A single-phase ideal transformer has a primary-to-secondary turns ratio of 4:1. If the secondary impedance is 2 $$Omega$$, the reflected impedance ($$Omega$$) as seen on the primary is most nearly:

A.  0.5
B.  2.0
C.  8.0
D.  32.0

Due to the presence of a transformer between a source and a load, the impedance of the load will appear different to the source compared to its actual value. The reflected impedance is a term used to describe the impedance of the load as seen from the source.

The reflected impedance depends on the turns ratio of the transformer, according the following equation:

$$\begin{gathered} R_P=R{\left(\frac{N_P}{N_S}\right)}^2 \\ {R}_{P = }Reflected Impedance \\ R = Actual Load Impedance \\ {N}_P = Number of turns of primary winding \\ {N}_S= Number of turns of secondary winding \end{gathered}$$

Since our primary-to-secondary turns ratio is given as 4:1:

$\frac{N_P}{N_S}=\frac{4}{1}=4$

Solving for the reflected impedance given a secondary impedance of 2Ω:

$$\begin{gathered} R_P=R{\left(\frac{N_P}{N_S}\right)}^2 \\ {R}_P=\left(2\right){\left(4\right)}^2 \\ {R}_P=32\Omega \end{gathered}$$

Therefore, our reflected impedance is equal to 32Ω, which is answer D.