Kirchhoff's Loop Rule Problem.

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In summary, the problem lied in the assumption that the cell with emf E is connected between A and D.
  • #1
Prashasti
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Homework Statement


Actually, the main question is to find out the potential difference between the points 'A' and 'D'.
The cell with emf 'E' has not been shown in the main question, but I have assumed it to be connected between 'A' and 'D'. Each resistor is of 1 Ohm.

2. The attempt at a solution
While solving the question, I applied Kirchhoff's Laws
Let 'I' be the current and let 'I1' be the current through the branch QR.
Since the total resistance offered by the branches (QR, PS, TU) is same (One ohm) , the currents through them will also be the same.
Assuming E > 3 V ( where 'E' is the emf of the cell)

Applying Kirchhoff's Current Law,
I = 3 I1
Applying KVL to the loop BAPQRDCB,
Δ V = 0
So, I(1) + I1(1) + 3 = E ...(1)
Similarly,
When KVL is applied to the loop BAPSDCB,
I(1) + I1 (1) + 2 = E ...(2)
(1) = (2), as E = E,
Which, on solving, gives 2 = 3!
Where did I go wrong?
uyy.png
 
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  • #2
Prashasti said:

Homework Statement


Actually, the main question is to find out the potential difference between the points 'A' and 'D'.
The cell with emf 'E' has not been shown in the main question, but I have assumed it to be connected between 'A' and 'D'. Each resistor is of 1 Ohm.

2. The attempt at a solution
While solving the question, I applied Kirchhoff's Laws
Let 'I' be the current and let 'I1' be the current through the branch QR.
Since the total resistance offered by the branches (QR, PS, TU) is same (One ohm) , the currents through them will also be the same.

That is not true. The potential difference is the same across the three branches: VP-VS. As there are different cells in every branch, the voltages across the resistors are also different.

You can denote the currents flowing in the branches I1, (upper branch) I2 (middle branch) and I3 (bottom branch). Then I=I1+I2+I3. Write the loop equation for the loops which contain E and the series 1 ohm and one of the branches.

kircloop.JPG
 
  • #3
Prashasti said:
The cell with emf 'E' has not been shown in the main question, but I have assumed it to be connected between 'A' and 'D'.
1) What was shown in the main question ? Nothing there? Or a short-circuit ? Was anything mentioned about E at all ?

Since the total resistance offered by the branches (QR, PS, TU) is same (One ohm) , the currents through them will also be the same.
No. There are voltage sources too !
2) so that gets rid of I = 3 I1. You will need to split up I in I1, I2, I3.

[edit] ah, ehild was faster !
 
  • #4
No, only the portion of the figure between the points A and D has been given.

But, I've assumed E > 3V.. So the current will pass according to the cell with emf 'E' only? Am I right?
 
  • #5
Prashasti said:
No, only the portion of the figure between the points A and D has been given.

But, I've assumed E > 3V.. So the current will pass according to the cell with emf 'E' only? Am I right?
No. You are not correct about that.

Use Kirchhoff's Rules, as suggested by the title of your thread.

Don't introduce anything extra into the circuit.
 
  • #6
Was the current I given? Or were you asked to give the potential difference in terms of I?
 
  • #7
ehild said:
Was the current I given? Or were you asked to give the potential difference in terms of I?
No. I added it myself..
 
  • #8
SammyS said:
No. You are not correct about that.

Use Kirchhoff's Rules, as suggested by the title of your thread.

Don't introduce anything extra into the circuit.
What would have happened if there was a question with the same circuit diagram and a cell with emf (E) 12 V was present there?
 
  • #9
If it is no indication that current flows through the leftmost resistor, you have to assume it zero. There is current in the three branches only, and find the potential difference between points A and D.
 
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  • #10
Prashasti said:
What would have happened if there was a question with the same circuit diagram and a cell with emf (E) 12 V was present there?
The question in your Original Post asks the potential difference between A and D. If you introduce an ideal voltage source, E, as you have the answer becomes trivial. The potential difference between A and D is just that of the voltage source.

It would help if you would state the problem with the wording as it was given to you.
 
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  • #11
So, the mistake lies in that "assumption" itself. It messed everything up.
Thanks a lot!
 

What is Kirchhoff's Loop Rule?

Kirchhoff's Loop Rule, also known as Kirchhoff's second law, states that the sum of all voltage drops in a closed loop circuit is equal to the sum of all voltage sources in that same loop. It is one of the fundamental laws in circuit analysis.

How do you apply Kirchhoff's Loop Rule in circuit problems?

To apply Kirchhoff's Loop Rule in circuit problems, you must first identify all the loops in the circuit. Then, starting from any point in the loop, choose a direction to travel and assign a voltage drop sign (+ or -) to each component based on the direction of current flow. Finally, write an equation using Kirchhoff's Loop Rule by setting the sum of all voltage drops equal to the sum of all voltage sources.

Can Kirchhoff's Loop Rule be applied to any type of circuit?

Yes, Kirchhoff's Loop Rule can be applied to any type of circuit, including series, parallel, and combination circuits. It is a fundamental law that applies to all circuits.

What is the difference between Kirchhoff's Loop Rule and Kirchhoff's Junction Rule?

Kirchhoff's Loop Rule deals with voltage drops and sources in a closed loop circuit, while Kirchhoff's Junction Rule, also known as Kirchhoff's first law, deals with current entering and leaving a junction in a circuit. Both rules are essential for analyzing complex circuits.

Can Kirchhoff's Loop Rule be violated?

No, Kirchhoff's Loop Rule cannot be violated. It is a fundamental law in circuit analysis and is based on the principle of conservation of energy. If the rule appears to be violated, it is likely due to an error in the circuit analysis or measurement.

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