Test: ABCD Parameters of Transmission Line - Electrical Engineering (EE) MCQ

Open circuiting a-a^‘ we get V₁ = I_y × 2 and I_y = I₂/2 and V₂ = I_x × 4 and I_x = I₂/2. On solving and substituting, we get h₂₂ = I₂/V₂ = 1/2 Ω.

Open circuiting a-a^‘ we get V₁ = I_y × 2 and
I_y = I₂/2 and V₂ = I_x × 4 and I_x = I₂/2. On solving and substituting, we get h₁₂ = V₁/V₂ = 1/2.

Short circuiting b - b^‘, h₂₁ = I₂/I₁ when V₂ = 0 and -I₂ = I₁/2 => I₂/I₁ = -1/2. So h₂₁ = -1/2.

h₁₁ = V₁/I₁ |V₂ = 0. So short circuiting b - b^‘, V₁ = I₁((2||2)+1) = 2I₁ => V₁/I₁ = 2. On substituting we get h₁₁ = V₁/I₁ = 2Ω.

h₂₁ = I₂/I₁ |V₂ = 0. So the hybrid parameter h₂₁ is called short circuit forward current gain.

h₁₁ = V₁/I₁ |V₂ = 0. So the hybrid parameter h₁₁ is called short circuit input impedance.

D is a transmission parameter and is given by D = -I₁/I₂ |V₂=0. Short circuiting b-b^‘, I₁ = 7/17 V₁ and -I₂ = 5/17 V₁. So we get I₁/I₂ = 7/5. So D = 7/5.

The transmission parameter B is given by B = -V₁/I₂|V₂ = 0. Short circuiting b - b^‘, -I₂ = 5/17 V₁ => -V₁/I₂ = 17/5. On substituting we get B = 17/5 Ω.

C = I₁/V₂|I₂ = 0. By open circuiting b-b^‘ we get V₂ = 5 I₁ => I₁/V₂ = 1/5. On substituting we get C = I₁/V₂ = 1/5 Ω.

Open circuiting b - b^‘, V₁ = 6 I₁, V₂ = 5I₁. On solving V₁/V₂ = 6/5. On substituting we get A = V₁/V₂ = 6/5.