Kirchhoff's Laws and Simultaneous Equations: Solving a Confusing Circuit

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In summary, the conversation discusses a circuit problem in which the individual is struggling to find the currents using Kirchhoff's laws. The circuit consists of 5 resistors with different values. The individual attempted to solve the problem using a Wheatstone bridge approach, but was advised by their tutor to separate it into 3 loops and use simultaneous equations. However, the values in the equations were mostly 0.5 and 1, making it difficult to solve. The conversation ends with a request for help from the expert.
  • #1
tommoturbo
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Homework Statement



I am struggling as usual with quite a confusing circuit i need to find i1 i2 and i3 in this circuit using kirchhoffs laws

i know i need 3 loops



loop 1 is V1...ABCD...V1

loop 2 is V1...ACD...V1

Loop 3 is V1...ABD...V1






Homework Equations







R1=.5 ohm R2=1 ohm R3=1ohm R4=.5ohm R5=1ohm



The Attempt at a Solution


i have worked out loop1 to be



I1R1-I3R5-(I2+I3)R4 = 1V



but the other loops look wrong and then i can't seem to use simultaneus equations effectively.



Any help much appreciated
 

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  • #2
I can't view your image until it's approved. If you'll upload it to a free image hosting site, and post the link here, I'll be able to view it sooner.
 
  • #4
The currents you've shown are branch currents, not loop currents.

The circuit is an unbalanced wheatstone bridge. See:

http://en.wikipedia.org/wiki/Wheatstone_bridge

If you redraw it to look like a wheatstone bridger, it will be easier to draw distinct loop currents.
 
  • #5
Hi Electrician

Thats how i initially approached the question i sent my workings out to my tutor who told me to separate it into the three loops and then use simultaneus equations to find the unknowns,

when i did it initially whith the wheatsone bridge approach i got values of 1.5 1 and 05 for the various currents when i do the loops the values in teh equations are mostly .5's and 1s
and are difficult to multiply out
 
  • #6
Post a new image with the circuit drawn as a wheatstone bridge with 3 loop currents indicated. Show the equations you derived for each loop. Then I can help you.
 
  • #7
ok will do!


thanks Ian
 

1. What is KCL and why is it important in circuit analysis?

KCL stands for Kirchhoff's Current Law, which states that the sum of currents entering and exiting a node in a circuit must equal zero. It is important in circuit analysis because it allows us to understand and analyze complex circuits by applying the principle of conservation of charge.

2. How do you apply KCL to a circuit with multiple nodes and branches?

To apply KCL to a circuit with multiple nodes and branches, you need to first identify all the nodes in the circuit. Then, write an equation for each node using KCL, setting the sum of currents entering the node equal to the sum of currents exiting the node. Finally, solve the resulting system of equations to find the unknown currents in the circuit.

3. Can KCL be used for both DC and AC circuits?

Yes, KCL can be used for both DC (direct current) and AC (alternating current) circuits, as long as the currents are all in the same direction. However, in AC circuits, the currents are represented as phasors instead of scalar values.

4. What are simultaneous equations and how are they related to KCL?

Simultaneous equations are a set of equations with multiple unknown variables that are solved simultaneously. In circuit analysis, applying KCL to a circuit with multiple nodes results in a system of simultaneous equations, which can be solved to find the unknown currents in the circuit.

5. Are there any limitations to using KCL in circuit analysis?

One limitation of KCL is that it assumes ideal conditions, such as negligible resistance in connecting wires and perfect conductivity in circuit components. In real-world circuits, these assumptions may not hold true, which can lead to errors in using KCL for analysis. Additionally, KCL is only applicable to circuits with linear components.

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