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

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G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Pdiss = P3-A + P12-V =279 + 6.17 = 285.17 W________________________________________________________________________Solution:Problem 2.25Known quantities:Schematic of the circuit shown in Figure P2.25.Find:The power delivered by the dependent source.Analysis:24V24i = = A = 2 A(7 + 5) Ω 12i source= 0.5i2= 0.5⋅( 4) = 2 AThe voltage across the dependent source (+ ref. taken at the top) can be found by KVL:− v D+ ( 2A)(15Ω)+ 24V= 0 = 54v VTherefore, the power delivered by the dependent source isPD= vDisource= 54 ⋅2= 108 W.________________________________________________________________________Solution:Problem 2.26DKnown quantities:Schematic of the circuit in Figure P2.26.Find:a) If V1= 12.0V, R1= 0.15Ω,RL= 2. 55Ω, the load current and the power dissipated by theloadb) If a second battery is connected in parallel with battery 1with V2= 12.0V, R2= 0. 28Ω,determine the variations in the load current and in the power dissipated by the load due to theparallel connection with a second battery.Analysis:a)ILPV=R + RLoad112=0.15 + 2.551=2LL= I R = 50.4 W.L122.7= 4.44 Ab) with another source in the circuit we must find the new power dissipated by the load. To do so, we writeKVL twice using mesh currents to obtain 2 equations in 2 unknowns:2.20
G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2I−V( I1+ I2)+ V− I22=2R21 2 1R10R + I RL= VSolving the above equations gives us:20.28⋅I2.55⋅2− 0.15⋅I( I + I )I1= 2.95 A, I2=1. 58 A = I1 + I2= 4. 532P I R = 52.33 WLoad=L L1I LA21= 0+ 0.28⋅I2= 12This is an increase of 1%.________________________________________________________________________Solution:Problem 2.27Known quantities:Open-circuit voltage at the terminals of the power source is 50.8 V; voltage drop with a 10-W load attachedis to 49 V.Find:a) The voltage and the internal resistance of the sourceb) The voltage at its terminals with a 15-Ω load resistor attachedc) The current that can be derived from the source under short-circuit conditions.Analysis:(a)v s50. 8V(b)(c)(49V)RL2= 10W R = 240Ω= , the open-circuit voltageRRRSvL240= 49 50.8 = 49R + 240LSS+ RLS(240)(50.8)= − 240 = 8. 82Ω49Rv =R + RSv=15LS50.8 32. 0L8.82 + 15vS50.8( RL= 0) = = = 5.76R 8.82i ACCS________________________________________________________________________Solution:Problem 2.28=Known quantities:Voltage of the heater, maximum and minimum power dissipation; number of coils, schematics of theconfigurations.V2.21
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