# Maximum VSWR that can be tolerated?

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.

• You are saying the accepted answer is D? Mar 3, 2019 at 13:02
• 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. Mar 3, 2019 at 13:07
• Your calculation says VSWR=1, but option D is 2.5. Mar 3, 2019 at 14:01
• Omg, I can't believe I made that mistake. Thanks for pointing it out. I'll fix it immediately Mar 3, 2019 at 14:10
• wait, is that official material? They're capitalizing the units incorrectly … Mar 3, 2019 at 14:20

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