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

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2a) R4= 100Ωb) R4= 1kΩc) R4= 10kΩd) R = 100kΩ4.Assumptions:The voltmeter behavior is modeled as that of an ideal voltmeter in parallel with a 120- kΩ resistor.Analysis:We develop first an expression forTherefore,IIIV RS= I1 S RS+ R1+ R2|| R 2= I4 R1 R2+ R3+ R4RR= I RRRS+ R1 + R2||VR 4in terms of R4 . Next, using current division:( R + R )34 ⋅ ( R ) + + 3+ R4 R2R34 R 4 SRSR4= IR444 RSR4= IS RS+ R1+ R2||66000 ⋅ R4=6R + 2.1352⋅10Without the voltmeter:a)b)c)d)VR 4VR 4VR 4= 3.08 V= 30.47 V= 269.91 VVR 4= 1260.7 V.( R + R )34 ⋅ RNow we must find the voltage drop across R4with a 120-kΩ resistor across R4. This is the voltage thatthe voltmeter will read.IV= I RR122R2R+ R23+ RRS ⋅ R2|| 4( R) + + Ω 3+ ( R4||120kΩ R2R3( R ||120 ) R 4 S+ R + RkSR4= IR4= I ISS4 R= 82.57319RS+ R1+ R2|| ( R3+ ( R4||120kΩ)( 120000 + R ) ⋅ R44RSR44+ 320.28⋅1064 ⋅ R2+ R3R2+ ( R4||120 )kΩ2.48