A single phase step-down transformer has a high voltage primary winding and a low voltage secondary winding. The secondary winding is connected to a load impedance \(Z_2\). The impedance \(Z_2\) when referred to the primary winding is denoted by \(Z_2'\). Which of the following conditions is true?
Show Hint
Recall that an impedance on the secondary is referred to the primary by multiplying by the square of the turns ratio, \(Z_2' = a^2 Z_2\) with \(a = N_1/N_2\).
Step 1: Understanding the Concept.
Instead of working with the general formula right away, it helps to test the claim with an actual numeric example, since a step-down transformer always has more primary turns than secondary turns.
Step 2: Key Formula or Approach.
Take a transformer with turns ratio $a = \frac{N_1}{N_2} = 4$, which is a valid step-down ratio (primary voltage is 4 times the secondary voltage). The rule for shifting an impedance from secondary to primary is $Z_2' = a^2 Z_2$.
Step 3: Detailed Explanation.
Suppose the load on the secondary is $Z_2 = 10\ \Omega$. Referred to the primary, this becomes
\[ Z_2' = a^2 Z_2 = (4)^2 \times 10 = 16 \times 10 = 160\ \Omega \]
Comparing the two: $Z_2' = 160\ \Omega$ is much larger than $Z_2 = 10\ \Omega$. This is not a coincidence of the numbers chosen. Since $a = N_1/N_2 > 1$ for any step-down transformer (more primary turns than secondary turns), the multiplying factor $a^2$ is always greater than 1, so $Z_2'$ always comes out bigger than $Z_2$, regardless of which specific step-down ratio is used.
Checking the other options against this example confirms they fail: $Z_2' \neq Z_2$ rules out (A), and $Z_2'$ is nowhere close to a fraction of $Z_2$ (it is 16 times bigger, not between $0.5Z_2$ and $Z_2$, and not less than $0.5Z_2$), which rules out (B) and (D).
Step 4: Final Answer.
For a step-down transformer, the impedance referred to the primary is always scaled up by a factor greater than 1, so $Z_2' > Z_2$, matching option (C).
\[ \boxed{Z_2' > Z_2} \]
