EE - Find Unknown Branch Current by Solving Simultaneous Equations

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DaveBoman
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Homework Statement
Find branch current by assigning direction arbitrarily and then assigning resistor polarity accordingly. Use KVL around loops and then KCL at nodes to get equations. Solve equations for unknown currents.
Relevant Equations
see picture
Handwritten procedure for circuit analysis: steps to assign current directions and resistor polarities, apply KVL to loops 1-3, apply KCL at nodes A-D, and solve for unknown currents, with loop and node equations written out.
Handwritten circuit schematic titled "1. Branch current Method" with a 12 V source, resistors R1 15 kΩ, R2 22 kΩ, R3 7.5 kΩ, R4 8.2 kΩ and RL 10 kΩ, nodes A, B and C, mesh loops 1, 2 and 3, and branch currents I1, I2, I3, I4, IL and IT labelled.
 
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I am at a complete loss on this problem. I have gotten the loop equations and I have gotten the KCL equations. I have tried to write all of this out thoroughly. I am at a loss as to what to do here.

I have 7 equations and 6 unknowns. The unknowns are I1, I2, I3, I4, IL, and IT, correct?
How can I plug in the KCL equations into the loop equations to get 3 equations and 3 unknowns?
 
Welcome to PF. :smile:

DaveBoman said:
I have 7 equations and 6 unknowns. The unknowns are I1, I2, I3, I4, IL, and IT, correct?
How can I plug in the KCL equations into the loop equations to get 3 equations and 3 unknowns?
I have 3 KCL equations to solve for the 3 node voltages. That is all you need to determine the currents, no?
 
The KCL equations are linked to the KVL equations by the V-I relationship of each component. So, you actually can create an equation for each component (Like V1 = I1⋅R1), and equation for each node (KCL), and an equation for each loop (KVL). You will have variables for the voltage and current of each component. Of course, as you have seen, you will have way more equations than unknowns. That's OK, use the ones you like, to arrive at a solution for simultaneous linear equations. I will often start by eliminating all of the voltages (in terms of currents), or vice-versa.

You might try it first with a simpler version? In any case if you don't show us what you've done it's hard to help.
 
DaveBoman said:
I am at a complete loss on this problem. I have gotten the loop equations and I have gotten the KCL equations. I have tried to write all of this out thoroughly. I am at a loss as to what to do here.
Don't let indecision paralyze you. Sometimes you just need to try something, and you'll see a path forward.

DaveBoman said:
I have 7 equations and 6 unknowns. The unknowns are I1, I2, I3, I4, IL, and IT, correct?
How can I plug in the KCL equations into the loop equations to get 3 equations and 3 unknowns?
Note that the KCL equations aren't independent. The KVL equations and any three of the KCL equations will leave you a system of six equations and six unknowns.

If you want to get down to three equations and three unknowns, just choose three currents to keep in the loop equations. Use the KCL equations to express the other currents in terms of the chosen three.
 
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See [typical]
 
Attachments
  • Handwritten circuit schematic for the branch current method with three loop currents marked in red: Icirc1 through the top loop with R1, R2 and R3, Icirc2 through the 12 V source loop, and Icirc3 through the lower-right loop with R4 and RL. Beside it is the KVL equation Icirc1 x (R1+R3+R3) - Icirc3 x R3 - Icirc2 x R2 = 0.
    Kirchoff Circuit 1 KVL.jpg
    33.7 KB · Views: 183
Wrong KVL
 
Attachments
  • Handwritten circuit schematic for the branch current method with three mesh loops marked in red, labelled Icirc1, Icirc2 and Icirc3, containing resistors R1 15 kΩ, R2 22 kΩ, R3 7.5 kΩ, R4 8.2 kΩ, RL 10 kΩ and a 12 V source; a typed note reads KVL in circuit 1 typical: Icirc1x(R1+R3+R2)-Icirc3xR3-Icirc2xR2=0.
    Kirchoff Circuit 1 KVL.jpg
    33.7 KB · Views: 178