Coils of transformer

The primary coil of the transformer has 400 turns. A current of 1.5 A passes through it and is connected to a voltage of 220 V. For the secondary coil, find the voltage, current, and number of turns if the transformation ratio k = 0.1.

Final Answer:

U₂ =  22 V
I₂ =  15 A
N₂ =  40

Step-by-step explanation:

$$\begin{gathered} N_1=400 \\ {I}_1=1.5\text{A} \\ {U}_1=220\text{V} \\ k=0.1 \\ k=\frac{N_2}{N_1}=\frac{U_2}{U_1}=\frac{I_1}{I_2} \\ {U}_2=k\cdot U_1=0.1\cdot 220=22\text{V} \end{gathered}$$

$I_2=I_1/k=1.5/0.1=15\text{A}$

$$\begin{gathered} N_2=k\cdot N_1=0.1\cdot 400=40 \\ \text{ Verifying Solution: } \\ {P}_1=U_1\cdot I_1=220\cdot 1.5=330\text{W} \\ {P}_2=U_2\cdot I_2=22\cdot 15=330\text{W} \\ {P}_1=P_2 \end{gathered}$$

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