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Here i am confused whether i have to use the formula for power is

  1. \$P_l=\frac{V_i^{2}}{Z_0}(1-\Gamma_{l}^2)\$

OR

  1. \$P_l=\frac{V_i^{2}}{2Z_0}(1-\Gamma_{l}^2)\$

if i am using the first formula answer is matching but power should be calculated for rms values so in my view power formula should be 2nd correct.

please help me with this confusion

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  • \$\begingroup\$ @pasabaporaqui, It's hard to make a load with imaginary impedance at DC. \$\endgroup\$
    – The Photon
    Commented Sep 22, 2017 at 16:08

1 Answer 1

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On an exam, I'd short-circuit this. Given \$Z_0\$ and \$Z_L\$, you can quickly calculate what the VSWR should be. Then check which of the four given answers is consistent with the expected VSWR.

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  • \$\begingroup\$ sir thanks for the answer....what you are saying i know this shortcut but this method will not give me exact value of voltage also its based on luck that only one option will be there that will satisfy that like this question that only option c is satisfying with the VSWR method as you said....according to no option should be correct if i use the formula 2. \$\endgroup\$
    – Rohit
    Commented Sep 22, 2017 at 16:35
  • \$\begingroup\$ @Rohit, The four given answers have different ratios \$v_{max}/v_{min}\$ so this method will not match more than one of the allowed answers. I would ignore the power information as not necessary to answer the question. \$\endgroup\$
    – The Photon
    Commented Sep 22, 2017 at 18:20

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