Esercizio 5

DiAlfredo Di Fiore

N_1=3720,\quad N_2=124,\quad V_1=300\ \text{V},\quad R=10\ \Omega,\quad P_2=\tfrac{9}{10}P_1\ (\eta=0{,}9).

Tensione al secondario
\frac{V_2}{V_1}=\frac{N_2}{N_1} \ \Rightarrow V_2=V_1\frac{N_2}{N_1} =300\cdot\frac{124}{3720} =10\ \text{V}.

Corrente e potenza al secondario (carico puramente resistivo)
I_2=\frac{V_2}{R}=\frac{10}{10}=1\ \text{A},\qquad P_2=\frac{V_2^2}{R}=\frac{10^2}{10}=10\ \text{W}.

Potenza al primario e corrente primaria
P_1=\frac{P_2}{\eta}=\frac{10}{0{,}9}=\frac{10}{9}\cdot 10=11{,}11\ \text{W}, \qquad I_1=\frac{P_1}{V_1}=\frac{11{,}11}{300}=0{,}037\ \text{A}\approx 37\ \text{mA}.

Risultati
I_2=1\ \text{A},\qquad I_1\approx 37\ \text{mA},\qquad V_2=10\ \text{V},\qquad P_2=10\ \text{W},\ P_1\approx 11{,}11\ \text{W}.

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