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

Percentage differential relay with harmonic restraint. Differential protection compares the currents entering and leaving a protected zone. For a transformer, CTs on the HV and LV sides, after ratio correction, give I_HV and I_LV. The operating current is I_op = |I_HV - I_LV|, and the restraint current is Iᵣ = (I_HV + I_LV)/2. A percentage relay trips when I_op > I_set + k Iᵣ, where k is the slope. The slope is usually 15-20% because it must tolerate CT saturation, ratio error, tap-changer position and external faults, while retaining sensitivity to internal faults. The pickup setting is normally a small value, such as 0.2-0.5 A or 5-10% of rated current, so that the relay is secure for through faults but sensitive to internal faults. Conceptually, the characteristic is a straight line starting from the pickup point on the I_op axis and rising with slope k; below the line is the restraint/no-trip region, above it the operating/trip region. A sketch would plot I_op vertically and Iᵣ horizontally.

A block diagram would show CTs feeding a differential current transformer or digital comparator; the difference current drives the operating coil, while the sum current drives the restraint coil or bias winding. A second-harmonic filter, analogue or digital, samples the CT current and forms the ratio I₂/I₁. The trip logic is: trip only if (I_op > I_set + k Iᵣ) and (I₂/I₁ < 15-20%). During transformer energisation, magnetising inrush is large, unbalanced and rich in second harmonic; without restraint the relay would maloperate. Harmonic restraint therefore blocks the trip during inrush but remains available for genuine internal faults. This is standard in NTPC and state electricity board transformer protection, especially at 220 kV and 400 kV levels.

Lead network. For the described circuit, R₁ is in parallel with C, and this combination is in series with R₂, with Vₒ across R₂. Thus Z₁ = R₁ || 1/(jωC) = R₁/(1+jωCR₁). The transfer function is H = Vₒ/Vᵢ = R₂/(R₂+Z₁) = R₂(1+jωCR₁)/(R₁+R₂+jωCR_1R₂). Hence the output leads the input by φ = tan⁻¹[ωCR_1R₂/(R₁+R₂)]. Equivalently, the lag of Vᵢ relative to Vₒ has tangent (R₁+R₂)/(ωCR_1R₂). Given f = 100 Hz, ω = 2π100 = 628.3 rad/s, C = 0.1 μF, R₁ = 100 kΩ, so ωCR₁ = 6.283. With tan 44.43° ≈ 0.980, 6.283 R₂/(100+R₂) = 0.980, giving R₂ ≈ 18.5 kΩ. The magnitude is |H| = R₂ sqrt[1+(ωCR₁)²] / sqrt[(R₁+R₂)²+(ωCR_1R₂)²] ≈ 0.709. Therefore the steady-state output amplitude is |Vₒ| = 10 × 0.709 ≈ 7.09 V.

Minimum weight spanning tree. A spanning tree of a connected graph G = (N, A) is a connected, acyclic subgraph containing all N nodes and exactly N-1 arcs. A minimum weight spanning tree (MST) is a spanning tree whose total arc weight is minimum among all spanning trees.

If all arc weights are distinct, the MST is unique. Suppose two MSTs T₁ and T₂ exist. Since they differ, choose an arc e in T₁ but not T₂. Adding e to T₂ creates a unique cycle. This cycle must contain some arc f in T₂ but not T₁; otherwise T₁ would contain the same cycle. Because weights are distinct, either w(e) < w(f) or w(f) < w(e). If w(e) < w(f), replacing f by e in T₂ gives a lighter spanning tree, contradicting minimality of T₂. If w(f) < w(e), replacing e by f in T₁ gives a lighter spanning tree, contradicting minimality of T₁. Hence T₁ and T₂ cannot be distinct, so the MST is unique. In practice, Kruskal or Prim algorithms find the MST, and the uniqueness theorem assures that the least-cost connected topology is unambiguous when costs are distinct. This result supports least-cost transmission network planning, where utilities seek a connected, loop-free grid at minimum cost.