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Erschienen in:
1994 | OriginalPaper | Buchkapitel
Chapter 15
verfasst von : L. A. A. Warnes
Erschienen in: Solutions to Problems
Verlag: Macmillan Education UK
Enthalten in: Professional Book Archive
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1 Using the e.m.f. equation for the side with the most turns (the secondary) we find <m:math display=’block’> <m:mrow> <m:msub> <m:mi>E</m:mi> <m:mn>2</m:mn> </m:msub> <m:mo>=</m:mo><m:mn>2200</m:mn><m:mo>=</m:mo><m:mn>4.44</m:mn><m:msub> <m:mi>B</m:mi> <m:mi>m</m:mi> </m:msub> <m:mi>A</m:mi><m:mi>f</m:mi><m:msub> <m:mi>N</m:mi> <m:mn>2</m:mn> </m:msub> <m:mo>=</m:mo><m:mn>4.44</m:mn><m:mo>×</m:mo><m:mn>0.55</m:mn><m:mo>×</m:mo><m:msup> <m:mrow> <m:mn>0.55</m:mn></m:mrow> <m:mn>2</m:mn> </m:msup> <m:mo>×</m:mo><m:mn>15000</m:mn><m:msub> <m:mi>N</m:mi> <m:mn>2</m:mn> </m:msub> </m:mrow> </m:math>]]</EquationSource><EquationSource Format="TEX"><![CDATA[$${E_2} = 2200 = 4.44{B_m}Af{N_2} = 4.44 \times 0.55 \times {0.55^2} \times 15000{N_2}$$ Whence <m:math display=’block’> <m:mrow> <m:msub> <m:mi>N</m:mi> <m:mn>2</m:mn> </m:msub> <m:mo>=</m:mo><m:mfrac> <m:mrow> <m:mn>2200</m:mn></m:mrow> <m:mrow> <m:mn>4.44</m:mn><m:mo>×</m:mo><m:mn>0.55</m:mn><m:mo>×</m:mo><m:msup> <m:mrow> <m:mn>0.015</m:mn></m:mrow> <m:mn>2</m:mn> </m:msup> <m:mo>×</m:mo><m:mn>15000</m:mn></m:mrow> </m:mfrac> <m:mo>=</m:mo><m:mn>267</m:mn><m:mtext> </m:mtext><m:mi>t</m:mi><m:mi>u</m:mi><m:mi>r</m:mi><m:mi>n</m:mi><m:mi>s</m:mi></m:mrow> </m:math>]]</EquationSource><EquationSource Format="TEX"><![CDATA[$${N_2} = \frac{{2200}}{{4.44 \times 0.55 \times {{0.015}^2} \times 15000}} = 267{\kern 1pt} turns$$ And the primary turns are then <m:math display=’block’> <m:mrow> <m:msub> <m:mi>N</m:mi> <m:mn>1</m:mn> </m:msub> <m:mo>=</m:mo><m:mfrac> <m:mrow> <m:msub> <m:mi>E</m:mi> <m:mn>1</m:mn> </m:msub> <m:msub> <m:mi>N</m:mi> <m:mn>2</m:mn> </m:msub> </m:mrow> <m:mrow> <m:msub> <m:mi>E</m:mi> <m:mn>2</m:mn> </m:msub> </m:mrow> </m:mfrac> <m:mo>=</m:mo><m:mfrac> <m:mrow> <m:mn>500</m:mn><m:mo>×</m:mo><m:mn>267</m:mn></m:mrow> <m:mrow> <m:mn>2200</m:mn></m:mrow> </m:mfrac> <m:mo>=</m:mo><m:mn>60.7</m:mn><m:mtext> </m:mtext><m:mi>t</m:mi><m:mi>u</m:mi><m:mi>r</m:mi><m:mi>n</m:mi><m:mi>s</m:mi></m:mrow> </m:math>]]</EquationSource><EquationSource Format="TEX"><![CDATA[$${N_1} = \frac{{{E_1}{N_2}}}{{{E_2}}} = \frac{{500 \times 267}}{{2200}} = 60.7{\kern 1pt} turns$$ The nearest whole number is 61.