# Current Direction and Voltage Polarity in Capacitor/Inductor Circuits

I believe that this question may be simple for those with a little more experience, but it is still puzzling me.

Studying capacitance/inductance, I understand that the capacitor in the DC case behaves as a signal blocker, i.e. an open circuit. On the other hand, still in direct current, inductors work as a short circuit (v = 0).

In my studies, in Irwin's book, this circuit was presented, and it asks for the calculation of the total stored energy. The calculation of the energy itself in I understand, using the current for this. My question is how to know the polarity of the voltages, and the direction of the current.

When I see this example, I don't understand why the current directions $$\I_{L_2}\$$ and $$\I_{L_3}\$$ are that way. however I understand that the current can be placed for each case and I can find it negative (in this case, opposite direction).

On the other hand, when doing the analysis using Kirchoff's Law of Voltage, I don't understand how the book puts that

$$6I_{L_1}+3I_{L_2}+6I_{L_2}=9$$

I mean, how do you know that the polarity of the resistors is opposite to the polarity of the voltage source?

If possible, I would also like to ask for a brief explanation for the polarity in the capacitor, although that seems a bit clear from the very definition of the capacitor.

I apologize for any writing/translation problems.

• Once you assume a direction arrow, voltage drops across the resistors are plus on the side the arrow enters. May 12 at 13:41
• Normal we assumed that the current flows from higher (more positive) to lower potential. In your example circuit, we see a 9V voltage source. Thus, we can assume that the current will flow out of the positive terminal of the voltage source. Next, we have a 3A current source with a marked current direction. And since the current is flowing from "+" to "-" we can assume that the 3A will flow "right" and it will add to the I_L1 current and "created" I_L2 current. I_L2 = I_L1 + 3A.
– G36
May 12 at 13:47
• As for the resistor, polarity look here at this picture electronics.stackexchange.com/questions/430513/…
– G36
May 12 at 13:48
• @G36, according to the passive sign convention, I understand that the current flows from + to -, only in this case wouldn't it have to be counterclockwise? Another point, assuming I_L2 = I_L1 + 3A, in this situation, I_L2 = -1.2 A, because it is a negative value, is it correct to say that, in fact, the direction of the current flows the opposite of what is in my diagram? May 12 at 14:23
• After reflecting on his answer, I was able to understand the question of direction. Actually, it is not relevant at first, because regardless of the direction chosen for the current, this will determine the polarity of the voltage caused in the resistors, and after the calculations, realizing that the current value is negative, using Ohm's Law, the resistor polarity is also changed in the multiplication of the signals. Thanks, @G36. That was very clear. Would it be convenient to post the full solution of the question here? May 13 at 14:57