0
\$\begingroup\$

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?

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

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:

schematic

simulate this circuit – Schematic created using CircuitLab

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

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.