# Source transformation - parallel resistors

In the book's circuit, we have 6 Ω and 4 Ω resistors in series, and we have a 10 Ω resistor in parallel to them. I don't understand how we sum these resistors and find 20 Ω for the equivalent resistor.

Are they not in parallel? Shouldn't the equivalent resistor be 5 Ω, like calculated below?

$$\frac{(6+4)\times 10}{(6+4)+10}=5\ \Omega$$

• XcellentEEE, Look very closely. Right at the two dots showing in the top diagram you have one series branch that includes 4 devices. They are all in-series. So you can freely swap them around. This means you can freely swap the lower 10 Ohm resistor, by 'passing in through' the voltage source, so that it is on the other end of the voltage source and is now more clearly in-series with the other two resistors. Just as you can add the 6 Ohm and 4 Ohm, you can now add the 6 Ohm, 4 Ohm, and newly moved 10 Ohm.
– jonk
Nov 20 at 11:45
• All three resistors are in series. Not in parallel. Because they are all in the same series branch together.
– jonk
Nov 20 at 11:45