Kirchhoff's Law Help: Solve Voltage & Current for Circuit

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SUMMARY

The discussion centers on applying Kirchhoff's Laws to solve for voltage and current in a circuit involving a total resistance of 4.5Ω and a supply voltage of 18V. The user successfully calculates the total current flow to be 4/3 A and determines the voltage drop across the first resistor to be 4V, resulting in a remaining voltage of 14V. The user further analyzes the parallel branches, concluding that the currents through each branch are 1/3 A and 1 A, respectively. The discussion emphasizes the importance of understanding potential differences and how they affect current calculations in circuit analysis.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Basic knowledge of Ohm's Law
  • Familiarity with series and parallel resistor combinations
  • Ability to perform circuit analysis using voltage and current equations
NEXT STEPS
  • Study Kirchhoff's Current Law (KCL) for further circuit analysis
  • Learn about Thevenin's and Norton's Theorems for simplifying circuits
  • Explore advanced circuit simulation tools like LTspice
  • Investigate the impact of varying resistor values on circuit behavior
USEFUL FOR

Students, electrical engineers, and hobbyists interested in circuit analysis and design, particularly those looking to deepen their understanding of Kirchhoff's Laws and their applications in practical scenarios.

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Homework Statement


http://imgur.com/DuBgK


Homework Equations


I1=I2+I3

Sum of the directed potential differences around any closed loop equals zero

The Attempt at a Solution


I worked out the total resistance which came out to be 4.5Ω [3 + 1/(1/6+1/2)]
But now I don't know which voltage to use to calculate the total current flow in the circuit. The confusion I have is the +18V, I don't understand where it comes from hence I am confused as to what amount of potential to use.. Thanks in advance.
 
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Remember that potential is, in some sense, arbitrary. The loop rule says that if you start at some point, and traverse one whole loop and come back to your starting point, the total change to your potential is zero. So taking any loop starting at 18 V and coming back to that point will leave you with a potential of 18 V, so the 18 V doesn't actually matter for figuring out the relative currents. What it is useful for is working out the absolute numbers; e.g., once you figure out the current i across the top left resistor, you can subtract iR from 18 V to get the potential at point A.
 
Oh thanks for the advice there :D
So I worked out the current flow to be 4/3 A (Vsupply/Rtotal) and found that the voltage drop across the first resistor to be 4V so 18-4 = 14V right? :P
Here is how I'm finishing it:
Since the supply has lost 4V it is left with 2V across the parallel section. Since the voltage across parallel is constant, I worked out the current for each of the branches to be 1/3 A and 1 A so now I can just subtract iR from the remaining electrical potential to find the solution to the rest of the points. Hope I am doing this right, thank you for your help.
 

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