Introduction / Context:
For a lossless transmission line, the open-circuit and short-circuit input impedances are related through the characteristic impedance. This question tests your ability to use that core relationship to compute the short-circuit input impedance from a known open-circuit input impedance.
Given Data / Assumptions:
- Z0 = 50 Ω (characteristic impedance).
- Open-circuit input impedance Zoc = 100 + j150 Ω.
- Line assumed lossless so classical relationships apply.
Concept / Approach:
For a lossless line at the same physical length and frequency, Zoc and Zsc satisfy Zoc * Zsc = Z0^2. Therefore Zsc = Z0^2 / Zoc. Complex division can be performed by multiplying numerator and denominator by the complex conjugate of Zoc.
Step-by-Step Solution:
Step 1: Compute Z0^2 = 50^2 = 2500 Ω^2.Step 2: Write Zsc = 2500 / (100 + j150).Step 3: Multiply by conjugate: 2500(100 − j150) / [(100)^2 + (150)^2] = (250000 − j375000) / 32500.Step 4: Real part = 250000 / 32500 ≈ 7.69 Ω; Imag part = −375000 / 32500 ≈ −11.54 Ω.Step 5: Thus Zsc ≈ 7.69 − j11.54 Ω.
Verification / Alternative check:
Check product: Zoc * Zsc ≈ (100 + j150)(7.69 − j11.54) ≈ 2500 (within rounding), confirming correctness.
Why Other Options Are Wrong:
50 Ω: This would imply a matched termination, not a shorted input.100 + j150 Ω: That is the open-circuit value, not short-circuit.7.69 + j11.54 Ω or 11.54 − j7.69 Ω: Wrong sign/magnitude from incorrect complex division.
Common Pitfalls:
Forgetting to use the complex conjugate in division; mixing up Zoc and Zsc roles; assuming resistive results when reactive parts are essential.
Final Answer:
7.69 - j11.54 Ω
Discussion & Comments