*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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