# Exam question superposition problem

#### midnightblack

Joined Feb 29, 2012
31 Hi.

I am trying to figure out if I'm wrong or the given answer is wrong.

I am to solve for I using the principle of superposition in figure 1.2.

This is my methodology:

Set all voltage sources=0.

Turn on right side voltage source. Circuit reduces to 12V voltage source and 4k resistance since the 6k is shorted. I is in the opposite direction to voltage.

V=IR==> I=V/R = 12V/4k=3mA (Now here the answer says -3mA but that would only happen when I and V are in the same direction no?)

Set all voltage sources=0.

Turn on left side voltage source. Circuit reduces to 12V voltage source and 6k resistance. This time I is in the same direction as V.

V=IR==> I=V/R = -12V/6k=-2mA (here the answer is given as 2mA but again, this would only happen when I and V are in opposite directions, no?)

Therefore by principle of superposition

I=[3+(-2)]mA= 1mA (But answer is -1mA)

Is there anything I'm missing or is the answer wrong? (I am inclined to believe the answer is not wrong as it was given in the mark scheme of an exam)

#### Jony130

Joined Feb 17, 2009
5,157
The answer looks good, but they use just like 90% of engineers conventional current flow. And conventional current is said to flow from positive to negative.

#### WBahn

Joined Mar 31, 2012
25,566
The question isn't whether the current is positive or negative due to its relationship to a particular voltage, the question is whether it is positive or negative relative to the direction indicated by the arrow associated with the current being sought. The unknown current I is positive if it is flowing in the same directijon as the arrow and negative if it is flowing in the direction opposite the arrow -- irrespective of signs of the voltages. In point of fact, there is no voltage that is directly related to this current because we are talking about a wire that is assumed to be resistanceless and, hence, has no voltage across it.

As Jony130 says, nearly everyone uses "conventional current" which flows out of the positive terminal of a source and returns to the negative terminal. Hence (as he shows in his diagram), the current from the source on the right flows out the bottom, up through the center wire (and hence in the opposite direction of the indicated direction for positive current for I and is, therefore, negative) while current from the source on the left flows up out of the top terminal and down through the center wire, and is therefore positive.