I am trying to solve the 2013 paper set by ISRO for electrical engineers. I wanted to verify if my answer to Question no 14 is correct.


The maximum RMS current is 20A, so maximum tolerable RMS voltage will be \$E_{max} = 20 \times Z_o = 20 \times 50 = 10^3\ \text{volts} \$, from this the power we get is also \$ 20 \times 10^3 \ \text{watts}\$.

Also total power transmitted into the load is given to us \$ P_T = (1 - |\Gamma|^2)P_i = 10 \times 10^3 \ \text{watts}\$

So we can find the reflection coefficient \$ \Gamma = \frac{1}{\sqrt{2}}\$ thus

$$ VSWR = \frac{1+|\Gamma|}{1-|\Gamma|} = 5.828$$

Actually I'm not getting any answer in the options.

I am not clear if I was right in thinking that the incident power

$$P_{incident} = \frac{E_{max}^2}{Z_o}$$

An alternative approach,

if I know max current flowing into the load is 20 A,

$$ 20^2Z_L = 10^4 $$

So we can find \$Z_L\$ = 25

$$\Gamma= \frac{Z_L - Z_o}{Z_L + Z_o}= \frac{25-50}{25+50} = -0.33$$

$$ VSWR = \frac{1+|\Gamma|}{1-|\Gamma|} = (0.5)^{-1} = 2$$

I really was not truly happy with any of my approaches to this problem.

  • \$\begingroup\$ You are saying the accepted answer is D? \$\endgroup\$ Mar 3, 2019 at 13:02
  • \$\begingroup\$ They have not released the answer key, so I don't know the correct answer. I just showed my attempt at it. I might be totally wrong. \$\endgroup\$
    – Aditya P
    Mar 3, 2019 at 13:07
  • \$\begingroup\$ Your calculation says VSWR=1, but option D is 2.5. \$\endgroup\$
    – Hearth
    Mar 3, 2019 at 14:01
  • \$\begingroup\$ Omg, I can't believe I made that mistake. Thanks for pointing it out. I'll fix it immediately \$\endgroup\$
    – Aditya P
    Mar 3, 2019 at 14:10
  • 1
    \$\begingroup\$ wait, is that official material? They're capitalizing the units incorrectly … \$\endgroup\$ Mar 3, 2019 at 14:20

1 Answer 1


Ifor and Iref constructively interferring \begin{equation} I_{max} = I_{for}+I_{ref} = 20a \end{equation} Power delivered to the load \begin{equation} P_{for}-P_{ref}=50I_{for}^2-50I_{ref}^2=10000w \end{equation} Two Equations and two unknowns, solve for Ifor and Iref interferring destructively = Imin \begin{equation} I_{for}=15a, I_{ref}=5a, I_{min}=I_{for}-I_{ref}=10a \end{equation} Calculate Imax/Imin \begin{equation} SWR_{max}=I_{max}/I_{min}=20/10=2:1 \end{equation}


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