed{50}$

So the secondary coil needs $50$ turns.

Part (c): Why the transformer needs a.c. not d.c.

A transformer works because a continuously changing magnetic field in the iron core links the secondary coil and induces an e.m.f. in it. With an a.c. supply, the current (and so the field) is constantly changing direction and size, which keeps inducing an e.m.f.

With a d.c. supply instead:

The current, and so the magnetic field it produces, would quickly settle to a constant value. A constant field is not a changing field, so (apart from a brief instant when the supply is first switched on or off) almost no e.m.f. would be induced in the secondary coil. The charger would not work.
Without the changing field to limit it (as happens with a.c.), a steady d.c. current in the primary coil would be restricted only by the coil’s low resistance, so it could grow very large, overheating and potentially damaging the coil.

Final answers

(a) Step-down: $N_s < N_p$ (fewer turns on the secondary than the primary).
(b) $N_s =$ $50$ turns.
(c) Any two of: no continuously changing field, so (almost) no e.m.f. is induced in the secondary; the transformer does not transform the voltage at all with d.c.; the primary coil could carry an excessive current and overheat, since there is no changing field to limit it.