Assignment5_Solutions

Assignment5_Solutions - P2.80 Firs‘r we write a node...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P2.80* Firs‘r, we write a node vol’rage equa’rion To solve for The open-circuit volTage: JOJL 4. 0V ’ 0 vs; V“ —10+VOC _1 10 5 _ Solving, we find V” = 6.667 V. Then zeroing The sources, we have This cir‘cui'r: iOJi. 1 Thus, R, = = 3333 Q. The Thévenin and Norfon equivalenTs are: 1 3.333JL P2.81* The equivalent circuit of The battery with The resisTance connec'red is CW 9 loan. ;" = 6.3100 2 0.06 A R, = a = 50 Q P2.82 WiTh open-circuiT condiTions: 6 32 W5, 0 u 24 Q 3 A 6 D O h Solving, we find vflb = —12 V. WiTh The source zeroed: (.3 5.2 U 24 El 6 Q ‘ Ob 1 = — = 5 Q R’ 1/6 +1f[6 + 24) The equivolenT circuits are: R,= 5 :2 v, = 12 v NoTice The source poloriTy reloTive To Terminals aond b. P2.84 The 7-9 resisTor has no effecT on The equivolenT circuiTs because The volToge across The 12-V source is independenT of The resisTor value. R, = 12 o P237 The equivoienT circuiT I.uiTh a load oTToched is: E4.» Vt a Rs. For a load of 2.2 kfl, we have i, = 4.4/2200 = 2 mA , and we can write V, = i4 —R,t‘,. SubsTiTuTing values This becomes 4.4 = V, — 000219, (1) Similarly, for The 10-141 load we ob’rain 5 = V; — 0.000519, (2] Soiving Equcrrions (1) and (2), we find V, = 5.2 V and R, = 400 Q. P2.88 Open-circui'r condi'rions: 5 Q 1'" 35 ll] £2 . 15—v, v, . . _ . . _ r, — 5 10 +10 r, + 0.5;, — O Solvmg, we find 1/, — 10 IWand 10 10+10: then we have V, = V = v, 5V. SC ShorT—circuiT condi'rions: 5 g; 1*, 10 $2 [SV ix 215;” 5% — I; + 0.51; = 0 Solving, we find VX = 7.5Vand Then we have {M = % = 0.75A. Then, we have 1% = V H, 26.679. Thus The DC .' .56 equivalenTs are: Ra, = 6.6? a P2.89 As is Problem P260, we find The Thévenin equivalent: 4.9m v 9 9L -- Rx Then maximum power is obTained for a load resisTance equal To The Thévenin resisTance. i’ '2 P zlvr’zl 23.333w R max P2.91* To maximize The power To R1, we musT maximize The volTage across iT. Thus, we need To have R = 0. The maximum power is 2 P :20 max =80W ...
View Full Document

{[ snackBarMessage ]}

Page1 / 4

Assignment5_Solutions - P2.80 Firs‘r we write a node...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online