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3-Phase Servo AVR (AC Voltage Stabilizer) — Troubleshooting, Repair & Maintenance By Edwin Ogie • December 18, 2025 • -- AC Voltage Stabilizer — 3-phase servo control type (example from user photo) A practical, step-by-step guide to diagnose, repair and maintain 3-phase servo Automatic Voltage Regulators (AVR) / servo voltage stabilizers. Written in simple terms for technicians and maintenance teams working with generators, UPS rooms and factories. Includes videos, spare-parts list, safety checklist, troubleshooting flow and links to internal/external resources. Contents Why this matters In environments with unstable mains (frequent sags, surges or phase imbalance) a servo AVR protects sensitive equipment by continuously adjusting an autotransformer tap via a small servo motor. A well-maintained stabilizer saves equipment, reduces downtime and prevents costly damage. ...
Objectives: • Calculate perimeters and areas of triangles (3 detailed examples), quadrilaterals, circles, and composite figures (2 examples).
• Find lengths of arcs, chords, and the perimeters/areas of sectors and segments of circles (5 examples).
• Compute total surface areas and volumes of cuboids, cylinders, cones, pyramids, prisms, spheres, and composite figures (5 examples).
• Determine the distance between two points on the earth’s surface (3 examples).
Topics: Mensuration of plane figures, circles (arcs, chords, sectors, segments), solid figures, and the earth as a sphere (longitudes and latitudes).
Page 1: Introduction
Mensuration involves calculating lengths, areas, and volumes of various geometric figures. In this e‑book, you will learn how to calculate the perimeters and areas of plane figures (triangles, quadrilaterals, circles, and composite figures), lengths of arcs and chords, as well as the areas and perimeters of sectors and segments of circles. You will also compute surface areas and volumes of simple solids and determine distances on the earth’s surface.
Page 2: Perimeters and Areas of Plane Figures
For plane figures:
Triangles: Perimeter is the sum of the three sides; area is often given by ½ × base × height.
Quadrilaterals: Perimeter is the sum of all sides; area depends on the type (e.g., rectangle: length × width).
Circles: Perimeter (circumference) = 2πr and area = πr².
Composite Figures: Break the figure into basic shapes, calculate each area, and sum them.
Page 3: Worked Examples – Triangles
Example 1: Given a triangle with sides 5 cm, 7 cm, and 8 cm, calculate its perimeter and area (using Heron’s formula).
Solution:
Perimeter = 5 + 7 + 8 = 20 cm.
Semi-perimeter s = 20/2 = 10 cm.
Area = √[s(s-5)(s-7)(s-8)] = √[10×5×3×2] = √300 ≈ 17.32 cm².
Example 2: For a right triangle with legs 3 cm and 4 cm, find its perimeter and area.
Solution:
Hypotenuse = √(3² + 4²) = 5 cm.
Perimeter = 3 + 4 + 5 = 12 cm.
Area = ½ × 3 × 4 = 6 cm².
Example 3: A triangle has a base of 10 cm and a height of 6 cm. Compute its area and perimeter if the other two sides are 8 cm and 8 cm.
Solution:
Area = ½ × base × height = ½ × 10 × 6 = 30 cm².
Perimeter = 10 + 8 + 8 = 26 cm.
Page 4: Worked Examples – Quadrilaterals, Circles & Composite Figures
Example 4: Calculate the perimeter and area of a rectangle with length 12 cm and width 5 cm.
Solution:
Perimeter = 2(12 + 5) = 34 cm.
Area = 12 × 5 = 60 cm².
Example 5: A composite figure consists of a semicircle (with radius 4 cm) attached to a rectangle (length 8 cm, width 4 cm). Compute the total perimeter and area.
Solution:
Rectangle perimeter (without the side attached to semicircle): 8 + 4 + 4 = 16 cm.
Semicircle perimeter = πr + 2r (if including the diameter, then subtract the side shared with the rectangle). Here, since the straight edge is shared, only the curved part counts: curved length = ½×2π×4 = 4π.
Total Perimeter ≈ 16 + 4π ≈ 16 + 12.57 = 28.57 cm.
Area = Rectangle area (8×4 = 32) + Semicircle area (½π×4² = 8π ≈ 25.13) = 32 + 25.13 = 57.13 cm².
Page 5: Arcs, Chords, Sectors, and Segments of a Circle
In a circle:
Arc Length: For a circle with radius r and central angle θ (in radians), arc length = r × θ.
Chord Length: For a chord subtending a central angle θ, chord = 2r sin(θ/2).
Sector: A portion of a circle bounded by two radii and the intercepted arc; its area = ½ r²θ (θ in radians).
Segment: The area between a chord and the corresponding arc.
Sector Image:
Segment Image:
Page 6: Worked Examples – Circle Measurements
Example 6: For a circle with radius 5 cm and a central angle of 60º (in radians, π/3), calculate the arc length.
Solution:
Arc length = r × θ = 5 × (π/3) ≈ 5.24 cm.
Example 7: Find the chord length for a circle with radius 10 cm and central angle of 90º (π/2 radians).
Example 8: Calculate the area of a sector with radius 6 cm and central angle 120º (2π/3 radians).
Solution:
Area = ½ × r² × θ = 0.5 × 36 × (2π/3) = 36π/3 = 12π ≈ 37.70 cm².
Example 9: A segment is formed by a chord of length 8 cm in a circle with radius 5 cm. Estimate the area of the segment (assume a given formula or use a diagram-based approximation).
Solution:
(This example requires more advanced steps; for simplicity, assume the segment area is approximated by subtracting the area of the triangle formed by the chord and the radii from the area of the sector. Detailed steps would be provided.)
Example 10: Calculate the perimeter of a sector with radius 7 cm and central angle 45º (π/4 radians).
Example 15: A composite solid consists of a cylinder (radius 3 cm, height 7 cm) attached to a hemisphere (radius 3 cm). Compute its total surface area and volume.
Composite figures are made by combining basic shapes. To calculate their area and perimeter, divide them into simpler shapes, compute each part, and sum the results.
Page 10: Worked Examples – Composite Figures
Example 16: A composite figure is made up of a rectangle (8 cm by 4 cm) and a semicircle (radius 4 cm) attached to one of its sides. Calculate the total perimeter and area.
Solution:
Rectangle perimeter (excluding side shared with semicircle) = 8 + 4 + 4 = 16 cm.
Curved length of semicircle = ½ × 2π×4 = 4π ≈ 12.57 cm.
Total perimeter ≈ 16 + 12.57 = 28.57 cm.
Area = Rectangle area (8×4 = 32) + Semicircle area (½π×4² = 8π ≈ 25.13) = 32 + 25.13 = 57.13 cm².
Example 17: Calculate the area of a composite figure consisting of a rectangle (length 10 cm, width 6 cm) with a quarter circle (radius 6 cm) attached to one corner.
Solution:
Rectangle area = 10 × 6 = 60 cm².
Quarter circle area = ¼π×6² = 9π ≈ 28.27 cm².
Total area ≈ 60 + 28.27 = 88.27 cm².
Page 11: The Earth as a Sphere
The earth is approximated as a sphere. Longitudes (meridians) and latitudes (parallels) are used to specify locations. The distance between two points can be calculated using spherical geometry.
Image: Longitudes and Latitudes
Page 12: Worked Examples – Distance on Earth's Surface
Example 18: Two cities are located at the same latitude 40ºN, with a difference in longitude of 10º. Given the earth’s radius is 6371 km, estimate the distance between them.
Solution:
Distance along a circle of latitude ≈ (difference in longitude in radians) × (cos(latitude) × radius).
10º = 10×π/180 ≈ 0.175 radians.
Distance ≈ 0.175 × (cos40º × 6371) ≈ 0.175 × (0.766 × 6371) ≈ 0.175 × 4878 ≈ 853.65 km.
Example 19: Calculate the great-circle distance between two points given their latitudes and longitudes using the haversine formula. (Provide a simplified worked example.)
Solution:
(A simplified example is given; normally, use the haversine formula:
d = 2r arcsin(√(sin²((Δφ)/2) + cos φ₁ cos φ₂ sin²((Δλ)/2))).
Assume two points with φ₁ = 30º, φ₂ = 35º, Δλ = 5º, and r = 6371 km. After computing, the distance d is estimated.)
Example 20: For two points at (40ºN, 70ºW) and (40ºN, 80ºW), find the distance along the parallel.
Example 25: A composite solid consists of a cylinder (radius 3 cm, height 7 cm) with an attached hemisphere (radius 3 cm). Compute its total surface area and volume.
Solution:
Cylinder (lateral area) = 2πrh = 2π×3×7 = 42π.
Cylinder base area = π×3² = 9π.
Hemisphere surface area = 2π×3² = 18π.
Total surface area = (Cylinder lateral + cylinder base + hemisphere) = 42π+9π+18π = 69π ≈ 216.77 cm².
Cylinder volume = π×9×7 = 63π.
Hemisphere volume = (2/3)π×27 = 18π.
Total volume = 63π + 18π = 81π ≈ 254.47 cm³.
Page 14: Distance on Earth’s Surface
The earth is approximated as a sphere. Longitudes and latitudes are used to locate points on the surface. Distances can be computed using spherical geometry.
Image: Longitudes and Latitudes
Example 26: Two cities at the same latitude 40ºN differ by 10º in longitude. Estimate the distance between them (r = 6371 km).
Example 27: Use the haversine formula (simplified) to calculate the great-circle distance between two points (example values provided).
Solution:
(A simplified version: Given latitudes 30º and 35º, and longitudes differing by 5º, apply the haversine formula to compute d. Detailed steps would be provided in a full solution.)
Example 28: For points at (40ºN, 70ºW) and (40ºN, 80ºW), compute the distance along the parallel.
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