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Why do we get a resistance $R_B(\beta+1)$ for so many of these examples?

Well, the exact situation is different for each, but in many there is a voltage drop $i_{\rm tot} R$ across some resistor $R$, where $i_{\rm tot}$ is the sum of $i_b$ and $\beta i_b$. This then gives a voltage drop $i_b(\beta+1) R$, which is where the $\beta+1$ factor comes from.



Kate Scholberg 2017-03-21