- #1

grekin

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## 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)