ok so I am a beginner and i ran across something i wasn't able to find the answer for yet...

Lets say i have a circuit like so....

V= DC Voltage
Ω= Resistor*

 ┌────┬────────┐
 │    ║        ║
 │    Ω        Ω
+│    ║   R1   ║
 V    ├──═Ω═───┤
-│    ║        ║
 │    Ω        Ω
 │    ║        ║
 └────┴────────┘

**my question is, for Resistor R1, which side would be designated (+) and which would be (-) ,
**and which way would current flow L»R or R»L?

 

Answers:

1) It is indeterminate which side is +ve with respect to the other side. You would have to specify the values of the other four resistors.

2) Conventional current flows from the higher potential to the lower potential. The two points do not necessarily have to be opposite polarity. They can both be +ve or they can both be -ve, so long as there is a potential difference. For example, one point can be -3V and the other can be -4V. Current will flow from the -3V potential to the -4V potential.

3) When analyzing circuits it does not matter which node you call +ve and which is -ve. If the result is a negative current then you made the wrong assumption.

 

ok Thanx, that actually makes alot of sense now.

 

You should follow the "passive sign convention", which means that you get to pick, randomly if you so choose, either the polarity of the voltage across R1 or the current through it, but once you pick the polarity of one, the polarity of the other should be consistent.

So if you pick the current so that I_R1 flows left to right, the V_R1 would be defined such that the left side of R1 is positive relative to the right side.

As MrChips pointed out, if the values turn out to be negative, then you simply made the wrong choice and the actual voltage and current are of the opposite polarity.

Note that you don't HAVE to use the passive sign convention. You could independently pick the polarity of each voltage and each current in the circuit randomly. But using the passive sign convention makes if far less likely that you will make mistakes that go uncaught.