Question

A 30 ohm load is connected to the 6 volt secondary of a transformer with a 120 voltprimary. What ohmic value does the load appear to be to the source

Answer 1

12000 ohm

Explanation:

We have the following:

`I_1\cdot V_1=I_2\cdot V_2`

Ohms law states that:

`\begin{gathered} I_1=(V_1)/(Z_1) \\ I_2=(V_2)/(Z_2) \end{gathered}`

We replace the previous equations in the first one, and the following remains:

`\begin{gathered} (V_1)/(Z_1)\cdot V_1=(V_2)/(Z_2)\cdot V_2\rightarrow\text{ we multiply} \\ (V^2_1)/(Z_1)_{}=(V^2_2)/(Z_2) \\ \text{solving for Z1} \\ Z_1\cdot V^2_2=Z_2\cdot V^2_1\rightarrow\text{ we multiply cross} \\ Z_1=Z_2\cdot\frac{V^2^{}_1_{}}{V^2^{}_2}_{}\rightarrow\text{ we divide by V2 squared} \\ \text{ replacing the determined values} \\ Z_1=30\cdot(120^2)/(6^2) \\ Z_1=12000 \end{gathered}`