Understanding the Current Division Rule: A Circuit Diagram Guide

In summary: A flow through 3Ω resistor? I think current would still flow through the 3Ω resistor because it's a parallel circuit. I don't think there would be a voltage difference between A and B because the voltage would be the same across the 3Ω resistor.
  • #1
Harsharma17
10
0
Homework Statement
I earlier asked this question on PF for equivalent resistance, but now I want to ask a different thing in the same question.
https://www.physicsforums.com/threads/how-to-solve-this-network-of-resistors-for-equivalent-resistance.1000068/
The question is to calculate the amount of current in AB, so I thought solving for the resistors would lead to the current value, but it won't be the case here. The assumption I used, was to convert AB into H shape wire, and then to solve for parallel resistors on both sides of AB. If I consider the 4 ohm answer, then 2 A current would flow through AB as it would be in series. But the correct answer is 0.66 A, which means we need to consider potential difference of A and B, and then apply Kirchhoff's law to find current in AB.
But the procedure is time-consuming, and I am supposed to solve such questions with shortcuts and tricks rather than direct formulas (it's for a competitive exam). So I googled it, and found a solution, whose screenshot I have provided alongwith circuit diagram.
Relevant Equations
It could be said I am basically asking for the short cut alternative for Kirchoff's law, but I know such an option exists, so I am asking for the explanation of the less time-consuming method.
Here's the diagram of the circuit:
Screenshot_2020-12-08-12-51-30-736_com.google.android.apps.docs_2.png


And here's the solution I found on a site on Google:
IMG_20210228_235511.jpg

I tried to follow through the solution given, but can't understand how it is applying current division rule to find current in QA and QB, and subtracting current values to find the current in AB is still confusing to me ( I also can't request the person on that site to explain it to me). Please explain it to me.
 
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  • #2
Please post the appropriate circuit diagram. Saying that you asked this question already without providing a link is of little help. You should also provide a link to the site you found using google if you want help with that.
 
  • #3
kuruman said:
Please post the appropriate circuit diagram. Saying that you asked this question already without providing a link is of little help.
Sorry for not providing the link and diagram, now I have edited the question to give these details.
 
  • #4
kuruman said:
You should also provide a link to the site you found using google if you want help with that.
Providing the link to that site would be useless, because the site allows it's answers to be viewed by only those who have an account there. It is a study course-type site, and most of the questions asked are neglected there, so I asked it's explanation here. Nevertheless, here's the link:
https://www.meritnation.com/ask-ans...t-as-shown-in-figure/academics/13267875?hcb=1
 
  • #5
Harsharma17 said:
I tried to follow through the solution given, but can't understand how it is applying current division rule to find current in QA and QB, and subtracting current values to find the current in AB is still confusing to me ( I also can't request the person on that site to explain it to me). Please explain it to me.
I do not understand this solution, so I cannot explain it to you. I did it in a way that I understand and can explain. I agree with the answer of 0.66 A.
Harsharma17 said:
Providing the link to that site would be useless, because the site allows it's answers to be viewed by only those who have an account there. It is a study course-type site, and most of the questions asked are neglected there, so I asked it's explanation here. Nevertheless, here's the link:
https://www.meritnation.com/ask-ans...t-as-shown-in-figure/academics/13267875?hcb=1
Thank you for providing the link. I asked for it because it because I thought it was about a different problem.

It would help to redraw the resistors conventionally as we already discussed. I have done some of that work and put in the obvious connections and labels on the left and on the right sides of the circuit. Can you complete it with only vertical and horizontal segments of connections in the middle space and then label points A and B? If so, you will see quite easily what's going on.

Four Resistors.png


My method consisted of assigning current ##I## through the 3Ω resistor in which case the 6Ω resistor would draw current 2A-##I##. I then applied Kirchhoff voltage and current rules to a loop in the redrawn circuit that included these two resistors and the segment labeled AB.
 
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  • #6
kuruman said:
View attachment 278896

My method consisted of assigning current ##I## through the 3Ω resistor in which case the 6Ω resistor would draw current 2A-##I##. I then applied Kirchhoff voltage and current rules to a loop in the redrawn circuit that included these two resistors and the segment labeled AB.
I think current division rule would give I current in 6ohm, and 2 I current in 3 ohm, because current is inversely proportional to resistance.
 
  • #7
Harsharma17 said:
I think current division rule would give I current in 6ohm, and 2 I current in 3 ohm, because current is inversely proportional to resistance.
That would be the case if there is nothing else beyond the parallel 3Ω and 6Ω resistors. It is certain however that if you have a ##2\text{A}## current splitting two ways and ##I## goes one way what goes the other way is what's left over, ##2\text{A} - I.## Did you complete drawing the circuit?
 
  • #8
I tried putting H-shape wire in place of AB, but that would mean all the current would flow from left part of circuit to right part through AB, but that's not the case. Current is going from A and B to Q, not only from either A or B. But I can't draw it in any other way.
I was unable to upload the image of circuit here, so please click the link below:
https://drive.google.com/file/d/1pgqzcI1Q0Ty8VgrojdYz3_p-jBO1XR_T/view?usp=drivesdk
I seem to contradict myself, but I know this diagram is wrong.
 
  • #9
Harsharma17 said:
How to find current in branch AB having no resistors but with voltage difference

1614588986652.png

Do you seriously believe that there is a voltage difference between A and B ?
 
  • #10
phinds said:
Do you seriously believe that there is a voltage difference between A and B ?
If there wouldn't be voltage difference between A and B, then how would current of 0.66 A(as given in the answer) would flow from A to B? If we don't consider the answer for a moment, it is clear from the figure that the circuit is not a balanced wheatstone bridge, as 3/6 is not equal to 6/3. Then it's implied that there exists a current flow from A to B.
 
  • #11
Harsharma17 said:
If there wouldn't be voltage difference between A and B, then how would current of 0.66 A
How does current flow through a wire? Does it require a voltage difference between one end of the wire and the other?
 
  • #12
phinds said:
How does current flow through a wire? Does it require a voltage difference between one end of the wire and the other?
Yes, voltage, or potential difference is required for flow of current, just like two terminals of battery provide the voltage to the circuit, and electrons move from one point to another till this difference is cancelled.
Here's the link for a page about voltage and current flow with various analogies to make the point more understanding:
https://www.allaboutcircuits.com/textbook/direct-current/chpt-1/voltage-current/
 
  • #13
Harsharma17 said:
Yes, voltage, or potential difference is required for flow of current ...
No. Not through a wire (well, and ideal wire anyway). Until you get that straight you're going to have a lot of trouble w/ circuits.

The current around a circuit requires a voltage and the current through a non-zero-ohm element requires voltage, but the current flows the same all the way around the circuit and there is no voltage drop across the wires that connect the elements.

Here's a simple example:

1614606231263.png


No voltage across the wires.
 
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  • #14
phinds said:
How does current flow through a wire? Does it require a voltage difference between one end of the wire and the other?
I sent you a PM about this.
 

1. What is the current division rule?

The current division rule is a principle in circuit analysis that states that the total current entering a junction in a circuit will be divided among the different branches based on their resistance. This rule is based on Ohm's law and can be used to calculate the current flowing through each branch in a circuit.

2. How is the current division rule applied in a circuit diagram?

The current division rule is applied by first identifying the junction in the circuit where the current is divided. Then, the resistance of each branch connected to the junction is determined. The current flowing through each branch can then be calculated using the formula I = V/R, where I is the current, V is the voltage, and R is the resistance of the branch.

3. What are some practical applications of the current division rule?

The current division rule is commonly used in circuit analysis and design. It can be used to determine the current flowing through different components in a circuit, such as resistors, capacitors, and inductors. This information is important in designing circuits for specific purposes, such as in electronic devices and power systems.

4. Are there any limitations to the current division rule?

Yes, there are limitations to the current division rule. It assumes that the resistors in the circuit are connected in parallel and that the voltage is constant throughout the circuit. In real-world circuits, these conditions may not always be met, which can affect the accuracy of the calculations using the current division rule.

5. How can I use the current division rule to troubleshoot a circuit?

The current division rule can be helpful in troubleshooting a circuit by allowing you to determine the current flowing through each branch. If the calculated current does not match the expected current, it can indicate a problem with a component or a connection in the circuit. By using the current division rule, you can identify the specific branch where the issue may be occurring and then focus your troubleshooting efforts on that area.

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