Solving Circuit Problems with Kirchhoff's Rules

  • Thread starter joej
  • Start date
  • Tags
    Rules
In summary, the current in each battery is 1 ohm and the terminal voltage of the 6V battery is 6. volts. Kirchhoff's Rules can be used to solve for the currents in the circuit. Loop 1 will be for I1, loop 2 will be for I2, and loop 3 will be for I3. The junction rule is used at point b on the diagram.
  • #1
joej
29
0
okay I got a question asking me to determine the currents in I1, I2, and I3 in a circuit assuming that the resistance of each battery is 1 ohm (I drew a separate 1ohm resistor after every battery to represent this) and the terminal voltage of the 6V battery

now I have drawn a diagram of the circuit and labeled every part of it, I am assuming I have to use Kirchhoff's Rules to solve this problem

now here is where I get stuck, do I make 3 loops since there are 3 batteries or will 2 loops be fine
(using the diagram loop 1 will probably be 'abefa' and loop 2 will be 'abcdefa', do I need loop 3 'ebcde')

also at which point should I apply Kirchhoff's junction rule?
 

Attachments

  • diagram.jpg
    diagram.jpg
    6.7 KB · Views: 455
Physics news on Phys.org
  • #2
You have 3 unknowns, so you need three equations. One of these is the junction rule for the three currents (at either of the two junctions...you'll see that they both give you the same equation). The other two will be two loop equations.

If you write the third loop equation, you'll find that it is actually just a combination of the two previous ones, and so, gives you nothing new.
 
  • #3
okay still seems I'm screwing up somewhere...

junction is at point b on my graph so... I1 = I2+I3

loop 1 is 'bcdefab' => +12-9(I1)-15(I3)+6-19(I3) = 0

loop 2 is 'befab' => +11(I2)-12-15(I3)+6-19(I3) = 0

I1 = 18-34(I3) / 9 ==> I1 = 2-3.8(I3)

I2 = -6-34(I3) / -11 ==> I2 = 0.55+3.1(I3)


I3 = I1 - I2

I3 = (2-3.8(I3))-(0.55+3.1(I3)) ==> I3 = 1.45 - 6.9(I3)


I'm guessing I'm screwing up somewhere in the veginning with either the loops of the junction cause that's th only place I can think of where I migth be going wrong unless I'm not noticing something
 
  • #4
What makes you think you're wrong ? Everything looks okay to me, so far.

You've got : I3 = 1.45 - 6.9(I3)

Or, 7.9(I3) = 1.45 ==>I3 = (1.45/7.9) A
 
  • #5
that gives me

I3 = 0.18A

the answer key tells me I am wrong as it says I3 should be 0.055A

so yeah basically I'm stuck and not quite sure where I'm going wrong
 
  • #6
When I solved it, i got i3 = 0.185 A. I believe the answer key is wrong.
 
  • #7
not that I don't trust you but is there someone else that could confirm this as well, I would like to have a few confirmations that they key is actually wrong before I decide to go against it, thnx ;)
 
  • #8
joej said:
not that I don't trust you but is there someone else that could confirm this as well, I would like to have a few confirmations that they key is actually wrong before I decide to go against it, thnx ;)

I get 0.18A for I3 also. Make sure that you've got the question right... You might have gotten the battery orientation wrong or something.
 
  • #9
okay tripple checked everything, I got nothing wrong so it seems that the answer key is bogus, thnx for the help
 

Related to Solving Circuit Problems with Kirchhoff's Rules

1. What are Kirchhoff's Rules?

Kirchhoff's Rules, also known as Kirchhoff's Circuit Laws, are two fundamental principles used to analyze electrical circuits. These rules are based on the principles of conservation of charge and energy.

2. What is the first rule of Kirchhoff's Rules?

The first rule, also known as Kirchhoff's Current Law, states that the algebraic sum of currents entering and leaving a junction in a circuit must equal zero. In other words, the total current flowing into a junction must be equal to the total current flowing out.

3. What is the second rule of Kirchhoff's Rules?

The second rule, also known as Kirchhoff's Voltage Law, states that the algebraic sum of voltage drops in a closed loop must equal the algebraic sum of voltage sources in that loop. This law is based on the principle of conservation of energy.

4. Why are Kirchhoff's Rules important?

Kirchhoff's Rules are important because they provide a systematic and accurate method for analyzing complex electrical circuits. They are fundamental principles that are used in various fields of electrical engineering, such as circuit design, power systems, and telecommunications.

5. How do I apply Kirchhoff's Rules to solve a circuit?

To apply Kirchhoff's Rules, follow these steps: 1) Identify all the junctions and loops in the circuit. 2) Apply Kirchhoff's Current Law at each junction to write equations for the currents. 3) Apply Kirchhoff's Voltage Law in each loop to write equations for the voltages. 4) Solve the resulting system of equations to find the values of currents and voltages in the circuit. 5) Check your solution by ensuring that Kirchhoff's Rules are satisfied in the circuit.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
442
  • Introductory Physics Homework Help
Replies
4
Views
792
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
724
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
28
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
602
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top