(a) A 400 V, 50 Hz, 3-phase star-connected cylindrical rotor synchronous motor has synchronous impedance of (0·5 + j 2·5) Ω per…

(a) Synchronous motor at maximum power

Per phase V = 400/√3 = 230.9 V and Z_s = 0.5 + j2.5 = 2.55∠78.7° Ω, so R_a = 0.5 Ω and Z_s² = 6.5 Ω². Taking "maximum power" as the maximum mechanical power developed, it occurs at load angle δ = θ_z = 78.7°, where P_max = 3[EV/Z_s − E²R_a/Z_s²].

Setting 50 000 = 3[230.9E/2.55 − 0.5E²/6.5] gives 0.0769E² − 90.58E + 16 667 = 0, so E = 228.2 V per phase (the other root, 949 V, is unrealistic). The line value is 395 V.

Current: I = (V∠0° − E∠−78.7°)/Z_s = 114.2∠−28.4° A, so the motor current is 114 A.

Input power factor = cos 28.4° = 0.879 lagging. Check: input power = 3 × 230.9 × 114.2 × 0.879 = 69.6 kW, which equals 50 kW developed plus copper loss 3 × 114.2² × 0.5 = 19.6 kW.

(b) Half-controlled converter feeding a battery

Battery current I₀ = 1800/100 = 18 A. Total load resistance is 2 + 1 = 3 Ω, so the average output voltage is V₀ = 100 + 18 × 3 = 154 V.

(i) Triggering angle. For a half-controlled bridge, V₀ = (V_m/π)(1 + cos α) with V_m = 240√2 = 339.4 V. So 1 + cos α = 154π/339.4 = 1.4254, cos α = 0.4254 and α = 64.8°.

(ii) Fundamental input current. The freewheeling diode makes the ac current a quasi-square wave of height 18 A and width (π − α) in each half cycle, so I₁(rms) = (2√2/π) I₀ cos(α/2) = 0.9003 × 18 × cos 32.4° = 13.7 A.

(iii) Input power factor. Total rms source current = I₀√((π − α)/π) = 18 × 0.8 = 14.4 A. The input power is V₀I₀ = 154 × 18 = 2772 W (1800 W to the battery plus 972 W in the 3 Ω). So pf = 2772/(240 × 14.4) = 0.80 lagging. The displacement factor is cos(α/2) = 0.844.

(c) AM signal

x(t) = 7 cos(170×10³πt) + (5/2)[cos(160×10³πt) + cos(180×10³πt)]. The sum in brackets is 2 cos(10×10³πt) cos(170×10³πt), so x(t) = 7[1 + (5/7) cos(10×10³πt)] cos(170×10³πt). This is the form A_c[1 + μ cos ω_m t] cos ω_c t with carrier 85 kHz, modulating frequency 5 kHz and modulation index μ = 5/7 = 0.714, so it is amplitude modulated.

(i) P_c = 7²/2 = 24.5 and P_s = 2 × (2.5²/2) = 6.25 in the same units, so P_s/P_c = μ²/2 = 25/98 = 0.255.

(ii) Efficiency = P_s/(P_c + P_s) = μ²/(2 + μ²) = 6.25/30.75 = 20.3 per cent.

(d) Unknown load impedance on a transmission line

With a short circuit, minima are λ/2 apart, so λ = 50 cm. With the load, a minimum appears 8 cm toward the load from a short-circuit minimum position. Measured from the load, the first minimum is therefore at d_min = 25 − 8 = 17 cm, i.e. β d_min = 2π × 17/50 = 122.4°, and tan 122.4° = −1.576. With S = 4:

Z_L = Z₀(1 − jS tan βd)/(S − j tan βd) = 50(1 + j6.30)/(4 + j1.576) = 37.7 + j63.9 Ω ≈ 74.2∠59.5° Ω.

The load is inductive, as expected because the minimum has moved toward the load.

(e) Transient current in the series RLC circuit

With the switch in position 1 for a long time, the inductor carries the steady current 10/5 = 2 A, which is the initial condition i(0) = 2 A (an inductor current cannot jump). The capacitor voltage is v_C(0) = Q₀/C = 1 mC/200 µF = 5 V, with its positive plate at the bottom, so the current i enters its positive plate.

After switching, R i + L di/dt + v_C = 0, and differentiating gives i'' + (R/L) i' + i/(LC) = 0, that is i'' + 50 i' + 50 000 i = 0. The roots are s = −25 ± j222.2, so the response is underdamped.

i(t) = e^(−25t)(A cos 222.2t + B sin 222.2t). From i(0) = 2, A = 2. From the loop equation at t = 0⁺: L di/dt = −R i(0) − v_C(0) = −15 V, so di/dt(0⁺) = −150 A/s. Then −25A + 222.2B = −150, so B = −0.45.

i(t) = e^(−25t)(2 cos 222.2t − 0.45 sin 222.2t) A for t ≥ 0.