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for the given schematic my professor had current flow from R2 to R1 when attempting to finding voltage gain. I'm just wondering why he goes from R2 to R1 and not the other way around. I know you can solve it that way but wanted to know if there were a specific reason for doing it like that. Thanks.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Not sure what you mean. That's how the circuit works, therefore that's how you solve it. \$\endgroup\$
    – Matt Young
    Feb 15, 2020 at 3:03

2 Answers 2

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The left-hand side of R1 is connected to ground (or more strictly, to the circuit's 0 V reference). Your professor has chosen to show current flowing from the output of the op-amp to ground, which is how we usually think of (conventional) current flowing.They also know that Vout = +2 V so current will be following in that direction. Although it's usual to try and draw circuit diagrams with signals flowing from left to right, feedback circuits (as R1 and R2 are) connect from the output of an amplifier back to its input and hence carry currents in the opposite direction to the forward signal flow.

As you become more familiar with circuit analysis you'll find parts of circuits where you can't state the direction of current flow just by inspection. You just assign a direction arbitrarily (but consistently). If, when you finish the analysis, the sign of the current is negative it just means the current is flowing in the opposite direction to your original choice.

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  • \$\begingroup\$ I see, that makes a lot more sense. So essentially it doesn't even matter which dierction I have current flow the answer will be either positive or negative, but if answer is negative then I know that current is actually flowing the other way. Thanks for clearing it up. \$\endgroup\$
    – Johanis
    Feb 15, 2020 at 3:34
  • \$\begingroup\$ Yes, it doesn't matter which way you assign current flow at the start of your analysis, as long as you have worked out the correct direction by the end of it. If your original question has now been answered please accept whichever answer helped you more. \$\endgroup\$
    – Graham Nye
    Feb 15, 2020 at 3:53
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It's arbitrary which way you consider the current to flow as long as you're consistent. So if you pick current flowing from ground through R1 to R2 to the output, that will work too.

The resulting values will be negative in this particular case (if Vin was < 0 the signs would be positive).

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