Introduction / Context:
Quarter-wave sections can transform impedances dramatically. An open circuit at one end becomes a short at a distance λ/4 away, producing voltage magnification at the open end. This is a classic standing-wave scenario in transmission-line theory.
Given Data / Assumptions:
- Line: Z0 = 300 Ω, length = λ/4.
- Source: Vs = 10 V (Thevenin), Rs = 50 Ω.
- Load end: open-circuited (ZL → ∞).
Concept / Approach:
A λ/4 open transforms to a short at the input: Zin(λ/4) = 0. Thus the source sees (almost) a short, setting a large current. Along the line, the standing wave has a voltage maximum at the open end of magnitude related to the forward-wave amplitude and Z0.
Step-by-Step Solution:
Step 1: Input impedance at the source-end of λ/4 open: Zin = 0 (short).Step 2: Source current: Is = Vs / (Rs + Zin) = 10 / 50 = 0.2 A.Step 3: Forward-wave voltage amplitude on the line is approximately V+ ≈ Is * Z0 = 0.2 * 300 = 60 V (for a shorted input, the standing-wave peak equals this at the open end).Step 4: Therefore the voltage magnitude at the open end is 60 V.
Verification / Alternative check:
Energy consistency: The source drives a low input impedance, resulting in high current but the λ/4 transformation yields high voltage at the open end—consistent with quarter-wave resonator behavior.
Why Other Options Are Wrong:
10 V or 5 V: Ignore the quarter-wave impedance transformation and standing-wave magnification.60/7 volt: Unrelated scaling; not supported by Z0/Rs ratio.30 V: Would correspond to Z0 = 150 Ω, not 300 Ω.
Common Pitfalls:
Forgetting that a λ/4 section transforms an open to a short; confusing voltage and current maxima positions along the line.
Final Answer:
60 V
Discussion & Comments