Kirchhoff's Current Law Example # 2: 9 Resistors join in a network

9 Resistors are connected in a network. 4 branch currents are known, Find the unknown values.
Solution: Solving such circuit for unknown values require the application of Kirchhoff's current law.

Step 1: Label all nodes
Step 2: Apply KCL at node e

∑Current In = ∑Current Out
2 A + 18 A = ∑Current Out
∑Current Out = 20 A

20 A current leaves through the node 'e'.

Step 3: Apply KCL at node d
∑Current In = ∑Current Out
20 A = 5 A + Unknown current
Unknown current = 15 A
This unknown current is the one which moves up from the node d.

Step 4: At node c
∑Current In = ∑Current Out
5 A = 3 A + Unknown
Unknown current = 2 A

Step 5: Now at node a
∑Current In = ∑Current Out
15 A + 2 A = ∑Current Out
∑Current Out = 17 A

Step 6: Finally, at node b
∑Current In = ∑Current Out
17 A + Unknown = 18 A
Unknown = 18 A - 17 A
Unknown = 1 A

Finally, our solution is:


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