Finding the magnitude and direction of current thro Kirchhoff law

In summary, the person is seeking help with finding the magnitude and direction of current through a 10 ohm resistor. They have used the superposition principle and have gotten a different answer from the one given in the book. They are asking for assistance in finding where they went wrong.
  • #1
logearav
338
0

Homework Statement



Revered Members,
Please go through my attachment. I have been asked to find the magnitude and direction of current through the 10 ohm resistor. The answer that is given in the book is 1/23 Ampere from A to B.
But i solved as given in the attachment but i got the answer as 5/37 Ampere which is I3...the current which I assumed to pass through 10 ohm resistor. Since Please help me where i went wrong.

Homework Equations





The Attempt at a Solution


 

Attachments

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  • #2
Your work and results look fine. Must be an error in the book's answer.
 
  • #3
I used the superposition principle as follows:

1. Short the 2V source & get voltage across the 10Ω = +2.1622V.
2. Short the 8V source and get same voltage = -0.8108V
3. Superpose the voltages to get +1.3514V across the 10Ω.
4. Divide by 10Ω to get i3 = 0.13514A which agrees with both of you!
 
  • #4
Thank you gneill and rude man for your help.
 
  • #5


Dear fellow scientist,

Thank you for sharing your work with us. First of all, it is important to understand that there are two laws in Kirchhoff's circuit analysis - Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL). In order to find the magnitude and direction of current through the 10 ohm resistor, we need to use both laws.

KCL states that the algebraic sum of currents entering and leaving a node (or junction) in a circuit is equal to zero. In this case, the node between the 10 ohm resistor and the 5 ohm resistor is our focus. Applying KCL at this node, we get the equation I1 + I2 = I3, where I1, I2, and I3 are the currents flowing through the 10 ohm, 5 ohm, and 15 ohm resistors, respectively.

KVL states that the sum of voltage drops across a closed loop in a circuit is equal to the sum of voltage rises. In this case, the loop we are interested in is the one containing the 10 ohm resistor, the 5 ohm resistor, and the 15 ohm resistor. Applying KVL to this loop, we get the equation -10I1 + 5I2 + 15I3 = 0, where the negative sign for the 10 ohm resistor indicates that the voltage drop is in the opposite direction of the current flow.

Solving these two equations simultaneously will give us the values of I1, I2, and I3. From your attachment, it seems that you have correctly solved the equations and obtained the value of I3 as 5/37 Ampere. However, the book's answer of 1/23 Ampere is also correct. This is because the direction of current flow is arbitrary and can be chosen in either direction. As long as the magnitude of the current is correct, the direction does not matter.

In conclusion, it seems that you have not gone wrong in your calculations. The difference in the answer is due to the direction of current flow, which is arbitrary. I hope this helps clarify your doubts. Keep up the good work!

Best regards,
 

Related to Finding the magnitude and direction of current thro Kirchhoff law

1. What is Kirchhoff's current law?

Kirchhoff's current law, also known as Kirchhoff's first law or the junction rule, states that the total current flowing into a junction must equal the total current flowing out of the junction. This is based on the principle of conservation of charge, where the amount of charge entering a junction must be equal to the amount leaving the junction.

2. How do I apply Kirchhoff's current law to find the magnitude and direction of current?

To apply Kirchhoff's current law to find the magnitude and direction of current, you must first identify all the junctions in the circuit. Then, for each junction, you can set up an equation stating that the sum of all currents entering the junction is equal to the sum of all currents leaving the junction. This will allow you to solve for the unknown currents and determine their magnitude and direction.

3. Can Kirchhoff's current law be applied to both series and parallel circuits?

Yes, Kirchhoff's current law can be applied to both series and parallel circuits. In series circuits, there is only one path for current to flow, so the current entering a junction must equal the current leaving the junction. In parallel circuits, the current entering a junction is split between multiple paths, but the sum of all currents entering the junction is still equal to the sum of all currents leaving the junction.

4. What is the difference between Kirchhoff's current law and Kirchhoff's voltage law?

Kirchhoff's current law deals with the conservation of charge at junctions in a circuit, while Kirchhoff's voltage law deals with the conservation of energy in a closed loop. Kirchhoff's voltage law states that the sum of all voltage drops in a closed loop must equal the sum of all voltage gains in the same loop. Both laws are important for analyzing and solving complex circuits.

5. Are there any limitations or exceptions to Kirchhoff's current law?

While Kirchhoff's current law is generally accurate and applicable to most circuits, there are a few limitations and exceptions. One limitation is that it assumes ideal conditions, such as no resistance in wires and perfect connections. Exceptions include circuits with rapidly changing magnetic fields or circuits with non-linear elements, such as diodes or transistors, which can cause deviations from the expected current flow.

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