How can I assign polarities for electric elements , say resistors..
You can explain on this circuit
I am in introductory circuit analysis so my response is not intended to be the definitive answer... in your diagram, aren't the resistor polarities already assigned?
Yes they are already assigned. But here's the general rule: If a current is entering a resistor (or any circuit device for which the voltage drops across it in the direction of current flow), then the point where the current enters is denoted + and the other -. Special case for voltage sources (denoted as cells here): the current enters at the -ve terminal and exits at the positive terminal.
You can assign polarities (to resistor voltages) any way you wish... That just means, "Put your voltmeter / multimeter probes like this across the resistor" (where the +ve is the red probe and -ve is the black one). The polarity (which end of the resistor has the higher potential) should come out the same regardless of what you assign, assuming your analysis is correct.
If you were to take a voltmeter or DMM (Digital MultiMeter) and measure the terminals of a battery one way, and then measure the terminals the other way, the only difference you should see would be the sign of the potential. This is a favorite question in basic circuit analysis classes because it's something very easy to overlook when you're in a rush.
EDIT: Unless the + and - signs are already on the resistor (as they are in this question, and as I alluded to earlier). Then the question says, "Measure the potential this way"
One thing I should add to my post:
Sometimes it's possible that the current would enter the positive terminal of the voltage source and exit from the negative. This happens for example, when you connect two voltage sources (or cells) with their positive points pointing away from each other in series. In this case the voltage polarities on that cell isn't an indication of which way the current flows, but which way it would flow if the other cell isn't connected backwards in series with this one.
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