Power split at a T-junction – Voltage at the node when a Z0 line feeds two parallel Z0 lines A transmission line of characteristic impedance Z0 feeds a lossless junction that splits into two identical lines, each also of Z0. What is the steady-state voltage at the junction (on each branch) relative to the incident voltage on the main line?

Introduction:
Power dividers implemented by directly paralleled transmission lines create impedance mismatch unless transformed. Understanding node voltages and reflections at such junctions is key for designing splitters and predicting mismatch loss and return loss.

Given Data / Assumptions:

• Main line characteristic impedance Z0.
• Junction splits into two identical Z0 lines in parallel (net input Z = Z0/2).
• Matched terminations far down the branches so no further reflections return.

Concept / Approach:

The effective load seen by the main line at the junction is Z_in = (Z0 ∥ Z0) = Z0/2. The reflection coefficient at the junction is Γ = (Z_in − Z0) / (Z_in + Z0) = (Z0/2 − Z0) / (Z0/2 + Z0) = −1/3. The node (transmitted) voltage equals incident plus reflected at the junction: V_node = V_inc * (1 + Γ) = V_inc * (1 − 1/3) = (2/3) V_inc. Each branch sees this same node voltage as its forward wave.

Step-by-Step Solution:

Verification / Alternative check:

Power conservation: reflected power fraction is |Γ|^2 = 1/9, so delivered is 8/9. Each branch takes 4/9 → total 8/9, matching conservation.

Why Other Options Are Wrong:

Zero or infinite are nonphysical; “equal to” ignores mismatch; one-third confuses node voltage with reflection magnitude.

Common Pitfalls:

Mixing voltage division with power split; forgetting that the junction is mismatched without a quarter-wave transformer or Wilkinson structure.

Final Answer:

two-thirds of the incident voltage.