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HW #1 Solutions

# HW #1 Solutions - P-1.9(2,4,5 This problem is similar to...

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Unformatted text preview: P-1.9 (2,4,5) This problem is similar to Exampie 1-7. 1-1032” 39 1:3“ ) . , ' in“) _ ‘ me e 0 W) - Figure 1.7: Circuit for Problem P- ' 1.9. (a) How many nodes are present in the circuit in Figure 1.7? ' (b) How many meshes are present in that circuit? (0) Write a KCL equation at every node in the circuit in terms of the indicated current and voltage variables. (d) Write a KVL equation for every mesh in the circuit. Write those equations using the indicated current variables by incorporating Ohm’s Law for each resistor. Solution: (a) The circuit has four nodes, three on the top row of the circuit and one at the bottom. (b) The circuit has three meshes. (c) Let node a be the connection between the positive terminal of the voltage source and the 2 Q resistor, node 5 be the connection where the three resistors are joined, node c be the connection joining the 39 resistor, 49 resistor, and the current source, and node 0'! be the one at the base of the circuit. Then, the four KCL equations are: node a: ig(t) — i105) = 0. node b: i105) — i2(t) + i305) = 0 node c: —2'3 (15) -— £46,) + z", (t) = 0 node d: —_z'o(t) + i2 (13) + \$402) — is (t) g 0. a (d) Denote the three meshes as the left, middle, and right mesh, respectively. Then the three KVL equations are: left mesh: —v_, (t) + 2i1(t) + £20?) = 0 middle mesh: ——i2(t) — 31303) + 4i4(t) = 0 right mesh: —4i4 (t) + '00 (t) = 0. P-1.10‘ (23,5) _ . All resistances are measured in Ohms. This problem is similar to Example 177. (3.1%ch many meshes are present in the circuit in Figure 1.8? (b) How many nodes are present in that circuit? (c) Write a KVL equation at every mesh in the circuit in terms of the indi- cated voltage and current variables. (d) Write a KCL equation for every node in the circuit. Write those equations using the indicated voltage variables by incorporating Ohm’s Law for each resistor. 1: Figure 1.8: Circuit for Problem P- 1.10. Solution: (a) This circuit hes three meshes. Call them the top mesh, left mesh, and right mesh. . (b) This circuit has four nodes. In this circuit each is marked by I one of the solid dots. Beginning. at the upper left node and ‘ proceeding clockwise, call these nodes a, b, c, and d. - (c) The KVL equations are: top mesh: —’Ug(t) + v3(t) - 0405) = 0 left mesh: v1(t) + v2'(t) — vs(t) : 0 right mesh: —v1(t) + 12403) - v6(t) = 0. (d) The KCL equations at all of the nodes are: “205) + 113“) node of 5 15 + i§(t) r: 0 _ ”165) _ ’02“) ”46) __ “Odeb' . 20 5 + 10' ’0 . “U303 __ 114(15)- _. __ node c. 15 10 35(t) — O i ‘1 node d: -— “12%) — i5(t) + is (t) = 0. P—1.24 (4,5) (a) Write the KCL equations that constrain the currents at all of the nodes " of the network in Figure 1.29. (b) Write the KVL equations that constrain. the voltages for all of the meshes in that same network. ________——____________ Solution: (a) The network contains four nodes. If we write the KCL equa- tions by summing the currents that enter; the nodes, then the R4 Figure 1.29: Circuit for Problem P« . 1.24. Z4U) four equations are: node a: —i1(t) — 722(t) — 2'5 (t) = 0 node I): +126) + 1012“) '- is (t) = 0 node c: +i1(t) — 10i2(t) — 2'46) = 0 node d: +z'3 (t) + £405) + is (t) = 0. (b) Let the resistor voltages be deﬁned so that the current arrows are deﬁned as pointing from the + to the - terminals. Then the KVL equations on the three meshes are: mesh 1: v1(t) — vs(t) — v2(t) = 0 mesh 2: —vs(t). + 112(15) + v3(t) = 0 mesh 3: -—v3(t) + '05 (t) + v4(t) = 0. Recall that the voltage for each resistor gets a + sign if the path goes in the direction of the current (i.e., downstream) and a w sign if the path is upstream. 124.34 (4,5,6) For the circuit of Figure 1.35, express I005) in terms of is (t). . —-—-~—-——._.___.______,___________ Solution: Let v1(t) be the voltage across the 129 resistor (top-to- bottom) and let v20?) be the voltage across the 29 resistor (left-to- right). 'Note that v(t) is the voltage across both the 69 and the 129 resistors. From KVL in the middle mesh ‘ ' —v1(t) + v2(t) + ’U(t) an. 2 \$2 + Mt) o 129 we) _ Figure 1.35: Circuit for Problem P- 1.34. From KCL at the top left node gem + gee) = am). From KCL at the top right node gems) — %v(t) — 11—24:) = 0. From the last equation, we learn that 1 ‘02“) = all“). If we substitute this result into the top equation, we see that v1(t) = gum. Substituting for 111(t) and 112(t) into the remaining equation gives 1 gm) + its) = i5(t) OI" 8 v(t) = Ease). P-1.42 (1,4,5,6) Determine the voltage v(t) in the circuit in Figure 1.41. 129 m) va(t) e iv(t) 0 see) Figure 1.41: Figure for Problem P- 1.42. ' Solution: In the ﬁgure below We indicate some additional vari- ables. 129 it) Notice that we have let the current ﬂowing through the 40 resistor be denoted by £105). The current ﬂowing downward through the 29 resistor is v03) / 2. We can set up one KCL equation (at the circled node), and two KVL equations to solve for the three variables é(t), v(t), and i1(t). ' KCL: i1(t) + 62‘(t) 53—00:) = o KVL 0:: 4516,) + v(t) = @505) KVL ﬂ: 1212(15) —- 43.10?) = O. From the third equation ﬁt) 2 3m). Substituting this fact into the ﬁrst equation gives 9i(t)—-1—v(t)= 0 => 13(t) = its). 2 18 Finally, substituting this result into the second equation gives gvﬁf) + 1105) = wt) or 3 v(t) n Eu, (t). P-1.44 (4,5,6) In the circuit in Figure 1.43, which contains a voltage-dependent current source, determine v(t). + at) _ at) o 0 200:) Figure 1.43: Circuit for Prol: 1.44. ‘ Solution: We could solve this problem by setting up and solving the usual set of KVL equations, KCL equations, and element rela- tions. There is nothing wrong with approaching this problem that way. An alternative, however, is to evaluate some of the voltages and currents in the circuit whose value can be immediately deter- mined and then use these values to evaluate others. That is the approach discussed here. The current ﬂowing left to right through the horizontal 19 resis- tor is v(t). Since a current 2v(t) comes from the dependent current source, KCL tells us that a current of 3v(t) must ﬂow through the rightmost 2Q resistor from top to bottom. This will induce 3. volt- age of 6v(t). This means that there is a voltage of 7v(t) acress the leftmost 2Q resistor (by KVL) and a current of £1,105) ﬂowing through it. Since the current through the horizontal resistor was v(t), by KCL We have ﬁnally vct) + gum = Mi) .5 or “005) = 3453(1). ...
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