My problem is in the VA, VB parts is the previous circuit equivalent to this (OPTION A):
Or to this (OPTION B):
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5\$\begingroup\$ Are you asking about the polarity of the battery symbol? \$\endgroup\$– evildemonicJan 7, 2021 at 18:00
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\$\begingroup\$ Think of the op-amp output as another (third) "battery". Thus, in total there are three voltage sources connected through resistors to a common point (the op-amp inverting input). The op-amp adjusts its output voltage so that to keep the voltage of this (summing) point equal to zero. The name of this circuit is "op-amp inverting summer". \$\endgroup\$– Circuit fantasistJan 7, 2021 at 21:06
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2 Answers
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Va and Vb are simply voltage sources of a specified value. They’re not batteries. A simulator package will include symbols for voltage sources with settable parameters, and include versions that are 1- or 2- terminals.
I don't know what tool you're using, but some simulators (Falstad, for example) show the 2-terminal voltage source as a battery graphic. As you saw, this leads to some confusion with polarity since the parameter can be any value, positive or negative.
Further, model voltage sources are ideal (infinitely low impedance, infinite current), which we know is not the case with real batteries.
As for your drawing, Option (B) is probably less confusing as the voltage source parameter will be referenced to ground. But it would be even better if you used two 1-terminal voltage sources.
answered Jan 7, 2021 at 18:33
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It can be both, depends on voltage. If your Va and Vb are positive then you have option B, if your Va an Vb are negative, then option A
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\$\begingroup\$ ...that is if the voltage is written (displayed on the print) with positive numbers. If negative voltages were indicated with negative numbers, then Option B is the choice. Unless you wanted to negi-fy the negative numbers, then obviously it's Option A. For a complete analysis see: youtu.be/rMz7JBRbmNo?t=42 \$\endgroup\$ Jan 7, 2021 at 20:49
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\$\begingroup\$ No, this is not true. Only Option B is correct. The polarity of a voltage source is independent of its value. \$\endgroup\$ Jan 7, 2021 at 21:14
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\$\begingroup\$ Elliot, I'm gonna have to disagree regardless of how many downvotes I get. I stated it could be both and it could. What Chris said is correct but I just didn't care to explain to detail. \$\endgroup\$– HedgehogJan 8, 2021 at 11:35