Step 1 — reduce the electrical length. A transmission line repeats its behaviour every half wavelength, so only the remainder after removing whole half-wavelengths matters:

\(2.25\lambda = 2\lambda + 0.25\lambda\)

The 2λ contributes nothing, leaving a quarter-wave section. This single observation is the heart of the problem.

Step 2 — apply the quarter-wave transformer relation. For a lossless line of length λ/4,

\(Z_{in}=\dfrac{Z_0^{2}}{Z_L}=\dfrac{50^{2}}{75}=\dfrac{2500}{75}\)

\(Z_{in}=33.33\ \Omega\)

So statement (c) is correct and statement (a) is wrong. Z_in = 50 Ω would only arise if the line were matched (Z_L = Z₀) or an exact multiple of a half wavelength with a 50 Ω load — neither is the case here.

Step 3 — find the voltage at the input of the line. The generator, its 50 Ω internal impedance and Z_in form a simple divider:

\(V_{in}=V_g\dfrac{Z_{in}}{Z_g+Z_{in}}=30\times\dfrac{33.33}{50+33.33}=30\times\dfrac{33.33}{83.33}\)

\(V_{in}=12\ \text{V}\)

That 12 V is the value quoted in statement (d), so (c) and (d) go together — option 3.

Step 4 — the check that makes the quarter-wave action visible. Power delivered to the line is

\(P=\dfrac{V_{in}^{2}}{Z_{in}}=\dfrac{144}{33.33}=4.32\ \text{W}\)

and since the line is lossless, all of it reaches the 75 Ω load. Notice how the quarter-wave section has transformed the 75 Ω load down to 33.33 Ω at the input: an impedance above Z₀ is reflected to one below Z₀, because the transformer relation is reciprocal about Z₀.

Why the λ/4 transformer matters in practice. Choosing \(Z_0=\sqrt{Z_{in}Z_L}\) matches any two real impedances at a single frequency — the standard way of matching an antenna feed. Here the reverse is happening: an unmatched load is being transformed, giving a reflection coefficient \(\Gamma=(75-50)/(75+50)=0.2\) and a standing wave ratio of 1.5.

Hence, the correct pair is (c) and (d).