Node Analysis Question: Solving for Current in a Two-Loop Circuit

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In a two-loop circuit analysis, the current at a node must adhere to Kirchhoff's Current Law, which states that the sum of currents entering a node equals the sum of currents leaving it. The 4A current mentioned flows through the resistor, indicating that it remains constant throughout that branch. When determining the current direction, the sign (positive or negative) depends on the assumed direction of flow. It is essential to clarify the reference point for ground to accurately analyze the circuit. Understanding these principles is crucial for solving current in complex circuits.
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This is first time i came across this problem. But let's say u have two loop circuit. in the middle u have a a resistor going down and a it shows let's say 4A current after the resistor. After i make my ground point let's say at bottom of this circuit. Would the equation be just +/- 6A based on which direction the current is flowing ?
 
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Your question isn't very clear. At a node, the sum of the currents flowing into the node must equal the sum of the currents flowing out of the node.

As for the 4A flowing "after" the resistor, it is actually 4A flowing through the resistor (between the two nodes that make up the central branch the current is 4A at any point on the branch).
 

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