# What is wrong with my KVL approach on a really simple circuit? Hello, so this is a really simple circuit question that has been bothering me. I know that by simply looking at the circuit and using logic it makes sense that the voltage drop across the resistor(as pointed in the picture) should be 0.6V. But when I actually write the Kirchoff Voltage Law(KVL) around the loop as shown in the picture, I somehow get -1.8V. I am probably messing up my convention. Would someone please explain how I would get the same 0.6V using KVL as well?

Edit: found the answer from my class notes for your refernce. Is the solution assuming opposite convention for how the current flows in the circuit so as opposed to positive to minus terminal is it assuming minus to positive?  • You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.) Jan 28, 2019 at 23:48
• Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
– SRR
Jan 29, 2019 at 0:15
• Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
– Chu
Jan 29, 2019 at 1:15
• IMHO, the polarity here doesn't even matter. That's the key in KVL - sum of voltages around the loop should be equal to zero, nothing said about polarity. It's of course different matter if you have another voltage source that would oppose the 1.2 source - then polarity would matter. As for the direction of current indicated in the answer, seems like your instructor opted in for electron flow instead of conventional flow. Still doesn't matter in this case, though - same concept different direction Jan 29, 2019 at 3:54
• Is that 3mA flowing through the 200R informative, or is it a non-standard symbol for a current source? The convention is that current flows from +ve to -ve, so it looks like the book has drawn its voltage source backwards (we normally draw +ve upwards anyway) if the 3mA arrow is informative and that supply is th eonly source of power in the circuit. Jan 29, 2019 at 7:16