Question

The Input impedance of a $\frac{\lambda }{8}$ section of a lossless transmission line of characteristic impedance 50 $\Omega$ is found to be real when the other end is terminated by a load $Z_L=(R+jX)\Omega$. If X is 30 $\Omega$, the value of R (in $\Omega$) is

• A. 40

Answer 1

The correct option is A 40
$Z_{in}=Z_0\left(\frac{Z_L+j{Z}_{0}tan\beta l}{Z_0+j{Z}_{L}tan\beta l}\right)$

at $l=\frac{\lambda }{8}$

$\Rightarrow \beta l=\frac{2\pi }{\lambda }\times \frac{\lambda }{8}=\frac{\pi }{4}$

$\therefore tan\frac{\pi }{4}=1$

$Z_{in}=Z_0\left[\frac{Z_L+j50}{Z_0+j{Z}_L}\right]$

Given, $Z_L=R+jX=R+j30and{Z}_0=50\Omega$

$Z_{in}=50\left[\frac{R+j30+j50}{50+j(R+j30)}\right]$

$=50\left[\frac{R+j80}{(50-30)+jR}\right]=50\left[\frac{R+j80}{20+jR}\right]$

$=50\left[\frac{(R+j80)(20-jR)}{{20}^2+R^2}\right]$

Since, only real part of $Z_{in}$ exists, so imaginary part of $Z_{in}=0$

$\therefore Z_{in}=50\left[\frac{20R+1600j-j{R}^2+50R}{{20}^2+R^2}\right]$

$Z_{imaginary}=0=\frac{1600-R^2}{{20}^2+R^2}$

$\therefore R^2=1600$

R = 40 $\Omega$