Engineering How can Kirchhoff's Laws be used to solve for current in a circuit?

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    Kcl Kvl Laws
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Kirchhoff's Laws can be effectively applied to analyze circuits by eliminating shorted components, such as R6 in this case. By redrawing the circuit with a common ground and labeling node voltages, users can set up KCL equations for the unknown voltages. The discussion emphasizes the importance of writing equations for each node and ensuring there are enough equations to solve for the voltages. Once the node voltages are determined, branch currents can be calculated accordingly. This methodical approach allows for a clearer understanding of current flow within the circuit.
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Homework Statement
Use KVL and KCL to determine all voltages and currents through each resistor, and determine current in source.
Relevant Equations
All given is the power source is 5V. All the other elements are resistors (in ohms).
XmB8Rbz.png


I've been trying to use KVL/KCL here but I can't get any. The only thing I accomplished is R6 has 0V, 0A (KVL with its loop). All the other ones I can't get any? Thanks
 
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Welcome to the PF.:smile:

You are correct that R6 is shorted out, so eliminate it from the drawing. You can re-draw it a little more cleanly with the 5V source vertical with its bottom considered a common "ground". When I do that it looks like a bridge circuit, so label the 3 voltages at the 3 nodes (not including ground) and write the KCL equations. Can you show us that work?
 
IMG_20190903_230918.jpg


Thank you. Is this how to do it? I still can't see any patterns here, sorry.
 
Label the unknown voltages on the 3 right nodes and write the KCL equations for them. Then see if you have enough equations to solve for the unknown voltages. Once you have the node voltages, that gives you the branch currents...
 
Kirchoff or nodal.jpg

Attach currents to all resistors -for instance I1 through R1 and so on.

For nodal analyze name X and Y the potentials as per attached sketch.

For Kirchhoff way you have 3 loops:

abdfa, abcefa, abdefa

and two current sum points: d and e

I1=0.03852;I2=0.015485;I3=0.0171638;I4=0.02135755;I5=0.032649
 

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