Problem With Nodal Analysis Question

  • Thread starter wenqin123
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  • #1
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Homework Statement



https://courses.edx.org/static/content-mit-802x~2013_Spring/html/ps3_p3_fig1_a.png [Broken]

$$ R_1 = 7 \Omega \quad
R_2 = 49 \Omega \quad
R_3 = 52 \Omega \quad
R_4 = 91 \Omega \quad
R_5 = 101 \Omega $$
$$ V_1 = 5V \quad V_2 = 39V \quad V_3 = 44V$$

Homework Equations



$$ I = \frac{V}{R} $$

The Attempt at a Solution



I tried writing KCL for nodes B and M.

I follow the direction of the current in the circuit diagram. Furthemore, current flows into node M in the branch with the R_5 resistor, and flows away from node M into the branch with the R_3 resistor on the side (I ignore node N because I assume that it is the same node as node M).
Also, currents leaving a node are negative while currents that enter a node are positive.

I ground node M.

When I first see the negative terminal of a battery source, I count that as a voltage lift, otherwise, it is a voltage drop.
$$
\\
\\
\frac{(D-(B+V_3))}{R_2} + \frac{((M-V_2)-B)}{R_2} - \frac{(B-A)}{R_2} = 0
\\
\\
\\
\frac{(G-M)}{R_5}-\frac{(M-D)}{R_3} - \frac{((M-V_2)-B)}{R_2} = 0
$$
The first equation was for node B and the second was for node M.

Some places where I might have messed up:
I assume that G is +5V because it is next to the positive terminal of the battery and A is -5V since it is next to the negative terminal of the battery.
For node M with the branch that has resistor R_4, I subtracted the voltage V_2 from node M since we hit the positive battery terminal which a represents voltage drop. While for node B, I added the voltage V_3 to B since we hit the negative battery terminal first, which represents a voltage lift from the battery.

I then simplify the equation with those inferences and plug in the values for the the voltage and resistance of each branch:
$$
\\
\\
\frac{(D-(B+44))}{49} + \frac{((0-39)-B)}{91} - \frac{(B-(-6))}{7} = 0
\\
\\
\\
\frac{(6-0)}{101}-\frac{(0-D)}{52} - \frac{((0-39)-B)}{91} = 0
$$
I plug in the two equations into wolfram alpha to solve the system, and I get that
$$B = -4.9V \quad D= -22.7V$$
http://www.wolframalpha.com/input/?...-6)/7)+=+0,++6/101+-+(-d)/52+-+(-39-b)/91+=+0

I checked to see if the node potentials for B and D were correct by solving for I_2.
$$ I_2 = \frac{(B-V_3)-D}{R_2}
\\
\\
I_2 = \frac{(-4.9-44)+22.7}{49} = -0.53A$$
While the correct answer is 0.449A.

Could someone help me see where I went wrong? Thank you.
 
Last edited by a moderator:

Answers and Replies

  • #2
23
0
Firstly, there is only two essential nodes, so once you set node (M,N) to ground that leaves node B as the only other essential node. This means you only need to write ONE KCL equation to solve for one unknown nodal voltage. I think the problem you have here is that you need to write KVL equations. Ill start you off,

Going from B->A->G->M, this is the branch current that connects the two essential nodes. We start at voltage VB, so

VB-VR1+V1-VR5=0
VB+V1=VR1+VR5
Here I have the voltage drop across the resistors in terms of the source and the unknown,
so for this branch of KCL i would have
[itex]\frac{Vb+V1}{R1+R2}[/itex]+...+...=0

Now do the same for B->M and B->D->M, I do my current direction from the Node VB to the reference node, and came out with -0.449A, so 0.449A is correct for I2.
 
  • #3
3
0
Firstly, there is only two essential nodes, so once you set node (M,N) to ground that leaves node B as the only other essential node. This means you only need to write ONE KCL equation to solve for one unknown nodal voltage. I think the problem you have here is that you need to write KVL equations. Ill start you off,

Going from B->A->G->M, this is the branch current that connects the two essential nodes. We start at voltage VB, so

VB-VR1+V1-VR5=0
VB+V1=VR1+VR5
Here I have the voltage drop across the resistors in terms of the source and the unknown,
so for this branch of KCL i would have
[itex]\frac{Vb+V1}{R1+R2}[/itex]+...+...=0

Now do the same for B->M and B->D->M, I do my current direction from the Node VB to the reference node, and came out with -0.449A, so 0.449A is correct for I2.
Is it possible to solve for the currents using KCL?
 
  • #4
gneill
Mentor
20,905
2,857
Is it possible to solve for the currents using KCL?
Sure, but slightly indirectly. The KCL node equation is comprised of the sum of the individual branch currents entering/leaving the node. Once you solve for the node voltage you can break out each of those terms individually to give the individual currents.
 

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