From the Rolls-Royce experimental archive: a quarter of a million communications from Rolls-Royce, 1906 to 1960's. Documents from the Sir Henry Royce Memorial Foundation (SHRMF).
Calculation to find the current distribution in an armature for minimum power loss.
| Identifier | ExFiles\Box 3\5\ 05-page086 | |
| Date | 30th July 1919 guessed | |
| To find the Current Distribution in the Armature which gives the Minimum C²R loss Let x amperes be the current delivered to the battery by the dynamo, and let z amperes be the current in the resistances, also let a amperes be the current in the bottom right hand section of the armature winding. The currents in the various circuits are then as shown above, in black. Since the current distribution is symmetrical in the armature, z - a = a - x + z ∴ x = 2a ∴ a = ½ x Substituting ½ x for a, the current distribution now becomes as shown in red. If r = resistance of 1 quadrant of the armature winding (res. of connections between main & auxiliary brushes is negligible) Then Loss y = 2r (z - ½x)² + 2r (½x)² = 2r [z² - zx + ¼x² + ¼x²] = 2r [z² + ½x² - zx] But x is to be constant ∴ dy{F R Danby}/dz = 2r [2z - x] P.T.O. | ||