My Try:

# $Z_{in}=Z_0*\frac{Z_L+jZ_0tan\beta l}{Z_0+jZ_L tan\beta l}$

Now

$\beta l=\frac{2\pi\times37.5\times10^6\times10}{3\times10^8}=\frac{5\pi}{2}$

so its becomes

$Z_{in}=\frac{Z_0^2}{Z_L}$

$Z_{in}=\frac{200\times200}{100}=400\Omega$

So for part $(a)$

$I_{i}=\frac{200}{600}=\frac{1}{3}A$ also $V_{i}=\frac{400\times200}{600}=\frac{400}{3}$

Now for part $(b)$ from the generator we know the equation for lossless line is $I(z)=\frac{-j}{Z_0}sin\beta z V_i+Cos\beta z I_{i}$

so current at the load, that means at $\lambda/4$ distance from the generator will be $I=\frac{-j}{200}sin(\frac{5\pi}{2})\times\frac{400}{3}+0\times\frac{1}{3}=\frac{2}{3}\angle-90^{\circ}$

but for part$(b)$ the actual answer is $\frac{1}{3}\angle-90^{\circ}$

What is the mistake i am doing can anyone help please?

In part (a) you calculated the current as 1/3 so why should it suddenly be any different to the load current in part (b). As for phase angle, you have shown the line to be a equivalent to a quarter wavelength long at 37.5 MHz so, given the load is resistive, the phase angle of the load current has to be 90 degrees lagging as you have also shown.

• Thanks for the answer sir, but why its not coming by TL equations? – Rohit Sep 13 '17 at 6:56
• @Rohit I have no idea. One method produces X and another method produces Y. Both should produce the same OR there is a math error or formula error. I have only tried to show that your method for calculating answer (b) is incorrect based on the fact that Iin = Iout (RMS values) – Andy aka Sep 13 '17 at 10:36

For these problems, remember that the $z=0$ location (highlighted in the image below) is defined at the terminals of the load. So you are looking for $I(0)$ in part b.

For the input impedance, you do use the length because it is some distance $l$ away from the load and towards the generator. This is $z=-l$.

Now, I don't know whether you're using the right equation (can't remember off the top of my head all the TL equations) but hopefully this points you in the right direction.