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For the following network from the load perspective: R2 WW 750 E1 5V = RLEIK R4 12V E2 R3 w 100 A. (7) Find R Thevenin B. (8)
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Ri .c А 75ou Ru {21 RLEIK G=50 TOO B R₂ Ę=12v A. Jo find Rim (R Thevinin): open loop at A&B and replace energy lounced with tA Br. C. ERARIA R, =lkor 1 IN 25.87X10 A RN = Rm = 596.50m B maximum power 1 D. At RL = Rih =596.52 transfer takes place. 4.

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