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

ANSWER:

12000 ohm

STEP-BY-STEP EXPLANATION:

We have the following:

$I_1\cdot V_1=I_2\cdot V_2$

Ohms law states that:

$\begin{gathered} I_1=\frac{V_1}{Z_1} \\ I_2=\frac{V_2}{Z_2} \end{gathered}$

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

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