Question

a source with a resistive impedance of $100\\ \omega$ is connected to the primary coil of a transformer, and a resistance $r$ is connected to the secondary coil. if the transformer has 500 turns in its primary coil and 100 turns in its secondary coil, the greatest power will be dissipated in the resistor if $r$ is equal to
$\bigcirc 4\\ \omega$.
$\bigcirc 0.25\\ \omega$.
$\bigcirc 20\\ \omega$.
$\bigcirc 0\\ \omega$.
$\bigcirc 100\\ \omega$.

Explanation:

Step1: Recall the formula for impedance transformation in a transformer

The impedance transformation formula for a transformer is \(Z_{eq}=(\frac{N_1}{N_2})^2Z_2\), where \(N_1\) is the number of turns in the primary coil, \(N_2\) is the number of turns in the secondary coil, \(Z_{eq}\) is the equivalent impedance seen from the primary side, and \(Z_2\) is the impedance in the secondary side. For maximum power transfer, the equivalent impedance \(Z_{eq}\) should be equal to the source impedance \(Z_s\).

Step2: Substitute the given values into the formula

Given \(Z_s = 100\ \Omega\), \(N_1=500\), \(N_2 = 100\). We know that \(Z_s=(\frac{N_1}{N_2})^2R\). Rearranging for \(R\), we get \(R=Z_s(\frac{N_2}{N_1})^2\).

Substitute the values: \(R = 100\times(\frac{100}{500})^2\).

First, calculate \((\frac{100}{500})^2=\frac{1}{25}\). Then \(R = 100\times\frac{1}{25}=4\ \Omega\).

Answer:

4 Ω.