Learning goal
After this lesson, you should be able to select a reference node, write correct KCL equations at unknown nodes, solve the simultaneous equations and verify that charge is conserved.
The practice circuit
Consider a resistive circuit with ground (0 V) as the reference node and two unknown node voltages, VA and VB:
- A 5 Ω resistor connects node A to ground.
- A 10 Ω resistor connects node B to ground.
- A 4 Ω resistor connects nodes A and B.
- A 3 A ideal current source injects current into node A from ground.
- A 1 A ideal current source draws current out of node B to ground.
Step 1: Write KCL at node A
Resistor current from node A to ground is VA/5. Current from A to B is (VA − VB)/4. Their total is the 3 A supplied to node A.
Multiplying by 20 gives:
Step 2: Write KCL at node B
Current from B to ground through its 10 Ω resistor is VB/10. Current from B to A is (VB − VA)/4. The current source takes 1 A out of B, so the injected source current is −1 A.
Multiplying by 20 gives:
Step 3: Solve for the node voltages
The pair of equations is:
-5 V_A + 7 V_B = -20
Eliminate VA or VB using substitution or the matrix method.
| Quantity | Value | Interpretation |
|---|---|---|
| VA | 8.4211 V | Voltage at node A relative to ground |
| VB | 3.1579 V | Voltage at node B relative to ground |
Step 4: Verify the answer
At A, current through the 5 Ω resistor is 8.4211/5 ≈ 1.6842 A; current from A to B through 4 Ω is (8.4211 − 3.1579)/4 ≈ 1.3158 A. Their sum equals the injected 3 A.
At B, 1.3158 A arrives from A. The 10 Ω resistor draws 3.1579/10 ≈ 0.3158 A, and the current source draws 1 A. Total leaving B = 0.3158 + 1 = 1.3158 A, matching the arriving current.
Common mistakes in GATE EE nodal problems
- Mixing the directions of resistor currents in adjacent KCL equations.
- Forgetting that an outward-directed current source contributes a negative injected current.
- Using resistor conductance 1/R incorrectly when forming matrix coefficients.
- Skipping the KCL check after solving, allowing arithmetic mistakes to go unnoticed.
Try it yourself
Change the source at node B so that it injects +1 A into B instead of drawing 1 A out. Keeping all resistors and the source at A unchanged, calculate the new node voltages.
Show the solution
The revised nodal equations are 9VA − 5VB = 60 and −5VA + 7VB = 20. Solving gives VA = 13.6842 V and VB = 12.6316 V.