**Problem C1**Design values for Ra, Rb, and Rc so that Vout is determined by the following equation. Rd = 10k\[ V_{out} = 8(V_a - V_b) \]**Problem C2**Design values for Ra, Rb, and Rc so that Vout is determined by the following equation. Rd = 10k\[ V_{out} = 10V_a - 4V_b \]**Circuit Explanation:**The diagram shows an operational amplifier (op-amp) configuration with four resistors: Ra, Rb, Rc, and Rd. The input voltages are Va and Vb, and the output voltage is Vout.- **Ra** and **Rb** are connected in series with Va entering Ra, and there's a junction between Ra and Rb.- **Rc** is connected from the junction between Ra and Rb to the inverting input (negative terminal) of the op-amp.- **Vb** is directly connected to the junction of Rc and Rd.- **Rd** connects from the inverting input to the output of the op-amp, forming a feedback loop.The non-inverting input (positive terminal) is directly connected to the ground. Vcc and -Vcc represent the positive and negative supply voltages for the op-amp, allowing it to operate with bipolar signals.

Design value for Ra, Rb and Rc so that Vout is determined by the following equation Vout=8(Va-Vb) and also same for the value by the eqaution Vout=10Va-4VbSince This is an ideal operational amplifier Current entering the operational amplifier is zero I1=0A Also op-amp is connected in negative feedback configuration apply virtual short concept V1=V2 also I1=0A use voltage division rule V1=Va×RbRa+Rb Apply KCL at node V2 V2-VbRc+V2-Vout10K=0 on rearranging V2Rc+V210K-VbRc=Vout10K since V1=V2=Va×RbRa+Rb Vout=[Va×RbRa+Rb][Rc+10KRc]-Vb(10KRc) Vout=Va[Ra(Rc+10K)(Ra+Rb)Rc]-Vb[10K(Ra+Rb)(Ra+Rb)Rc] Vout=1(Ra+Rb)Rc[Rb(Rc+10K).Va-10K(Ra-Rb).Vb].................................(1)From given C1 Problem Vout=8(Va-Vb)....................................................(2) On comparing equation (1) and (2) we get, 1(Ra+Rb)Rc=8.....................................(3) Rb(Rc+10K)=1....................................(4) 10K(Ra+Rb)=1 (Ra+Rb)=110K.....................................(5) From equation 3 and 5 we get 1(110k)Rc=8 Rc=10K8 ohm from equation 4 Rb(10K8+10K)=1 Rb=890K ohm from equation 5 Ra+890K=110K on solving we get Ra=190K ohmNow from the given C2 problem Vout=10Va-4Vb Rewrite the eqaution as Vout=2[5Va-2Vb]...............................(6) Now similarly comparing equations (1) and (6) 1(Ra+Rb)Rc=2..........................(7) similarly, we can solve for the other two equations. Ra(Rc+10K)=5............................(8) 10K(Ra+Rb)=2...............................(9) Now from equations 7 and 9 1(210K)Rc=2 Rc=10K4 ohm From equation 8 Rb(10K4+10K)=5 on solving we get Rb=410K ohm From equation 9 10K(Ra+410K)=2 on solving we get Ra=-210K ohm

Engineering

Electrical Engineering

Design values for Ra, Rb, and Rc so that Vout is determined by the following equation. Rd = 10k Ra V+ O- Vo Va Vou = 8(Va-V½) Rb а Vcc Rc Rd Vb "roblem C2 esign values for Ra, Rb, and Rc so that Vout is etermined by the following equation. Rd = 10k V =10V – 4V, -

Design values for Ra, Rb, and Rc so that Vout is determined by the following equation. Rd = 10k Ra V+ O- Vo Va Vou = 8(Va-V½) Rb а Vcc Rc Rd Vb "roblem C2 esign values for Ra, Rb, and Rc so that Vout is etermined by the following equation. Rd = 10k V =10V – 4V, -

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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