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Unformatted text preview: The two node voltage equations are ‘01 91 Til—1’2
—15 —— — = 0
+60+15+ 5 ﬁg ’Ug—Ul
5 —
+2+ 5 Place these equations in standard form:
t: (1 + 1 +1) + ( 1)  15
1 60 15 5 “2 5 " +v(1+1) 5
“I 5 225— Solving, cl = 60 V and v2 = 10 V;
Therefore, i1 = ('01 — v2)/5 = 10 A [b] Plea: “(15 A)’Ul = “( 15 A)(60 V) = —900 W i 900 W(delivered)
[c] pm = (5 Am = (5 A)(10 V) = 50 W: —50 W(delivered) = 0 AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown: in“ + V1 so i 29 V2 49 4.5A 1;: v . 12:: 30%! The two node voltage equations are: ‘1}—
—4..5+1—”l+ 1 v2 = 0
1 6+2
’Ug 02*31 U2""30 _
1.2+ 6+2 +. 4 _ 0 Place these. equations in standard form: 1 Solving, '01 = 6 V '02 = 18 V To ﬁnd the voltage 1;, ﬁrst ﬁnd the current 21 through the series—connected 6 Q and 2 Q
resistors: i_’v1——*v2 6+2 8 = ——1.5A Using a KVL equation, calculate an: v=2i+v2=2(—1.5)+18=15V ————I——_ . —'I—I———.___— —__—_ _
—"———' ————_. P 4.6 Note that we have chosen the lower node as the reference node, and that the voltage
at the upper node with respectito the reference node is 1;... Write a KCL equation the equation: . 
on + 25 + 511., + 5 = 0 6‘00 = {—30 so on = —30/6 = 5 V P 4.7 [a] From the solution to Problem 4.6 we know on = —5 V; therefore
P401114 = (*5)(0.04) = 0.2'W
The power developed by the 40 mA source is 200 mW , [b] The current into the negative terminal of the 25 V source in the ﬁgure of
Problem 4.6 is ' 3'9 = (—5 + 25)/125 = 160 mA The power in the 25 V source is 19251; = —(25)(0.16) = —4 W The power develoPed by the 25 V source is 4 W [c] 3359 = ((116)2(5) = 128 mw
mm = (016)2(120) =3.072w
1925:: =. (5)2/25 =1 W Zpdi. = 0.123 + 3.072 + 1 = 4.2 w
229444 = 0.2 + 4 = 4.2 w (checks!) ...
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This note was uploaded on 08/12/2010 for the course ENGR 117 taught by Professor Nickarnold during the Fall '08 term at Santa Barbara City.
 Fall '08
 NickArnold

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