Solving for current between 2 nodes

[grekin]

Homework Statement

Assume that I = 22mA , V = 6.0V , and R = 350Ω. Determine the current between B and D using the superposition principle.

Homework Equations

V=IR
Superposition Principle
G=1/R
Voltage and Current dividers

The Attempt at a Solution

I think I'm understanding the superposition just fine, but having issues when actually solving for the current.

I_BD = I_BD' + I_BD''

Replacing the current source with an open circuit, I used mesh analysis to find I_BD'. With the current source removed, there were 2 meshes left in the circuit:

Mesh 1 (on the left):

400*i_1 - 300*i_2 = 6

Mesh 2 (on the right):

1050*i_2 - 300*i_1 = 0

i_1 = 0.019, i_2 = 0.0055

Using the voltage divider principle, v=6*(100/1050)=0.57.

V=IR, I_BD'=V/R=0.57/100=0.0057

Now removing the voltage source and replacing it with a short circuit, I used nodal analysis to solve for the voltages at each node, then used (v_d-v_b)/100 = I_BD''. The node between node A and node C will be addressed as node E.

Node A:

(1/100+1/350+1/300)*v_A - (1/100+1/300)*v_B - v_E/350 = 0

Node B:

-(1/100+1/350)*v_A + (1/100+1/300+1/100)*v_B - v_D/100 = 0

Node C:

(1/200+1/100)*v_C - v_D/100 - v_3/200 = 0

Node D:

-v_B/100 - v_C/100 + (1/100+1/100)*v_D = -0.022

Node E:

-v_A/350 - v_C/200 + (1/350+1/200)*v_E = 0.022

Solving, I got: v_A=0, v_B=-0.6, v_C=0, v_D=-1.4, v_E=28.I_BD'' = (v_d-v_b)/100 = -0.008

I_BD = 0.0057 - 0.008 = -0.23 (incorrect)

 

[FOIWATER]

I do seriously hate to answer your question with a question and no help directly to what you asked. It seems to me that a more straight-forward approach to this problem (and one that you may or may not have considered) would be to simplify the drawing with the current source opened. With the current source opened, the resistors from A to D to C to B could be added to give one series resistance, the current through which is the current through the resistor you are interested in due to the voltage source. Which would be easy to find by simplifying that circuit to find the total resistance and current, then applying ohms law to find the current.

You could next short the voltage source and simplify that circuit by combining the 100 ohm and 300 ohm resistors in parallel where a simple current divider would give you the current through the resistor due to the current source. It would be less work.

What do you think of this approach to the problem, does what I said make sense? or does mesh and nodal feel more 'comfortable' to you after reading this? Is there any part I might be able to clarify if it does not make sense as well.

Thanks, again sorry for not looking through your work, I will do so if this does not make sense.

 

[grekin]

I understand what you're saying, and it does seem like a much simpler approach than what I had come up with. However, I'm still not getting the correct answer. I believe I solved the part where the voltage source is shorted correctly:

100*300/(100+300) = 75. Applying the current divider, I got:

I_BD'' = -0.022 * 300/825 = -0.008, which is what I got using my method.

Now when solving with the current source removed, I simplified the resistors to get an equivalent resistance of 750 Ohms. I then combined the 300 and 750 Ohm resistors in parallel, then combined that equivalent with the 100 Ohm resistor to get a total resistance of 314.3 Ohms. Solving with V=IR, I got I_total = 0.019 A. However, this is the part where I'm not entirely understanding what you're saying. I tried applying the current divider on the following circuit:

I_BD' = 0.019 * 300/1050 = 0.0054 A

Combining the two currents, I_BD = -0.0026 A which is incorrect. However, I now received the following feedback:

"Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

 

[FOIWATER]

I like what you did.. maybe not round the equivalent resistance as much? take 314.25 instead of 314.3

What you are doing is how I would do it as well so I am not sure why it is not right. Maybe someone else could weigh in