Electronics and Communication Engineering - Exam Questions Papers

The time constant t is given by

As R_eq = 4 || 4 = 2 Ω

∴ i₁(t) = i_F - (i_F - i₁)e^-t/t

i_final = = 2.5 A

∴ i₁(t) = 2.5 - (2.5 - 1.25)e^-t/1/2

i₁(t) = 2.5 - 1.25e^-2t.

Apply KVL to first to first loop

6i₁ - 2i₂ = 10

(4 + R)i₂ - 2i₁ - 2i₃ = 0

4i₃ - 2i₂ = 0

but i₃ = 0.5 A

2 - 2i₂ = 0

∴ i₂ = 1 A, 6i₁ - 2 = 10

∴ i₁ = 2 A, (4 + R) - 2 X 2 -1 = 0

∴ R = 1 Ω

Using Nodal analysis for loop 2

At node A,

V_A - 20 + V_A + V_A - V_B = 0

3V_A - V_B - 20 = 0

3V_A - V_B = 20 ...(i)

At node B,

2V_B - 2V_A + 2V_B + V_B = 0

5V_B - 2V_A = 0 ...(ii)

Multiplying (i) by 2 and (ii) by 3

6V_A - 2V_B = 40

- 6V_A - 15V_B = 0

13V_B = 40

V_B ≅ 3 V

∴

M₁ = Z XOR R

M₁ = (X, Y) XOR R

= M₁ = [P.Q.(P + Q)]XOR R

= (PQ + PQ) XOR R = (P XOR Q) XOR R.

Minimum clock freq. = 2f_m

Where f_m is highest freq. component = 2 x 10 kHz = 10 kHz.

F = Σm (2, 3, 5, 7, 8, 9, 15).