Kirchoffs current law. Someone check to make sure its correct?

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Homework Help Overview

The discussion revolves around Kirchhoff's Current Law (KCL) in the context of a circuit analysis problem involving current values in a series configuration. Participants are examining the relationships between different currents at a node.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are analyzing the current values and their directions, questioning the signs and magnitudes of the currents based on KCL. There is a focus on the implications of the series configuration and how it affects the current relationships.

Discussion Status

There is an active exchange of ideas regarding the application of KCL, with some participants agreeing on the signs of the currents while others present alternative interpretations of the current relationships. The discussion reflects a productive exploration of the problem without reaching a definitive consensus.

Contextual Notes

Participants are working under the constraints of the problem as presented, with specific current values given and the requirement to apply KCL correctly. There is an emphasis on ensuring the signs of the currents are accurately represented in their equations.

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Due to the series configuration, Ic and Id are equal in magnitude but opposite in direction.
if Ia = 3A, Ic = 1A, then Id = -1A
Ia + Ic - Id - Ib = 0
3A + 1A - 1A - Ib = 0
as a result, Ia=Ib = 3A
 
data1217 said:
Due to the series configuration, Ic and Id are equal in magnitude but opposite in direction.
if Ia = 3A, Ic = 1A, then Id = -1A
Ia + Ic - Id - Ib = 0
3A + 1A - 1A - Ib = 0
as a result, Ia=Ib = 3A


I agree with the Ic Id sign. However, Using the Node with Ia and Ic entering and Ib leaving and using KCL. I get Ia+Ic-Ib=0 therefore Ia+Ic=Ib or 3A + 1A =4A
 
Valhalla said:
I agree with the Ic Id sign. However, Using the Node with Ia and Ic entering and Ib leaving and using KCL. I get Ia+Ic-Ib=0 therefore Ia+Ic=Ib or 3A + 1A =4A

You are right.
for the node with Ib and Id entering, Ia leaving
-Ia + Ib + Id = 0
-3 + Ib + -1 = 0
Ib =4A
 

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