Problem With Nodal Analysis Question

[wenqin123]

Homework Statement

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

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

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.

 

[bennyq]

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
$\frac{Vb+V1}{R1+R2}$+...+...=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.

 

[wenqin123]

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
$\frac{Vb+V1}{R1+R2}$+...+...=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?

 

[gneill]

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.

 

[HalfLostAndSearching]

I would first like to commend you on your thorough attempt at solving this problem. Your application of KCL and consideration of current direction are correct. However, I believe there are a few errors in your equations and assumptions that may have led to the incorrect values for B and D.

1. In the first equation for node B, you have written V_3 instead of V_2 in the first term. This should be corrected to:
$$
\frac{(D-(B+39))}{49} + \frac{((M-V_2)-B)}{91} - \frac{(B-A)}{7} = 0
$$
2. In the second equation for node M, you have written G instead of A. This should be corrected to:
$$
\frac{(A-M)}{101}-\frac{(M-D)}{52} - \frac{((M-V_2)-B)}{91} = 0
$$
3. Your assumption that G is +5V and A is -5V may not be correct. The voltage at these nodes should be determined by the voltage divider rule, taking into account the resistors R_1 and R_5. This would give G a voltage of 3.41V and A a voltage of -3.41V.
4. Your application of voltage drops and lifts may also be incorrect. The voltage drop across R_2 should be (M - V_2) - (B - V_2), as the current is flowing from M to B. Similarly, the voltage drop across R_4 should be (0 - V_2) - (D - V_2), as the current is flowing from 0 to D. This would give a different set of equations:
$$
\frac{(D-(B+39))}{49} + \frac{((M-V_2)-B)}{91} - \frac{(B-A)}{7} = 0
$$
$$
\frac{(G-M)}{R_5}-\frac{(M-D)}{R_3} - \frac{((M-V_2)-B)}{91} = 0
$$
5. Finally, when solving for I_2, you should use the corrected values for B and D in the equation:
$$
I_2 = \frac{(B-V_2)-(D