Consider a 300 Ω, quarter wave long at 1 GHz transmission line as shown in figure. It is connected to a 10 V, 50 Ω source at one end is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is

γ = a + jβ a = 0 ⟹ γ = jβ Propagating wave v = VI e-rz + VR erz No change in the voltage at the open end it will same, vr = 10 volt.

Correct Answer: 10 V

Explanation:

γ = a + jβ a = 0 ⟹ γ = jβ

Propagating wave

v = V_I e^-rz + V_R e^rz

No change in the voltage at the open end it will same,

v_r = 10 volt.

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