ECE-320 CH02.pdf - Unix.eng.ua.edu

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G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2J aP = VI = 110 V= 53a sb)( 23 a) = 2.53 K 2. KWJsW = Pt = 2 .53 K 24 hr 3600 = 218. 6 MJs hrc)centsCost = ( Rate)W = 6cents =KW hr( 2.53 KW )( 24 hr) = 364.3 $3. 64________________________________________________________________________Solution:Problem 2.10Known quantities:Current through elements A, B and C shown in Figure P2.10; voltage across elements A, B and C.Find:Which components are absorbing power, which are supplying power; verify the conservation of power.Analysis:A absorbsB absorbsC supplies( 35 V )(15 A)= 525 W( 15 V )(15 A)= 225 W( 50 V )(15 A)= 750 W= P C= 750 W= PB + P = 225 W + 525 W = 750 WTotal power suppliedTotal power absorbedATotal power supplied = Total power absorbed, so conservation of power is satisfied.________________________________________________________________________Solution:Problem 2.11Known quantities:Circuit shown in Figure P2.11 with voltage source,R s= 5 kΩ ; and resistance of the load, R L= 7 kΩ.V s= 12V; internal resistance of the source,Find:The terminal voltage of the source; the power supplied to the circuit, the efficiency of the circuit.Assumptions:Assume that the only loss is due to the internal resistance of the source.2.8
G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Analysis:KVL : −VS+ ITRS+ VT= 0VTOL : VT= ITRL∴ IT=RLVT−VS+ RS+ VT= 0RLVS12 VVSRL12 V 7 KΩVT= = = 7 V or VD : VT= == 7 V.RS5 KΩR R 5 K 7 K1 1S+LΩ + Ω+ +RL7 KΩ2 22VRVT( 7 V )PL = = = = 7 mWR R VL L7 Ka2P PoutR ITRL7 KΩLη = = === 0.5833 or 58.33% .2 2P P + P I R + I R 5 KΩ + 7 KΩinRSRLTS________________________________________________________________________Solution:Problem 2.12TLKnown quantities:Circuit shown in Figure P2.12; Current through elements D and E; voltage across elements B, C and E.Find:a) Which components are absorbing power and which are supplying powerb) Verify the conservation of power.Analysis:a)By KCL, the current through element B is 5 A, to the right. By KVL,v = v = V (positive at the top)D E10vA+ 5 −10−10= 0Therefore the voltage across element A isv A= 15 V (positive on top)A suppliesB suppliesC absorbsD absorbsE absorbsb)( 15 V )(5 A)= 75 W( 5 V )(5 A)= 25 W( 10 V )(5 A)= 50 W( 10 V )(4 A)= 40 W( 10 V )(1 A)= 10 WTotal power supplied= PA + PB= 75 W + 25 W = 100W2.9
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ECE-320 CH02.pdf - Unix.eng.ua.edu

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