Question

A transmission line of characteristic impedance $50\Omega$ is terminated in a load impedance $Z_L$. The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of $\lambda /4$ from the load. The value of $Z_L$ is

• A. $(19.23-j46.15)\Omega$
• B. $250\Omega$
• C. $(19.23+j46.15)\Omega$
• D. $10\Omega$

Answer 1

The correct option is D $10\Omega$

Ans : (a)


$VSWR=5=\frac{1+\left|\Gamma \right|}{1-\left|\Gamma \right|}$

$\therefore \left|\Gamma \right|=\frac{2}{3}\Rightarrow \Gamma =\pm \frac{2}{3}$

If the maximum is at $\lambda /4$ from load than minimum will be at load itself.

For the phase of $\Gamma$,

$2\beta x_{max}=2n\pi +\theta$

where $x_{max}$ is distance of $V_{max}$ from load

$2\cdot \frac{2\pi }{\lambda }\cdot \frac{\lambda }{4}=\theta$

$\Gamma =-\frac{2}{3}$

Now, $-\frac{2}{3}=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{Z_L-50}{Z_L+50}$

$\Rightarrow Z_L=10\Omega$