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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.

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