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Comment by periblepsis on Calculating voltage drop

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@TIMDRIPCAN If you mean over on the left side then it's just a simple resistor divider that can be converted from three parts (ideal voltage source, plus two resistors) to just two parts -- a new ideal voltage source and a series resistance. \$V_{_\text{TH}}=V\cdot\frac{R_2}{R_1+R_2}\$ and \$R_{_\text{TH}}=\frac{R_1\,\cdot\,R_2}{R_1+R_2}\$. Are you not aware of that? If not, burn it into your memory. Or is the issue something else?

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