Open circuited quarter wave transmission line

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reddvoid
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Homework Statement



Scanned exam question Q.85 with four answer options (A) 10 V, (B) 5 V, (C) 60 V, (D) 60/7 V, beside a diagram of a 10 V, 50 Ω source feeding a line labelled Z0 = 300 Ω, length λ/4, open at the far end.


The Attempt at a Solution


I m getting 10V as one side is O.C so the whole voltage come across it...correct me where i m going wrong
 
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  • Printed exam question Q.85 with a diagram of a quarter-wave transmission line, Z0 = 300 Ω, fed by a 10 V, 50 Ω source and open-circuited at the far end, with answer choices 10 V, 5 V, 60 V and 60/7 V.
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You're going about it wrong. First calculate the impedance at the open end of the transmission line. Hint, it's not infinite. A mismatched 1/4 wave transmission line acts like a transformer. After you calculate the impedance at the open end, then calculate the ratio of that impedance to the 50 ohms of the generator. Lastly, how does the ratio of voltages compare to the ratio of impedances of a transformer? If you first find the ratio of impedances then from that you can find the ratio of voltages. Once you've done that you would just multiply the input voltage by the voltage ratio.
 
I would use ABCD parameters, concatenating the ABCD parameters for the 50 ohm source resistor with the abcd parameters for the xmsn line. This method can handle any kind of oddball mismatches, such as in this case the 300 ohm xmsn line to the 50 ohm source, and any desired length of line (here λ/4) and far-end termination impedance (here ∞).

I'm sure that's overkill here & that there's an esier way, as skeptic2 is suggesting. Just pointing out that abcd parameters can handle the most general cases.