Ideal transformer

Problem

An ideal transformer feeds a $24 \ \mathrm{W}$ lamp at $12 \ \mathrm{V}$. The mains voltage on the primary is $230 \ \mathrm{V}$. Calculate: (a) the turns ratio $N_1/N_2$; (b) the secondary current $i_2$; (c) the primary current $i_1$.

Transformer: primary winding ($N_1$) and secondary winding ($N_2$) on the same ferromagnetic core.

Solution

(a) Turns ratio. In the ideal transformer the voltages are proportional to the turns: $\frac{N_1}{N_2} = \frac{V_1}{V_2} = \frac{230}{12} \approx \ev{19{,}2}$

(b) Secondary current. The lamp draws the whole power on the secondary: $i_2 = \frac{P}{V_2} = \frac{24}{12} = \ev{2{,}0 \ \mathrm{A}}$

(c) Primary current. In the ideal transformer power is conserved, $P_{\text{in}} = P_{\text{out}}$: $i_1 = \frac{P}{V_1} = \frac{24}{230} \approx \ev{0{,}104 \ \mathrm{A}}$

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Topics: Induzione elettromagnetica Concepts: Induttanza e autoinduzione · Corrente alternata · Legge di Faraday-Neumann-Lenz