Electronics and Communication Engineering - Exam Questions Papers

E = 6.82a_x + 6.82a_y + 32.89a_z V/m

r = a_x + a_y + a_z

r₁ = a_x + a_y

r₂ = - a_x + a_y

r₃ = - a_x - a_y

and r₄ = a_x - a_y

∴ r - r₁ = a_z;

r - r₂ = 2a_x + a_z;

r - r₃ = 2a_x + 2a_y + a_z

r - r₄ = 2a_y + a_z

= 6.8a_x + 6.8a_y + 32.9a_z .

Since 12A current is coming from one source and it is also known that 60V source is absorbing power i.e. current is flowing inside 60V source.

12 = x + 1 => I = 12 - x.

V_GS = V_G - I_DR_S

- 2 = 5.44 - 2.22 X 10^-3 X R_S

R_S = 3.35 KΩ.

It has only zeros at origin. Hence it is minimum phase system.

Use Y - Δ transformation

R_AB= (9 || 18) || (9 || 18) = 3 Ω.