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Unformatted text preview: P 3.10 Req = 63020 = 49
”30A = U49 2 (30 A)(4Q) = 120 V Therefore, since the three original resistors are in parallel with the current
source: ”300 = W30A = 120 V , 2 2 2
Thus, P3052 = —*U309 =1 1 0 _ =4
30 30 80 W P 3.18 = 80, Therefore, R1 + R2 + R3 = 7.2 Q R1+R2+R3 (R1 + R2)24 _ 12
(R1 + R2 + R3) — Therefore, 2(R1 + R2) = R1 + R2 + R3 Thus, R1 + R2 = R3; 2R3 = 7.2; R3 = 3.69 R2(24) _ 5
R1 + R2 + R3 4.8R2 2: R1 + R2 + 3.6 = 7.2
Thus, R2 = 159, R1 = 7.2  R2 ~ R3 = 210 P327 54Q27§2=18Q; 189+29=209; 20]](10+15+35):15§2; 675 = r
30+15 10A Therefore, 2'9 2 . __ 20HGO _ . _ ___ 2754
229 .. 20 (15) _ 11.25 A, 2.,  $411.25)  7.5 A P 4.1 [a] There are six circuit components, ﬁve resistors and the current source. Since the current is known only in the current source, it is unknown in
the ﬁve resistors. Therefore there are ﬁve unknown currents. [b] There are four essential nodes in this circuit, identiﬁed by the dark black dots in Fig. P44. At three of these nodes you can write KCL equations
that will be independent of one another. A KCL equation at the fourth
node would be dependent on the ﬁrst three. Therefore there are three
independent KCL equations. Sum the currents at any three of the four essential nodes a, b, c, and (1.
Using nodes a, b, and c we get —z‘g+z'1+z'2=0
“21+i4+23=0 i5—i2M’i3=0 [d] There are three meshes in this circuit: one on the left with the components 2'9, R1, and R4; one on the top right with components R1,
R2, and R3; and one on the bottom right with components R3, R4, and
R5. We cannot write a KVL equation for the left mesh because we don’t
know the voltage drop across the current source. Therefore, we can write
KVL equations for the two meshes on the right, giving a total of two
independent KVL equations. [e] Sum the voltages around two independent closed paths, avoiding a path that contains the independent current source since the voltage across the current source is not known. Using the upper and lower meshes formed
by the ﬁve resistors gives Rlil + R3’i3 " RQ’iQ : 0
Rgig 'i" R5’i5 ‘— R4’é4 = 0 P 4.6 Use the lower terminal of the 5 S2 resistor as the reference node. 110—60 1),, _
10 +~5~+3—0 Solving, 7),, = 10 V ...
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 Summer '06
 Milenkovic

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