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

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2IRFS= ITFS− Imˆ FS= 8.5 mA −100µ A = 8. 4 mAVRFS = VTFS= Vmˆ FS= 250 mVVRFS250 mVOhm law: R = = = 29. 76 Ω .I 8.4 mARFSc) from sensor characteristic: 220 °C –410 °C.________________________________________________________________________Solution:Problem 2.57Known quantities:Meter resistance of the coil; meter voltage at full scale; max measurable temperature.Find:a) The circuit required to meet the specifications of the new sensor.b) The value of each component of the circuitc) The linear range of the system.Assumptions:Sensor characteristics follow what is shown in Figure P2.57Analysis:a) A parallel resistor is required to shunt (bypass) the excess current.b) At full scale, meter:VmFSˆ = 250 mVrmˆ = 2.5 kΩ0. L.:VmFSˆImFSˆ = = 100 µ A.rmˆat full scale, sensor (from characteristics):T FS= 400 °CI TFS= 8. 5 mAKCL : − ITFS+ IRFS+ Imˆ FS= 0IRFS= ITFS− Imˆ FS= 8.5 mA −100µ A = 8. 4 mAVRFS = VTFS= Vmˆ FS= 250 mVVRFS250 mVOhm law: R = = = 29. 76 Ω .I 8.4 mARFSc) from sensor characteristic: 220 °C –410 °C.________________________________________________________________________Solution:Problem 2.58Known quantities:Schematic of the circuit shown in Figure P2.58; voltage at terminals with switch open and closed for freshbattery; same voltages for the same battery after 1 year.2.44