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Unformatted text preview: P2.80* Firs‘r, we write a node vol’rage equa’rion To solve for The opencircuit 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 opencircuiT 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 79 resisTor has no effecT on The equivolenT circuiTs because The volToge across The 12V 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 10141 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 Opencircui'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 ...
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 Spring '08
 DABBAGH

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