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Electrical Engineering
A truck weighing station incorporates a sensor whose resistance varies linearly with the weight: Rs = Ro+aw. Here Ro is a constant value, a a proportionality factor, and W the weight of each truck. Suppose Rs plays the role of R2 in the noninverting amplifier. Also, Ap=co, Vin = 1 V, R1=9 ko, Ro=1 ko, a=0.001 kn/kg. Calculate the output voltages for the following cases; 1) when the weighing station is empty II) when a truck of 3500 kg is on the weighing station III) recalculate the II), if an opamp with an open loop gain of A0=50 is employed in the circuit LoVout Vin Ao R₁ HI Rs O a. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,220 V O b. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,400 V O c. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,920 V O d. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout = 1,400 V O e. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,220 V O f. 1) Vout=8,786 V, II) Vout =1,666 V, III) Vout =1,220 V O g. 1) Vout-7,029 V, II) Vout=1,666 V, III) Vout=1,920 V O h. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout =1,920 V

A truck weighing station incorporates a sensor whose resistance varies linearly with the weight: Rs = Ro+aw. Here Ro is a constant value, a a proportionality factor, and W the weight of each truck. Suppose Rs plays the role of R2 in the noninverting amplifier. Also, Ap=co, Vin = 1 V, R1=9 ko, Ro=1 ko, a=0.001 kn/kg. Calculate the output voltages for the following cases; 1) when the weighing station is empty II) when a truck of 3500 kg is on the weighing station III) recalculate the II), if an opamp with an open loop gain of A0=50 is employed in the circuit LoVout Vin Ao R₁ HI Rs O a. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,220 V O b. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,400 V O c. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,920 V O d. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout = 1,400 V O e. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,220 V O f. 1) Vout=8,786 V, II) Vout =1,666 V, III) Vout =1,220 V O g. 1) Vout-7,029 V, II) Vout=1,666 V, III) Vout=1,920 V O h. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout =1,920 V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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A truck weighing station incorporates a sensor whose resistance varies linearly with the weight: Rs = Ro+aw. Here Ro is a constant value, a a proportionality factor, and W the weight of each truck. Suppose Rs plays the role of R2 in the noninverting amplifier. Also, Ap=co, Vin = 1 V, R1=9 ko, Ro=1 ko, a=0.001 kn/kg. Calculate the output voltages for the following cases; 1) when the weighing station is empty II) when a truck of 3500 kg is on the weighing station III) recalculate the II), if an opamp with an open loop gain of A0=50 is employed in the circuit LoVout Vin Ao R₁ HI Rs O a. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,220 V O b. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,400 V O c. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,920 V O d. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout = 1,400 V O e. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,220 V O f. 1) Vout=8,786 V, II) Vout =1,666 V, III) Vout =1,220 V O g. 1) Vout-7,029 V, II) Vout=1,666 V, III) Vout=1,920 V O h. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout =1,920 V
Transcribed Image Text:A truck weighing station incorporates a sensor whose resistance varies linearly with the weight: Rs = Ro+aw. Here Ro is a constant value, a a proportionality factor, and W the weight of each truck. Suppose Rs plays the role of R2 in the noninverting amplifier. Also, Ap=co, Vin = 1 V, R1=9 ko, Ro=1 ko, a=0.001 kn/kg. Calculate the output voltages for the following cases; 1) when the weighing station is empty II) when a truck of 3500 kg is on the weighing station III) recalculate the II), if an opamp with an open loop gain of A0=50 is employed in the circuit LoVout Vin Ao R₁ HI Rs O a. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,220 V O b. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,400 V O c. 1) Vout=8,786 V, II) Vout=1,666 V, III) Vout=1,920 V O d. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout = 1,400 V O e. 1) Vout-7,029 V, II) Vout=1,469 V, III) Vout=1,220 V O f. 1) Vout=8,786 V, II) Vout =1,666 V, III) Vout =1,220 V O g. 1) Vout-7,029 V, II) Vout=1,666 V, III) Vout=1,920 V O h. 1) Vout=8,786 V, II) Vout=1,469 V, III) Vout =1,920 V
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