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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?

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

enter image description here

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    \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Nov 20 at 11:45
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    \$\begingroup\$ All three resistors are in series. Not in parallel. Because they are all in the same series branch together. \$\endgroup\$
    – jonk
    Nov 20 at 11:45

2 Answers 2

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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.

No, the 10 Ω resistor is also in series with them: -

enter image description here

All I've done is swap the position of the 32 volt source and the 10 Ω resistor to make it easier on your eyes. There is no functional change to the circuit in making this swap.

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It's called the source transformation. A voltage source can be converted to a current source by changing the net series resistance associated with that voltage source to a shunt resistance (in shunt with the voltage source) and then replacing the voltage source with a current source whose value is given by Vsource/Rnet. In your case, the net series resistance associated with the voltage source on the right is 6+4+10 = 20 ohms. The value of the current source will thus be 32/20 = 1.6 A.

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