RC circuits - Current and electric potential

1. Oct 25, 2012

msemsey

1. The problem statement, all variables and given/known data
Given the drawing, calculate the current going through the 2Ω Resistor and then calculate the difference in electric potential from point a to b.

2. Relevant equations

V = IR
juction rule:
ƩI = 0 = Iin - Iout
ƩV = 0

3. The attempt at a solution
Well, I attempted to draw all the currents. The Green is from the 3Volts on the bottom, the orange is from the 12V on the top. I would think that I could use a sum of the different V=IR equations (12/4 + 3/2 + 3/2), but 6 is not the answer. I'm pretty lost clearly.

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2. Oct 26, 2012

Basic_Physics

For starters use just one current in each "branch", that is the current will change to another value if you continue through a junction. Label each of these i1, i2....
Lets try it with a new drawing then.

3. Oct 26, 2012

msemsey

Here is my updated drawing. I'm confident that I found all the currents and that the junction equations are right, but the loop rule equations confuse me... I'm not sure where the currents that don't pass through resistors go in the loop equations.

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4. Oct 26, 2012

aralbrec

You have too many unknowns in your KCL equations. You should recognize, eg, that i4=i5. Note with the two equations you have, you can only solve for two unknowns so the first step you would need to take with your equations is to eliminate all the duplicate currents. Make current a property of the branch.

For your loop equations, there are two loops and two currents. You have them identified as i1 and i2. i1 flows around the bottom loop and i2 flows around the top loop. Notice, eg, that for the i1 loop, the same current i1 must flow through the battery e, the resistance R and the 2 ohm resistor. The 2 ohm resistor is special because the current i2 must also flow through it, but it is the net current that causes the voltage drop. What justifies this net current calculation is KCL.