(a) A 400 V, 3-phase, 50 Hz, star-connected synchronous motor has per phase synchronous impedance Zₛ = (0·5 + j 3·5) Ω. It is…

(a) Synchronous motor as a synchronous condenser

Per phase V = 400/√3 = 230.9 V. With zero input power, the current is purely reactive and, for a condenser supplying reactive power, it leads the voltage by 90°.

Motor current: Q = √3 V_L I, so I = 100 000/(√3 × 400) = 144.3 A, taken as I = j144.3 A.

Excitation voltage: E = V − I Z_s = 230.9 − j144.3 (0.5 + j3.5) = 230.9 − j72.2 + 505.2 = 736.1 − j72.2 V. Hence |E| = 739.6 V per phase, which is about 1281 V line-to-line, at a load angle of −5.6°.

E is about 3.2 times V, so the motor is strongly over-excited, as it must be to deliver leading reactive power.

(b)(i) Sum of two independent Poisson variables

Let P(X₁ = k) = e^(−λ₁)λ₁^k/k! and P(X₂ = m) = e^(−λ₂)λ₂^m/m!. Since they are independent, for Y = X₁ + X₂:

P(Y = n) = Σ from k = 0 to n of P(X₁ = k) P(X₂ = n − k) = e^(−(λ₁+λ₂)) Σ λ₁^k λ₂^(n−k) / [k!(n − k)!].

Multiply and divide by n!: P(Y = n) = [e^(−(λ₁+λ₂))/n!] Σ C(n, k) λ₁^k λ₂^(n−k) = e^(−(λ₁+λ₂)) (λ₁ + λ₂)^n / n!, using the binomial theorem.

This is the Poisson probability with parameter λ₁ + λ₂, for n = 0, 1, 2, … Check with the moment generating function: M_Y(t) = M_X₁(t) M_X₂(t) = exp[(λ₁ + λ₂)(e^t − 1)].

(b)(ii) Poisson limit of the binomial

For a binomial variable with n trials and success probability p, P(k) = C(n, k) p^k (1 − p)^(n−k). Let n → ∞ and p → 0 with np = λ held fixed, so p = λ/n.

Then C(n, k) p^k = [n(n − 1)…(n − k + 1)/k!] (λ/n)^k = (λ^k/k!) (1)(1 − 1/n)…(1 − (k − 1)/n) → λ^k/k!. Also (1 − λ/n)^(n−k) → e^(−λ), because (1 − λ/n)^n → e^(−λ) and (1 − λ/n)^(−k) → 1.

Hence P(k) → e^(−λ)λ^k/k!, the Poisson distribution with parameter λ = np. The mean np = λ is kept, and the variance np(1 − p) tends to λ, as it is for a Poisson variable. The approximation is good for large n and small p (for example n ≥ 50, np ≤ 5).

(c) Thyristor commutation circuit

Before triggering, no dc current flows in the C-L branch, so the capacitor charges to the source voltage, V_C(0) = 200 V. At t = 0 the triggered thyristor connects the capacitor and inductor in a closed loop and, in parallel, the source drives 200/4 = 50 A through the thyristor and the 4 Ω resistor.

The LC loop: ω₀ = 1/√(LC) = 1/√(3 × 10⁻³ × 1.5 × 10⁻³) = 471.4 rad/s and Z₀ = √(L/C) = √2 = 1.414 Ω. The capacitor discharges through the thyristor as i_LC = (V_C(0)/Z₀) sin ω₀t = 141.4 sin 471.4t A.

The thyristor current is i_T(t) = 50 + 141.4 sin 471.4t A.

(i) Maximum current. i_T,max = 50 + 141.4 = 191.4 A.

(ii) Instant of maximum. sin ω₀t = 1 at ω₀t = π/2, so t = π/(2 × 471.4) = 3.33 ms.

(iii) Conduction time. The LC current reverses after a half period and the thyristor current reaches zero when sin ω₀t = −50/141.4 = −0.3536, so ω₀t = π + 0.3614 = 3.503 rad and t = 3.503/471.4 = 7.43 ms. With zero holding current the thyristor turns off there, so it conducts for about 7.43 ms.