This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1902 Excerpt: ...can be no iron in the path of the leakage flux from the outside coil, so the reluctance will be twice as great. The value that is represented by A for the inner coil becomes A + 2X + 2g for the outer. Likewise B is replaced by B + 2X + 2g, g being the space occupied by insulation between the coils. If the secondary circuit is open the secondary coil is idle, equivalent to so much air, and all the flux set up by the primary is leakage flux. As the secondary resistance can be replaced by an equivalent primary resistance, =--„" for purposes of calculation, so also the secondary inductance can be replaced by an equivalent inductance in the primary, /.,=--'. These values, Lp and L3, are to be used in the formula at the end of the last section for determining the regulation of a transformer. 54. Exact Solution of a Transformer In the treatment of regulation, efficiency, etc., heretofore, certain small errors have been allowed, due to neglecting the effects of the core, eddy currents, and hysteresis losses. The following graphic solution, adapted from Steinmetz, takes account of all these effects, and is general in all respects. It must first be understood that there are three fluxes to be considered: (1) The useful flux that links both coils. It is not in any definite phase with either / or /,. It is, however, always at right angles to the E.M.F. it induces, the direct in the secondary, ami the counter in the primary. (2) The leakage flux of the primary coil. This links the primary only, and being independent of /, is always in phase with /,,. (3) The leakage flux of the secondary coil. This is similarly in phase with /,. Let E,--E.M.F. induced in secondary, V, = difference of potential at secondary ter-minals, Ep = impressed primary pressure, E„...
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