(a) Two sources M1 and M2 emit messages x1, x2, x3 and y1, y2, y3 with the joint probability P(X, Y) as shown below in the matrix…

(a) Entropies of the two sources

Check. The nine joint probabilities sum to 1 (5/40 + 5/20 + 5/8 = 0.125 + 0.25 + 0.625 = 1). All logarithms are to base 2, so the results are in bits.

Marginal probabilities (row and column sums):

• P(x₁) = 3/40 + 1/40 + 1/40 = 1/8; P(x₂) = 1/20 + 3/20 + 1/20 = 1/4; P(x₃) = 1/8 + 1/8 + 3/8 = 5/8
• P(y₁) = 3/40 + 1/20 + 1/8 = 1/4; P(y₂) = 1/40 + 3/20 + 1/8 = 3/10; P(y₃) = 1/40 + 1/20 + 3/8 = 9/20

H(X) = −Σ P(x) log₂ P(x) = 0.125(3) + 0.25(2) + 0.625 log₂(1.6) = 0.375 + 0.500 + 0.4238 = 1.299 bits/message

H(Y) = 0.25 log₂4 + 0.3 log₂(1/0.3) + 0.45 log₂(1/0.45) = 0.500 + 0.5211 + 0.5184 = 1.539 bits/message

Joint entropy. H(X, Y) = −ΣΣ P(x, y) log₂ P(x, y):

• 3/40 (once): 0.2803
• 1/40 (twice): 2 × 0.1330 = 0.2661
• 1/20 (twice): 2 × 0.2161 = 0.4322
• 3/20 (once): 0.4105
• 1/8 (twice): 2 × 0.3750 = 0.7500
• 3/8 (once): 0.5306

H(X, Y) = 2.670 bits

Conditional entropies (chain rule H(X, Y) = H(Y) + H(X/Y) = H(X) + H(Y/X)):

• H(X/Y) = 2.670 − 1.539 = 1.130 bits
• H(Y/X) = 2.670 − 1.299 = 1.371 bits

Direct check of H(Y/X). Each row of P(y/x) is a permutation of {3/5, 1/5, 1/5}: row 1 is (3/5, 1/5, 1/5), and so on. Each row therefore has entropy 0.6 log₂(5/3) + 2 × 0.2 log₂5 = 0.4422 + 0.9288 = 1.371 bits, so H(Y/X) = 1.371 bits. Working the columns of P(x/y) directly gives H(X/Y) = 0.25(1.485) + 0.3(1.325) + 0.45(0.803) = 1.130 bits, which agrees.

Interpretation. Both conditional entropies are smaller than the corresponding source entropies, as they must be. The mutual information is I(X; Y) = H(X) − H(X/Y) = 1.299 − 1.130 = 0.169 bit per symbol. Observing one source removes only a small part of the uncertainty about the other.

(b) Current setting of the differential relay (Δ–Y, 50 MVA, 66/11 kV)

Connections. In a Merz–Price scheme on a Δ–Y transformer, the CTs on the delta (66 kV) side are connected in star. The CTs on the star (11 kV) side are connected in delta. This cancels the 30° phase shift and keeps zero-sequence current out of the relays.

Rated line currents.

• 66 kV side: I₁ = 50×10⁶/(√3 × 66×10³) = 437.4 A
• 11 kV side: I₂ = 50×10⁶/(√3 × 11×10³) = 2624.3 A

Pilot (CT secondary) currents at rated load.

• 66 kV CTs (star): 437.4 × 5/600 = 3.645 A in the pilot
• 11 kV CTs (delta): CT secondary = 2624.3 × 5/3000 = 4.374 A; the delta connection delivers √3 × 4.374 = 7.576 A to the pilot

Out-of-balance (spill) current. The given CT ratios are not matched to the transformer ratio. So even under healthy load the relay coil carries I_spill = 7.576 − 3.645 = 3.931 A at rated current.

Setting for faults up to 400% of rated current. The relay must not operate for a through (external) fault drawing up to 4 × rated current. Spill current rises in proportion: I_spill (400%) = 4 × 3.931 = 15.72 A

The relay pick-up must therefore be set above about 15.7 A, i.e. about 315% of a 5 A relay (say a 16 A, or 320%, plug setting). This setting seems very high, which shows that the CT ratios are mismatched. For balance, the 66 kV CTs would need a ratio of about 289:5 (437.4/7.576 × 5), or an interposing CT would have to correct the mismatch. With matched CTs, the spill would come only from CT errors and tap changing, and a normal setting of 20–40% could be used.

If the phase shift is corrected separately. Some texts compensate the 30° shift with interposing CTs, or inside a numerical relay, so that the two CT secondary currents are compared directly. The spill at rated load is then 4.374 − 3.645 = 0.729 A. At 400% it is 4 × 0.729 = 2.92 A, which calls for a setting of about 58% of a 5 A relay. The connection assumed above, with the 11 kV CTs in delta, is the conventional Merz–Price arrangement, and it is the basis of the main answer.

(c) Restriking voltage when a small current is chopped

Given. L = 15 H, C = 3000 pF = 3×10⁻⁹ F. The current is chopped at 20% of its 20 A peak: i = 0.2 × 20 = 4 A.

Energy transfer. At the instant of chopping, the magnetic energy ½Li² in the inductance has no current path. It oscillates into the stray capacitance, and the voltage across the breaker contacts peaks when all the energy is in C (ideal lossless LC circuit):

½Li² = ½CV² ⇒ V = i√(L/C)

Surge impedance. √(L/C) = √(15/3×10⁻⁹) = √(5×10⁹) = 70 711 Ω

Peak voltage. V = 4 × 70 711 = 2.83×10⁵ V ≈ 283 kV

The oscillation frequency is 1/(2π√(LC)) ≈ 750 Hz. If "20 A" were an rms value, the chopped current would be 5.66 A and the peak about 400 kV.

In practice, losses and the voltage already on the capacitance reduce or modify this figure. It still shows why interrupting small inductive currents (transformer magnetising currents, reactor currents) can cause dangerous overvoltages. Resistance switching or surge arresters are used to limit them.