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

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The discussion focuses on verifying Kirchhoff's Current Law (KCL) using a circuit analysis involving currents Ia, Ib, Ic, and Id. It is established that in a series configuration, Ic and Id are equal in magnitude but opposite in direction. The calculations show that if Ia is 3A and Ic is 1A, then Id must be -1A, leading to the conclusion that Ia equals Ib at 3A. However, an alternative analysis suggests that when applying KCL at a node with Ia and Ic entering and Ib leaving, the total current should equal 4A. The conversation confirms the correct application of KCL, highlighting the importance of consistent current direction and magnitude in circuit analysis.
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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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