(a) Discuss the percentage differential Relay with harmonic restraint with the help of diagram and also draw the conceptual…

(a) Percentage differential relay with harmonic restraint Principle. A differential relay compares the currents entering and leaving the protected zone (a transformer, for instance). CTs on both sides, with ratio and phase correction, give I₁ and I₂. The operating current is I_op = |I₁ − I₂| and the restraint (bias) current is I_r = (|I₁| + |I₂|)/2. The relay trips when I_op > I_set + kI_r, where k is the slope (typically 20–40%, higher for transformers with tap changers). The slope lets the relay ride out CT ratio errors, CT saturation and tap-changer mismatch on heavy through faults, while staying sensitive to light internal faults. Harmonic restraint. On energising a transformer, magnetising inrush flows on one side only, so it looks like a large differential current, but it is rich in second harmonic (typically above 15–20% of the fundamental) which a genuine fault current lacks. A filter extracts the second-harmonic current I₂ₙd, and the trip is blocked when I₂ₙd/I₁ exceeds the set ratio. A fifth-harmonic restraint can similarly block operation during over-excitation. Conceptual representation (block diagram). CTs on the two sides, through interposing CTs/phase and ratio matching, feed (i) a summing element giving I_op for the operating coil, (ii) an averaging element giving I_r for the restraint coil, and (iii) a harmonic filter and comparator giving the block signal. The trip output is an AND of "I_op above the biased characteristic" and "second-harmonic ratio below the set value". The characteristic plots I_op against I_r: a sloping straight line from the pickup point, trip region above it, restraint region below.

(b) Lead network R₁ in parallel with C forms the series arm and R₂ is the shunt arm, so V_o/V_i = R₂/(R₂ + Z₁) with Z₁ = R₁/(1 + jωCR₁), giving V_o/V_i = R₂(1 + jωCR₁)/[(R₁ + R₂) + jωCR₁R₂]. The phase lead is φ = tan⁻¹(ωCR₁) − tan⁻¹[ωCR₁R₂/(R₁ + R₂)]. ω = 2π × 100 = 628.3 rad/s, so ωCR₁ = 628.3 × 0.1 × 10⁻⁶ × 10⁵ = 6.283 and tan⁻¹(6.283) = 80.96°. Then tan⁻¹[6.283R₂/(100 + R₂)] = 80.96° − 44.43° = 36.53°, so 6.283R₂/(100 + R₂) = 0.7407 (R₂ in kΩ), which gives R₂ ≈ 13.4 kΩ.

|V_o/V_i| = R₂√(1 + 6.283²)/√[(R₁ + R₂)² + (6.283R₂)²] = 13.36 × 6.362/√(113.36² + 83.95²) = 85.0/141.1 = 0.603. |V_o| = 10 × 0.603 ≈ 6.03 V (amplitude), leading V_i by 44.43°.

(c) Minimum weight spanning tree (i) A spanning tree of a connected graph G = (N, A) is a connected, acyclic set of N − 1 arcs that touches every node. A minimum weight spanning tree is a spanning tree whose total arc weight is the least among all spanning trees. (ii) Uniqueness for distinct weights. Suppose T₁ ≠ T₂ are both minimum. Among arcs lying in exactly one of the two trees, let e be the lightest, and say e ∈ T₁, e ∉ T₂. Adding e to T₂ creates a cycle. That cycle cannot lie wholly in T₁ (a tree has no cycle), so it contains an arc f ∈ T₂ with f ∉ T₁. Since f also lies in exactly one tree and weights are distinct, w(f) > w(e). Then T₂ − f + e is a spanning tree of smaller weight than T₂, contradicting the minimality of T₂. Hence T₁ = T₂ and the MST is unique.