Question

. Use Ohm's law and voltage division to find the power supplied by the source and power dissipated in 60 and 8Ω resistors. 1.2Ω 4Ω 20 A 2Ω 8Ω 12Ω

Answer 1

8 omega and 12 omega are parallel = times 12/8 + 12 = 96/20 = 4.8 3 omega and 6 omega are parallel = 3 times 6/3 + 6 = 18/9 = 2 omega 4 omega and 2 omega are series = 4 + 2 = 6 omega 4.8 omega and 1.2 omega are series = 4.8 + 1.2 = 6 omega 6 omega and 6 omega are parallel = 6 times 6/6 + 6 = 36/12 = 3 omega 3 omega and 1 omega are series = 3 + 1 = 4 omega 2 omega and 4 omega are parallel = 2 times 4/2 + 4 = 8/6 = 1.333 omega ohm's law V = IR voltage v = 20 times 1.333 = 26.666 v power supplied by source P = VI P = 26.666 times 20 P = 533.333 w Apply current division rule I_3 omega = 20[2/2 + (1 + 3)] I_3 omega = 40/6 = 6.6667