Circuit with voltage source, resistor (two times) in series

I need to find the voltage drop between R1 (4 Ohm) and R2 (2 Ohm). I tried applying Kirchhoff's law in this way: \$-10+4I-8-2I=0\$ and then solving by \$I\$ I have \$I=9 A\$ but that's not correct. Can someone help me?

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    \$\begingroup\$ What is total voltage? What is total resistance? \$\endgroup\$ – Dwayne Reid Feb 26 '17 at 1:01

Why did you change the sign for the voltage sources and for the resistor drops? Imagine first there was only one source and one resistor (R10). You would have +10 - 4I = 0.

Now add the second resistor, you get +10 - 4I - 2I = 0. If you aren't comfortable with that, imagine first combining both resistors into one, it would be 6 Ohms, right? So you would have +10 - 6I = 0, which is the same as above.

Per KVL, the order of the the components doesn't matter, so you can apply the same logic to the sources. A 10V and 8V source in series and in the same polarity produce a 18V source. Don't be fooled by the fact the 8V source is on the right side of the circuit, upside-down; it's still adds to the total voltage source. So you would have +10 + 8 -4I - 2I = 18 - 6I = 0

You might find it easier to rearrange the drawing like this:


simulate this circuit – Schematic created using CircuitLab

Remember though this doesn't hold once you add any parallel item or branched circuit.


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