Open-circuit reflection at a transmission-line termination: Which statement correctly describes the voltage and current at the open end?

Difficulty: Easy

Correct Answer: The voltage at the termination is twice the incident voltage

Explanation:


Introduction / Context:
When a traveling wave reaches a mismatched termination, part or all of it reflects. An open circuit presents infinite impedance, resulting in a specific reflection behavior that sets the boundary conditions at the load.



Given Data / Assumptions:

  • Lossless transmission line for clarity.
  • Open-circuit load ZL → ∞.
  • Incident voltage wave Vi with reflection coefficient Γ = +1 at the load.


Concept / Approach:

At an open, the reflected voltage is in phase with the incident voltage. Therefore, at the very end of the line V_total = Vi + Vr = 2Vi. Current must be zero at the open because no conduction path exists: I_total = Ii + Ir = 0. This is the standard boundary condition used in standing-wave analysis.



Step-by-Step Solution:

Compute Γ = (ZL − Z0) / (ZL + Z0) → +1 for ZL → ∞.Add phasors at the termination: V_end = Vi + Vr = 2Vi.Current phasors cancel: I_end = Vi/Z0 − Vi/Z0 = 0.Hence voltage doubles while current becomes zero at the open.


Verification / Alternative check:

Smith chart analysis places the open at the rightmost point (Γ = +1) with voltage maxima and current minima at the load, matching the derived result.



Why Other Options Are Wrong:

  • Current doubling or both doubling violate the open-circuit boundary condition.
  • Zero voltage corresponds to a short circuit, not an open.
  • Unchanged values ignore reflections entirely.


Common Pitfalls:

Mixing up the signs for current reflection and confusing open- versus short-circuit cases.



Final Answer:

The voltage at the termination is twice the incident voltage

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