Question

5. (35 points) (a) (10 points) Calculate the equivalent resistance of the diagrammed network of resistors. (b) (10 points) Calculate the current through each resistor. (c) (10 points) Calculate the power dissipated by each resistor. (d) (5 points) Show that the power supplied by the battery equals the sum of the power dissipated by the resistors. (e) (5 extra credit points) What is Vbc = Vc - V.? (f) (5 extra credit points) What value of R2 will make Vbc = 0, assuming R1, R3, and R4 remain unchanged? 5. V= 10.0V R, = 10.00 Re 20.02 R₂ = 15.00 Ry = 12.02 R b Ru wo

Answer 1

PAR7- Equivalent Resistance Regu (R,+R₂) || (R2 + Rq) Ri a d I Rq Rq 1. Regu Reque (16+15) 11 (20+12) Q511 32 25832 = 14.0352 25+32 Regu= 14.0350 Ans - PART B supply.cassent, I= V Rear -0.71 25 A 14.035 current through Ri I, IX (Rp+RA) = 0.7125 (32) Rit Rat Rzthe I = 0.44 S7 current through Re Iq= IX (Ritra) = 0.7125 x 25 Ret Ret Rytra 5 Ig-0.3125 A

current through R3 13= 1 = O. HA dve to series commission of R&R3 current through RA 14 I 2 0.3125 A (due to series combination of R28 eq) PART-C power dissipated in RA = 1.6W P, = 1R1 = (0.2710 power dissipated in R2 lg- I R2 = (0.3125)²x20 = 1.953w - power dissipated in Rz P= I3 R3 - (0.4)2x15=2.44 power dissipated with Rq Pc = (12)?R4 -(0.3125 x 12 =1-181835W part-d power supprenthe Battery by Py = V. I loxo-7125 = 7.125w Sum of power dissipated in all Resistor PR - Pithet 3+ 4 = 1.67 1.953 +2.4 + 1.14875 pv proved PR = 7.1256 =

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