(a) (i) For the circuit shown below, steady state is achieved with switch K open. At t = 0, switch K is closed. Find the time…

(a)(i) Switching transient in the R_L circuit (10 marks)

Before t = 0 (K open). R_L is disconnected. The inductor acts as a short in steady state, so the 50 V source drives the 5 Ω resistor and the 20 Ω resistor in series: i_L(0⁻) = 50/(5 + 20) = 2 A.

After t = 0 (K closed). The inductor current cannot change instantly, so i_L(0⁺) = 2 A. The 5 Ω resistor, the 20 Ω load R_L and the source now form a network that feeds the 20 Ω + 1 H branch. Its Thevenin equivalent is

V_th = 50 × 20/(5 + 20) = 40 V and R_th = 5 ∥ 20 = 4 Ω.

For the inductor branch, R = R_th + 20 = 24 Ω, so the time constant is τ = L/R = 1/24 s, and the final inductor current is i_L(∞) = 40/24 = 5/3 A. Thus

i_L(t) = 5/3 + (2 − 5/3) e^(−24t) = 5/3 + (1/3) e^(−24t) A.

Current in R_L. The voltage across the branch (and hence across R_L) is v = 20 i_L + L di_L/dt. With di_L/dt = −8 e^(−24t) this gives

v(t) = 20 (5/3 + (1/3) e^(−24t)) − 8 e^(−24t) = 100/3 − (4/3) e^(−24t) V.

Therefore

i_RL(t) = v/20 = 5/3 − (1/15) e^(−24t) A ≈ 1.667 − 0.0667 e^(−24t) A, for t ≥ 0.

Initial and final values. i_RL(0⁺) = 5/3 − 1/15 = 1.6 A (check by KCL at 0⁺: (50 − v)/5 = v/20 + 2 gives v = 32 V and 32/20 = 1.6 A). Just before closing, i_RL = 0, so the current jumps to 1.6 A. i_RL(∞) = 5/3 ≈ 1.667 A (check: the total current 50/(5 + 10) = 10/3 A splits equally between the two 20 Ω paths).

(a)(ii) y-parameters of the two-port (10 marks)

Define I₁ = y₁₁V₁ + y₁₂V₂ and I₂ = y₂₁V₁ + y₂₂V₂. Node A (port 1) has a 1 Ω resistor to ground and a 2 Ω series resistor to node B (port 2). Node B has a 2 Ω resistor to ground and the dependent current source 3I₁ drawing current downward.

Port 1. I₁ = V₁/1 + (V₁ − V₂)/2 = 1.5V₁ − 0.5V₂.

Port 2. KCL at B: I₂ + (V₁ − V₂)/2 = V₂/2 + 3I₁, so

I₂ = V₂/2 + 3(1.5V₁ − 0.5V₂) − (V₁ − V₂)/2 = 4V₁ − 0.5V₂.

Hence

• y₁₁ = 1.5 S, y₁₂ = −0.5 S,
• y₂₁ = 4 S, y₂₂ = −0.5 S.

Check by short-circuit tests: with V₂ = 0, y₁₁ = 1 + 1/2 = 1.5 S; with V₂ = 0 the current 0.5V₁ enters node B from the series resistor and 3I₁ = 4.5V₁ leaves through the source, so I₂ = 4V₁ and y₂₁ = 4 S. The network is non-reciprocal (y₁₂ ≠ y₂₁) because of the dependent source, and y₂₂ is negative because the source sinks 3I₁ = 4.5V₁ − 1.5V₂.

(b) Diode in a resistive bridge (20 marks)

Circuit and diode model. The source is across the left and right corners, positive on the left, and the diode joins the top node T (junction of 2 Ω and 6 Ω) to the bottom node B (junction of the two 8 Ω resistors), conducting from T to B. Taking the source polarity as the one that forward-biases the diode (the reverse polarity would give a diode current of zero), the diode is an ideal 0.7 V drop plus a slope resistance: from Fig. (2), it starts conducting at 0.7 V and reaches 1.0 A at 0.9 V, so r_d = 0.2/1.0 = 0.2 Ω and v_d = 0.7 + 0.2 i_d for 0 ≤ i_d ≤ 1 A.

Thevenin equivalent seen by the diode. With the diode removed and the left corner at 20 V and the right at 0 V:

• V_T = 20 − 20 × 2/(2 + 6) = 15 V and V_B = 20 × 8/(8 + 8) = 10 V, so V_th = 15 − 10 = 5 V (T positive, so the diode is forward-biased);
• R_th = (2 ∥ 6) + (8 ∥ 8) = 1.5 + 4 = 5.5 Ω.

(i) Diode current. i_d = (V_th − 0.7)/(R_th + r_d) = (5 − 0.7)/(5.5 + 0.2) = 4.3/5.7 = 0.754 A, which lies within the linear range (below 1 A), and v_d = 0.7 + 0.2 × 0.754 = 0.851 V (below 0.9 V), confirming the region assumed.

(ii) Power loss in the diode. P_d = v_d i_d = 0.851 × 0.754 = 0.642 W.

(iii) Power supplied by the source. With V_B = 10 + 4 i_d = 13.02 V and V_T = V_B + v_d = 13.87 V, the source current is the sum of the currents in the two left-hand resistors: I_s = (20 − 13.87)/2 + (20 − 13.02)/8 = 3.066 + 0.873 = 3.939 A. The power supplied is P_s = 20 × 3.939 = 78.8 W.

Check. The resistors dissipate 3.066² × 2 = 18.80 W, 2.311² × 6 = 32.05 W, 0.873² × 8 = 6.09 W and 1.627² × 8 = 21.18 W. With the diode's 0.64 W the total is 78.8 W, equal to the source power.

(c) Resistance per metre of a coaxial copper line (10 marks)

At 400 MHz with conductors much thicker than the skin depth, the current flows in a thin surface layer and the resistance per unit length comes from the surface resistance. For copper with μ = μ₀ and ρ = 1.724 × 10⁻⁸ Ω·m:

R_s = √(πfμ₀ρ) = √(π × 400 × 10⁶ × 4π × 10⁻⁷ × 1.724 × 10⁻⁸) = 5.22 × 10⁻³ Ω per square.

The skin depth is δ = ρ/R_s = 3.30 μm, far below the conductor thicknesses, which justifies the surface-resistance approach.

The inner conductor has radius a = 1 mm and the outer conductor has inner radius b = 3.5 mm. The resistance per metre is the sum of the two conductors' contributions, each R_s divided by the circumference:

R = (R_s/2π)(1/a + 1/b) = (5.22 × 10⁻³/2π)(1/0.001 + 1/0.0035) = 8.30 × 10⁻⁴ × (1000 + 285.7) = 1.07 Ω/m.

The inner conductor contributes about 0.83 Ω/m and the outer about 0.24 Ω/m, since current density is highest on the smaller inner conductor.