# Combining resistors connected solely at one node, and also a question about parallel/series I'm learning how to combine resistors for Thévenin equivalency circuits, but I am really puzzled as to how I would calculate the Thévenin resistance for this circuit. In my mind, none of the resistances are either parallel or in series with one another, does this mean they can't be combined?

Originally, this circuit had a current source connected to the left of $R_{5\Omega}$, which I zeroed out.

My second question involves a different circuit. To find the Thévenin resistance for this circuit, I started out by zeroing the sources. So the $10\,V$ source is replaced by a closed circuit, and the $1\,A$ source is replaced by a open circuit.

• Following this, I see that $R_{20\Omega}\parallel R_{5\Omega}$. This yields the combined resistance $4\,\Omega$, taking the old $R_{5\Omega}$'s place. The second step is to add the new resistance $R_{4\Omega}$ to the resistance in series, $R_{6\Omega}$. This combined resistance is in parallel with our last resistance $R_{10\Omega}$ and together that's a Thévenin resistance of $5\,\Omega$.

• What makes me wonder though is if we in the second step notice that $R_{6\Omega}$ is also in series with $R_{10\Omega}$, and add those together, we get the resistance $R_{16\Omega}$ in old $R_{10\Omega}$'s place. Combining this, with it's parallel resistance $R_{4\Omega}$ yields us $3.2\,\Omega$, not $5\,\Omega$ as the earlier steps.

Why does it only work in one direction? And how am I to know which direction is the right one? Am I not correct in the statement that $R_{6\Omega}$ is in series with both $R_{4\Omega}$ and $R_{10\Omega}$? And if not, why?