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hw03solution - P 3.10 Req = 6||30||20 = 49 ”30A = U49...

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Unformatted text preview: P 3.10 Req = 6||30||20 = 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 54Q||27§2=18Q; 189+29=209; 20]](10+15+35):15§2; 675 = r 30+15 10A Therefore, 2'9 2 . __ 20HGO _ . _ ___ 27||54 229 .. 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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