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:

So that was all about this example. Also, check other topics on Basics of Electrical Engineering:

1. Basic properties of electrical engineering
2. SI units and derived units for electrical engineers
3. Electrical symbols
4. Ohm's law
5. Resistors in series
6. Resistors in parallel
7. Series-Parallel Circuits
8. Kirchhoff's current law
9. Kirchhoff's voltage law
10. Voltage divider
11. Current divider
12. Branch current analysis
13. Superposition principle
14. Thevenin theorem
15. Norton theorem