How to solve for v_o of this weighted summer circuit?

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SUMMARY

The discussion focuses on deriving the output voltage expression (v_o) for a weighted summer circuit using operational amplifiers. The correct formula for v_o is established as v_{o} = \frac{v_{1} R_{a}}{R_{1}}\frac{R_{c}}{R_{b}} + \frac{v_{2} R_{a}}{R_{2}}\frac{R_{c}}{R_{b}} - \frac{v_{3} R_{c}}{R_{3}} - \frac{v_{4} R_{c}}{R_{4}}. The negative signs for the last two terms are clarified as a result of the inverting configuration of the operational amplifiers in the circuit. Understanding the relationship between the currents and resistances in the circuit is crucial for accurate calculations.

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Electrical engineering students, circuit designers, and anyone involved in analog circuit design and analysis will benefit from this discussion.

dla
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Homework Statement


I am trying to derive the expression for v_o for this weighted summer circuit. I've attached the circuit diagram.

Homework Equations





The Attempt at a Solution



I get that the first op-amp output is just the total current x R_a, and then you take that output and find the current of R_b + current of v_3/R_3 and v_4/R_4 and multply all that by R_c to get the v_o, this is what I got

\huge<br /> v_{o} = \frac{v_{1} R_{a}}{R_{1}}\frac{R_{c}}{R_{b}} +\frac{v_{2} R_{a}}{R_{2}}\frac{R_{c}}{R_{b}} + \frac{v_{3} R_{c}} {R_{3}} +\frac{v_{4} R_{c}} {R_{4}}

However the answer says that the last two terms are negative, how come?

\huge v_{o} = \frac{v_{1} R_{a}}{R_{1}}\frac{R_{c}}{R_{b}} +\frac{v_{2} R_{a}}{R_{2}}\frac{R_{c}}{R_{b}} -\frac{v_{3} R_{c}} {R_{3}} -\frac{v_{4} R_{c}} {R_{4}}
 

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dla said:
I get that the first op-amp output is just the total current x R_a

output = -1[/size]x(total current x R_a)

The input goes to each OP_AMP's inverting input
 
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Thank you! I get it.
 

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