By Charles Kerridge
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Then 180 + 200 − 600 = −220 (CHECKS) If v3 = −100 V, then v1 = −120 V and v2 = −200 V. 14 [a] Write a KVL equation clockwise aroud the right loop, starting below the 300 Ω resistor: −va + vb = −0 so va = v b Using Ohm’s law, va = 300ia and vb = 75ib Substituting, so 300ia = 75ib ib = 4ia Write a KCL equation at the top middle node, summing the currents leaving: −ig + ia + ib = 0 so ig = ia + ib = ia + 4ia = 5ia Write a KVL equation clockwise around the left loop, starting below the voltage source: −200 V + v40 + va = 0 From Ohm’s law, v40 = 40ig and va = 300ia CHAPTER 2.
6 A. 6 A) = 120 V. 4 A)] = −(200 V)(2 A) = −400 W Thus the voltage source delivers 400 W of power to the circuit. 16 [a] Write a KVL equation clockwise around the right loop: −v60 + v30 + v90 = 0 From Ohm’s law, v60 = (60 Ω)(4 A) = 240 V, v30 = 30io , v90 = 90io Substituting, −240 V + 30io + 90io = 0 Thus io = so 120io = 240 V 240 V = 2A 120 Now write a KCL equatiohn at the top middle node, summing the currents leaving: −ig + 4 A + io = 0 so ig = 4 A + 2 A = 6 A [b] Write a KVL equation clockwise around the left loop: −vo + v60 = 0 so vo = v60 = 240 V [c] Calculate power using p = vi for the source and p = Ri2 for the resistors: psource = −vo ig = −(240 V)(6 A) = −1440 W p60Ω = 42 (60) = 960 W p30Ω = 30i2o = (30)22 = 120 W p90Ω = 90i2o = (90)22 = 360 W Pdev = 1440 W Pabs = 960 + 120 + 360 = 1440 W CHAPTER 2.
5 [a] Yes, independent voltage sources can carry whatever current is required by the connection; independent current source can support any voltage required by the connection. 6 Write the two KCL equations, summing the currents leaving the node: KCL, top node: 25A − 20A − 5A = 0A KCL, bottom node: − 25A + 20A + 5A = 0A Write the three KVL equations, summing the voltages in a clockwise direction: KVL, left loop: KVL, right loop: − v25 + v20 = 0 60V − 100V − v5 − v20 = 0 Problems 2–13 60V − 100V − v5 − v25 = 0 KVL, outer loop: Note that since v5 , v20 , and v25 are not speciﬁed, we can choose values that satisfy the equations.