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I have a cicuit here that's giving me trouble when finding out what V1 and V2 equal.The book says that I1 is traveling through the 8 ohm resistor (R1) towards ground. But when I build the circuit on multi-sim it says that I1 is traveling through (R1) upwards "clockwise. I'm confused why the book says I1 is counter clockwise and my simulator says clockwise.

R1 = 8 ohm R2 = 4 ohm R3 = 10 ohm I = 2A E = 64V

I thought I2 and I add up to I1. I'm so confused. How do I determine the direction of the current so I can build my equation. I always seem to be confused on how to determine the direction of the current. Does it matter? I"m still very confused on node analysis.

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    \$\begingroup\$ It would be helpful if you draw your schematic to show which resistor is R1, which is R2, and which is R3, and which node is 1 and which is 2. \$\endgroup\$ – The Photon Feb 3 '13 at 5:51
  • \$\begingroup\$ Also, what branches are "I2", "I1", and "I" going through? \$\endgroup\$ – The Photon Feb 3 '13 at 5:57
  • \$\begingroup\$ Finally, for pen-and-paper solutions, you'll want to remember the superposition principle to solve this. \$\endgroup\$ – The Photon Feb 3 '13 at 5:59
  • \$\begingroup\$ You mention V1 and V2 in your first sentence, but they aren't defined anywhere in your schematic! -1 \$\endgroup\$ – Olin Lathrop Feb 3 '13 at 16:38
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The book says that I1 is traveling through the 8 ohm resistor (R1) towards ground. But when I build the circuit on multi-sim it says that I1 is traveling through (R1) upwards "clockwise. I'm confused why the book says I1 is counter clockwise and my simulator says clockwise. ... I always seem to be confused on how to determine the direction of the current.

You don't need to know the direction of the current before you solve the circuit. You can guess whichever way you want. If the current is actually flowing the other way, you'll just get a negative number for the value of the current.

The book says that I1 is traveling through the 8 ohm resistor (R1) towards ground. But when I build the circuit on multi-sim it says that I1 is traveling through (R1) upwards "clockwise.

In the example in the book, what was the value they found for I1? If it was negative, there's no contradiction with the simulation result.

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If I1 is the current from the voltage source there is no contradiction. All paths to ground from the 64V source flow - upwards - through the 8 ohm resistor.

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Nodal analysis is the application of Kirchoff's Current Law (KCL) to solution of circuit node voltages. KCL basically says that all currents into a node (any branch or junction between components) equals the current out of the node. It has to be this way because otherwise the charges would pile up in the node, and the circuit would just explode :^) . It may seem counter-intuitive that by using KCL you end up focused on and solving for voltages, but that's the way it is.

To use KCL you need to choose a current flow convention. There are only 2 possible:

  • Positive current flow out of the node, and negative current flow into the node. You could state this as: \$I_{\text{out-of }\text{node} }\$ = \$I_{\text{into } \text{node}}\$, or 0 = \$I_{\text{out-of }\text{node} }-I_{\text{into } \text{node}}\$
  • Or, negative current flow out of the node, and positive current flow into the node. You could state this conventions as: \$I_{\text{into } \text{node}}\$ = \$I_{\text{out-of }\text{node} }\$, or 0 = \$I_{\text{into } \text{node}}\$ - \$I_{\text{out-of}\text{node} }\$ .

It doesn't matter which of these conventions you choose as long as you are consistent. Consistency may be the hobgoblin of little minds, but those hobgoblins are essential for Nodal analysis to work.

After choosing a current convention, make a process to follow.

  • Define the circuit nodes.
  • Discount or substitute any degenerate nodes. A degenerate node is any node which you can tell by inspection what it's voltage would be. Usually that will be a voltage source connected between a node and the ground node. Voltage at that node is just the value of the voltage source.
  • Write a current equation for all the non-degenerate nodes using Ohms Law.
  • Solve the set of equations simultaneously.

Let's use this process to solve your circuit. First I'm going to make a couple of name changes: calling the battery voltage \$V_B\$ (not E), and calling the current source \$I_S\$ (not I). I'm going to choose the convention of current out of the node being positive. (I like the positive out convention because it makes it obvious at a glance which node the equation is covering. For example, if the equation is for node 2 ( \$v_2\$ ) then in each term of the equation \$v_2\$ will be first and positive.)

Now, start the process:

  • Define the circuit nodes. Node 1 is the junction of \$V_B\$ and \$R_1\$. Node 2 is the junction of \$R_1\$ , and \$R_2\$ and \$I_S\$ . Node 3 is the junction of \$R_2\$ and \$I_S\$, and \$R_3\$.
  • Degenerate nodes. Node 1 is a degenerate node, so no equation is for node 1, just substitute \$V_B\$ in where ever \$v_1\$ would be in any node equations.
  • Write the current equations.
    Node 2: 0 = \$\frac{v_2-V_B}{R_1}+\frac{v_2-v_3}{R_2}+I_S\$
    Node 3: 0 = \$\frac{v_3}{R_3}+\frac{v_3-v_2}{R_2}-I_S\$
  • Solve the equations.
    \$v_1\$ = \$-\frac{-R_2 V_B-R_3 V_B+R_1 R_2 I_S}{R_1+R_2+R_3}\$ = 37.82V \$v_2\$ = \$\frac{R_3 \left(V_B+R_2 I_S\right)}{R_1+R_2+R_3}\$ = 32.73V

    You would get the same end result if you used the other current convention. That is basic Nodal analysis. Practice until you can do it without thinking much about it.

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