Kirchoff's Rules?

Question

Find the magnitude and direction of the current in the 2.0 resistor in the drawing if R = 3.0 and V = 2.5 V.

http://i21.photobucket.com/albums/b272/Cajunboiler/p20-76alt.gif

I know the direction is left. That's obvious. The thing that is confusing me about this problem is the extra voltage. I'm not sure what to do with it in helping me find the current. If there was only one, then I would definitely know how to do it.

Answer 1

Sorry to disappoint you, but multi-source circuits are common; in electronics circuits, at least; and, certainly, in textbooks.

There are a lot of ways you can tackle this problem, but since you mentioned Kirchhoff, let's use KCL to solve it. This simple network has only two nodes.

First, you choose a "datum" (reference) node. Let it be the one on the right. You now apply KCL to the left node. KCL states that the algebraic sum of all currents leaving (or entering) the node is zero. There are three branches in the network, and thus three currents "leave" the node. Let's identyfy them as I₁, I₂, I₃, bottom-to-top.

Assume node voltage of this node (relative to reference node) is V. Moreover, assume this unknown voltage is positive and greater than any other voltage in the network. Thus, all currents flow from the left node to the right.

On this assumption, we can write, starting bottom-to-top,

I₁ = (V + 4)/3
I₂ = (V − 2)/2
I₃ = V/1,

where 4 is added to V because positive of this source points downwards, aiding V. Since KCL states that the algebraic sum of these 3 currents is zero,

(V + 4)/3 + (V − 2)/2 + V/1 = 0.

Multiply this by 6 and simplify. You'll get

11 V = -0.5, ∴ V = - 1/22 = -0.04545... V.

Now, since you are interested in I₂, just subtitute V in the proper equation above; it should yield -1.2727... A. The negative sign means this current flows in opposite direction to our earlier assumption, i.e., flows INTO the node, instead of away from it.

Yep, you were right, this current flows to the left.

Answer 2

If you think of electron current flow, then the current is left. For engineers, current flow is positive charge flow - from the plus side of the battery, thru the components, and back to the battery at the negative terminal.

Draw 2 Kirchoff loops:
I1 from the bottom right corner of the circuit, thru the 4V,3Ohm,2Ohm,2.5V and back to the bottom right.
I2 from the bottom right corner of the circuit, thru the 4V,3Ohm,1Ohm and back to the bottom right.
(I don't like this font - I1 is current 1)

Then write the equations
1st loop: 4 -3*(I1+I2) -2*I1 -2.5 = 0
rearrange:
4 -5*I1 -3*I2 -2.5 = 0
I1 = (1.5 - 3*I2)/5

2nd loop: 4 -3(I1+I2) -1*I2 = 0
rearrange:
4 -3*I1 - 4*I2 = 0

Plug in I1 from rearranged 1st loop equation:
4 - 3(1.5 - 3*I2)/5 -4*I2 = 0
4 -.9 +1.8*I2 -4*I2 = 0
2.2*I2 = 3.1
I2 = 1.41 amps
a positive current in the direction I had you draw the 2nd loop. (Check my algebra and arithmetic)