Question

A 30-Volts battery with zero source resistance is connected to a coaxial line of characteristic impedance of 50 Ohms at t = 0 second and terminated in an unknown resistive load. The line length is such that it takes 400$\mu s$ for an electromagnetic wave to travel from source end to load end and vice-versa. At t = 400 $\mu s$. the voltage at the load end is found to be 40 Volts.

The steady-state current through the load resistance is

Answer 1

Ans : (b)



10 V reflects to add to 30 V in phase and gives 40 V


$\Gamma =\frac{V_r}{V_i}=\frac{10}{30}=\frac{1}{3}$

$\Gamma =\frac{Z_L-Z_0}{Z_L+Z_0}$

$\Rightarrow \frac{1}{3}=\frac{Z_L-50}{Z_L+50}$

$\Rightarrow Z_L=100\Omega$

In steady state, $V_L=30V$

$I_L=\frac{30}{100}=0.3A$