[Circuits] Solving a KCL Problem

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In summary, the conversation discusses a circuit with a potential contradiction at the V2 node due to inconsistencies in the direction of current arrows. The expert explains that as long as the math is consistent, the values will be correct. The confusion is resolved when calculating V2, which is found to have a higher voltage than both 60V and V1. This means that a current of 0.29A flows from V2 to 60V across a 10Ω resistor, and 4.90A flows from V2 to V1 across a 2Ω resistor. The expert also clarifies that it is not always the case that all currents are flowing into a given node.
  • #1
ainster31
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Homework Statement



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Homework Equations





The Attempt at a Solution



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Here is what I don't understand: how can this circuit exist? If you look at the V2 node, there are 3 currents going in but none going out. Isn't this a contradiction? From what I've heard, you can arbitrarily draw current arrows and if the direction of the arrows is wrong, you'll just get a negative current. But don't you have to at least be consistent in the rotation? If you look at the bottom left mesh, the currents are going clockwise and then there is a current that is going counter-clockwise.
 
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  • #2
ainster31 said:
Here is what I don't understand: how can this circuit exist? If you look at the V2 node, there are 3 currents going in but none going out. Isn't this a contradiction? From what I've heard, you can arbitrarily draw current arrows and if the direction of the arrows is wrong, you'll just get a negative current. But don't you have to at least be consistent in the rotation? If you look at the bottom left mesh, the currents are going clockwise and then there is a current that is going counter-clockwise.

So long as you're consistent in writing the math according to your assumed currents, the math will always take care of itself and present you with the right values (positive or negative) for each.

In fact, when writing node equations it's actually easier to maintain consistency by always assuming that all currents are flowing either into or out of a given node. That way you never trip up by forgetting which one or ones you had going in and which ones going out. Later, if you need the value of a particular current, you use solved-for node voltages and Ohm's law for a given branch.
 
  • #3
ainster31 said:
Here is what I don't understand: how can this circuit exist? If you look at the V2 node, there are 3 currents going in but none going out. Isn't this a contradiction?

No...there isn't any contradiction .The confusion clears as soon as you calculate V2 ,which you haven't .V2 = 62.89 V which means it is higher than both 60V and V1 .So,in reality a current of magnitude 0.29A flows from V2 to 60V across the 10Ω resistor and 4.90A from V2 to V1 across 2Ω resistor .

In other words at node V2 ,current of magnitude 5.19A enters whereas 0.29A and 4.90A leave .

It is your assumption that all currents are flowing into node V2.But in reality that is not the case .

Hope that helps
 
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1. How do you solve a KCL problem?

To solve a KCL (Kirchhoff's Current Law) problem, you first need to identify all the currents entering and leaving a node or junction in a circuit. Then, you can apply the KCL equation, which states that the sum of all currents entering a node must equal the sum of all currents leaving the node. You can use this equation to set up a system of equations and solve for the unknown currents.

2. What is the purpose of KCL in circuit analysis?

The purpose of KCL is to ensure that the conservation of charge is maintained in a circuit. It helps in analyzing and predicting the behavior of complex circuits by applying the principle of charge conservation at each node. KCL is an essential tool in circuit analysis and is used to solve various problems related to current flow in a circuit.

3. Can KCL be applied to any circuit?

Yes, KCL can be applied to any circuit, regardless of its complexity. It is a fundamental law in circuit analysis and is applicable to both DC and AC circuits. KCL can be used to analyze circuits with multiple sources, resistors, and other components. However, it is important to ensure that the circuit is in a steady-state condition before applying KCL.

4. How do you handle current sources in KCL problems?

When dealing with current sources in KCL problems, you need to consider the direction of the current. If the current source is entering a node, it is considered as a positive value, and if it is leaving a node, it is considered as a negative value. You can then use these values in the KCL equation to solve for the unknown currents.

5. Can KCL be used to calculate voltage in a circuit?

No, KCL cannot be used to directly calculate voltage in a circuit. KCL only deals with current flow in a circuit, and voltage is a measure of potential difference. However, once you have solved for all the unknown currents using KCL, you can use Ohm's law (V=IR) to calculate the voltage drop across each component in the circuit.

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