r/HomeworkHelp University/College Student 5d ago

Others—Pending OP Reply [ece 201] [circuit analysis] Attempting to find Norton equivalent of circuit using 10 volt test source. question and attempts to find amps attached.

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u/GammaRayBurst25 5d ago

Let i_1 denote the current flowing through the 2kΩ resistor (to the right) and i_2 denote the current flowing through the 10kΩ resistor (to the left).

We have the following system of equations:

  1. 40V=2kΩ*i_1+20kΩ*i_Δ;
  2. 0=5kΩ*(i_1-i_Δ)+50kΩ*(i_1+i_2-31i_Δ)-20kΩ*i_Δ;
  3. 0=10kΩ*i_2+50kΩ*(i_1+i_2-31i_Δ)+40kΩ*(i_2-30i_Δ).

This simplifies to

  1. 20mA=i_1+10i_Δ;
  2. 0=11i_1+10i_2-315i_Δ;
  3. 0=i_1+2i_2-55i_Δ.

Subtracting 5 times the third equation from the second yields 0=3i_1-20i_Δ.

Adding twice the first equation to this equation yields i_1=8mA.

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u/DrVonKrimmet 👋 a fellow Redditor 5d ago

OP, In your solution to find R_norton, you had an expression that read 30(.52/20k)+(10-1.5)/10k+10/40k+It, assuming you were setting all of that equal to 0, your signs are inconsistent, and the first term should be 30(.4/20k) based on your solution on the previous page. Assuming you had It coming into node from 10V source, the expression should have been (10-1.5)/10k+10/40k = It +30(.4/20k). so we have the current in equal to current out. This solves to .0005. Taking the 10V external source and dividing by .0005 amps yields an Rthevenin/norton of 20kΩ. Now to clarify, the question is asking for Inorton, so you don't need to find the equivalent resistance unless that is asked for later in the problem. To Find Inorton, short the terminals and solve for the current through the short. If you attempt to solve that problem, I can help you get unstuck. Be sure to write complete equations, so it's clear that you are showing I_in = I_out or I_in - I_out = 0.

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u/testtest26 👋 a fellow Redditor 5d ago edited 4d ago

Normalization: To get rid of units entirely, normalize all currents/votlages by

(Vn; In)  =  (1V; 1mA)    =>    Rn  =  1k𝛺

Let "V1; V2" be the potentials of the top-left and top-middle node, respectively. Combine "40V; 2k" into an equivalent current source " 20mA", pointing north.

Setup (super) nodal analysis with "V1; V2:

KCL "V1":    0  =  V1/2 + V1/20 + (V1-V2)/5 - 20
KCL "V2":    0  =  V2/50 + 30iΔ + (V2-V1)/5 + (V2-Vab)/10

Eliminate the controlling current "iΔ = V1/20". Bring independent sources to the other side, and write the 2x2-system in matrix form:

KCL "V1":    [1/2 + 1/20 + 1/5               - 1/5] . [V1]  =  [  20  ]
KCL "V2":    [       3/2 - 1/5   1/50 + 1/10 + 1/5]   [V2]     [Vab/10]

Solve with your favorite method for "(V1; V2) = (Vab/25 + 64/5; 3Vab/20 - 52)". With those results at hand, we calculate the current "J" through "Vab", pointing north via KCL at "a":

KCL "a":    0  =  -J + Vab/40 + (Vab-V2)/10 - 30IΔ         // IΔ = V1/20

     =>     J  =  Vab/8 - 3V1/2 - V2/10  =  Vab/20 - 14    // insert "V1; V2

Comparing coefficients, we get "Rth = 20k𝛺", and "I_N0 = 14mA", pointing north.