5210-5 Kirchhoff's Laws
Q: I need clarification for the following:
Lesson 5210-2 Topic 1 states "This means that a current flowing away from the node must be considered negative, since its direction is opposite to the assumed current direction specified by the law. To use the current law, remember that current entering are positive and current leaving are negative".
Compared to Apperception Exercise 1 Question "...If the law were stated that way, currents entering the junction would be considered (a) (positive) (negative) and currents leaving the junction (b)."
Answer key is negative, positive. Related comments includes "If a negative value is obtained when solving for an unknown current, it now means that this current is flowing toward the junction."
This appears to be contradictory. Please clarify so I can determine what I am missing if anything.
One thing that I must point out that your information was taken out of context and does not show what the intention of the question was. In our topic, it is true we stated that current entering a node is positive, and current leaving the node is negative.
This is due to our opening topic that the algebraic sum of the currents entering a node is zero. This information is extremely critical of the polarity involved as we must add or subtract the values from each other to arrive at the final sum of zero (as in the example given in the expression Ia + Ib + Ic + Id + Ie = 0. Both currents Ia and Id were negative, and when their combined value of -18A was added to the positive currents (4 + 10 + 2 = 18), this does give the algebraic sum of 0 (-18 + 18 = 0).
However, in the question from the Apperception Exercise, we pose the question as its converse effect: the algebraic sums of currents leaving the node equals zero. It does not contradict the original statement, but asks that you examine it from the counterpoint.
In this case, the currents entering a node would be negative, and the currents leaving a node would be positive. It still confirms that the algebraic sums of all currents to/from a junction must give us a total of zero, but now we simply change the signs of the values.
If we were to apply this to the example, now the currents that would be negative would be Ib, Ic, and Ie, while currents Ia and Id would be considered positive. It still gives us the values of -18 and + 16, but now we would write them as +18 – 18 = 0.
We definitely need to be careful as sometimes there is a previous statement that asks us to look at things the opposite way of the original presentation. It is also something that can vary if we are talking conventional versus electron flow of current, so this also needs to be taken into consideration.
This is why we also point out that in the introduction to the Apperception Exercise conventional current does flow opposite of previous lessons, and also that the battery polarity should not be paid attention to so long as we give the current direction in the figures.