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Can somebody explain to me how to calculate the total resistance of the circuit in the picture?

I do not understand how to deal with the potentiometer and the red line. Does the current flow over \$R_{2.1}\$, bypassing \$R_{2.2}\$ and going on to \$R_3\$ and \$R_{\text{B}}\$? So would it be \$R_{\text{total}} = R_{2.1} + (R_{\text{B}} \parallel R_3)\$?

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    \$\begingroup\$ Which 2 points do you wish to know the resistance? \$\endgroup\$ Commented Jun 21, 2016 at 7:37

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Assuming you want the resistance at the two wires, yes, R2.2 is shorted out (in parallel with zero). It's completely bypassed.

Therefore total resistance (when switch is closed) is:

Rt = R2.1 + R3//RB [ // symbol for 'in parallel'] Rt = R2.1 + (R3*RB/(R3+RB))

If you want the resistance across UB, then just ignore R2.1 and work out R3//RB.

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If the red line is a conductor, and in some variable resistance circuits there is often a wire from the wiper to one end of the pot, then yes, your description is correct.

Unfortunately you don't have your equation represented correctly, as the parallel combination of R3 and RB is (RB*R3)/(RB+R3). This is often written informally as RB||R3, with || pronounced 'in parallel with'. You may therefore have a typo in your Text/Mathjax, having intended to write RB||R3 rather than RB/R3.

Another way to think of it, staying with 'proper' resistance calculations, is that the resistance between R3 and the wiper is R2.2||Rwire, which is around zero because Rwire is roughly zero.

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