Problem — Ideal transformer

Problem

An ideal transformer has $N_1 = 1000$ turns on the primary and $N_2 = 50$ turns on the secondary. The primary is fed from the Italian mains, $V_1 = 230\;\text{V}$ rms, and the secondary is closed on a resistive load $R = 1{,}3\;\Omega$. Calculate the voltage $V_2$ on the secondary and the current $i_2$ flowing in the load.

Solution

Setup. In an ideal transformer the ratio between the voltages equals the ratio between the numbers of turns (a consequence of Faraday-Neumann’s law applied to the flux common to the two windings):

$\frac{V_2}{V_1} = \frac{N_2}{N_1}$

Secondary voltage.

$V_2 = V_1\,\frac{N_2}{N_1} = 230\cdot\frac{50}{1000} = 11{,}5\;\text{V}$

Current in the load. Applying Ohm’s law to the resistor:

$i_2 = \frac{V_2}{R} = \frac{11{,}5}{1{,}3} \approx 8{,}8\;\text{A}$

$\ev{V_2 = 11{,}5\;\text{V}, \qquad i_2 \approx 8{,}8\;\text{A}}$

The transformer steps down the voltage (turns ratio $20:1$) and, correspondingly, allows more intense currents in the secondary than in the primary.

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Topics: Onde elettromagnetiche Concepts: Legge di Faraday-Neumann-Lenz · Corrente alternata