Topic: Electromagnetics
A transmission line having characteristic impedance of 50 $\Omega$ has to deliver 10 kW power at 100 kHz to a load. Maximum permissible value of rms current anywhere along the line is 20 A. What is the maximum VSWR that can be tolerated on this line?
- A. 2
- B. 1
- C. 3
- D. 2.5
Answer: A. 2
Explanation: Power delivered $P = I^{2} Z_{L}$. Given $P = 10 \times 10^{3} \text{ W}$ and maximum $I_{rms} = 20 \text{ A}$, we can find $Z_{L} = \frac{10 \times 10^{3}}{20 \times 20} = 25 \text{ } \Omega$.
The reflection coefficient $\Gamma = \frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}} = \frac{25-50}{25+50} = -\frac{1}{3}$.
The Voltage Standing Wave Ratio (VSWR) is $\frac{1+|\Gamma|}{1-|\Gamma|} = \frac{1+1/3}{1-1/3} = \frac{4/3}{2/3} = 2$.