IRWIN 10e 5_109 - e11 ~ s: lot. n-,,, loJ::.5l_...

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-- Irwin, Basic Engineering Circuit Analysis, 10lE 5.1°9 Calculate the maximum power thatcan be transferred to R£in Fig. P5.109. 6V 6kO 2kil 3mA 10 k!l 10kO 9V Figure P5.109 SOLUTION: \jDe A B 0-----r---Mi'vL--~ -t"1/ - 6k. .JL. . iDl:. ..Q. ... VDC- t- 21.Sl. . GV t 10 k ,n, 4·sv 3mA YOe, .:: \81'"6- 4·0'= 14.5V ~11 ! , ---- ek:s: A B "H ·v t 11 21:. .52 f
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Unformatted text preview: e11 ~ s: lot. n-,,, loJ::.5l_ ->: I ,,'col .;",1 :::>~l i I;: r" Ril-j ::. Gt. 1"" ( IO~)(IDl:.) . -/Ol+ I Dt -II t.n. . .. Chapter 5: Additionai Analysis Techniques Problem 5.109 Irwin, Basic Engineering Circuit Analysis, 10/E 2 . i ! A + I 4.75 V II t.Q. .. I I -IS B. bYrnW -Problem 5.109 Chapter 5: Additional Analysis Techniques...
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IRWIN 10e 5_109 - e11 ~ s: lot. n-,,, loJ::.5l_...

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