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:

So that was all about this example. Also, check other topics on Basics of Electrical Engineering:
- Basic properties of electrical engineering
- SI units and derived units for electrical engineers
- Electrical symbols
- Ohm's law
- Resistors in series
- Resistors in parallel
- Series-Parallel Circuits
- Kirchhoff's current law
- Kirchhoff's voltage law
- Voltage divider
- Current divider
- Branch current analysis
- Superposition principle
- Thevenin theorem
- Norton theorem
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