Nodal analysis: practice problems with step-by-step solutions
Also called: node voltage method · KCL
Free Nodal analysis practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: 1 unknown node, 2 unknown nodes, 2 nodes, 3 sources.
Free practice problemEasy
Use nodal analysis to find the current IR1 through R1 in the direction of the arrow.
Nodal analysis is a standard method to find every voltage in a circuit. It works even when series and parallel rules are not enough.
A node is a point where two or more parts of a circuit meet. A whole stretch of plain wire counts as one node, because a wire has no voltage drop along it.
You choose one node as the reference, also called ground, and give it 0 V. Every other node voltage is measured from that node.
Think of heights on a map, measured from sea level. Water flows from high to low, and the flow in a pipe depends on the difference in height between its ends.
Once you know the node voltages, Ohm’s law gives every current. So the node voltages are the only unknowns you need.
Self-check Which node has a voltage of 0 V by definition?
Every node without a source
The node next to the largest source
The node with the most resistors
The reference node (ground) that you chose
Answer: The reference node (ground) that you chose. You pick the reference node yourself and call it 0 V. All other node voltages are measured from it.
2Kirchhoff’s current law
Charge cannot pile up in a node. So all the current that flows into a node must flow out of it again. This is Kirchhoff’s current law (KCL).
ΣIin = ΣIout
For example, if 3 A and 2 A flow into a node, then 5 A in total must leave it through the other branches.
You write the current in each resistor with Ohm’s law, using the voltages at its two ends:
I = (Va − Vb) / R
This is the current from a to b. If the result is negative, the current really flows from b to a. That is fine: keep the sign.
Self-check Node a is at 9 V and node b is at 4 V. A 5 Ω resistor connects them. What current flows from a to b?
1 A
1.8 A
2.6 A
−1 A
Answer: 1 A. I = (9 − 4)/5 = 1 A. Adding the voltages gives 2.6 A, swapping them gives −1 A, and 1.8 A ignores the voltage at b.
3The method step by step
Choose the reference node. The bottom wire is the usual choice. Mark it with the ground symbol.
Find the nodes whose voltage you already know. A node joined to ground by a voltage source sits at that source voltage.
Name every other node (A, B, …). Their voltages VA, VB, … are the unknowns.
At each unknown node, write KCL: the currents leaving through the resistors add up to the current pushed in by current sources.
Write each resistor current as (Vnode − Vother end)/R, then solve the equations.
Use Ohm’s law to find any branch current the question asks for.
Writing every resistor current as leaving the node makes the signs automatic. You do not have to guess the real directions first.
You need one equation per unknown node. Never write an equation for the reference node: its voltage is already known.
4A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
Find the node voltage VA and the current IR1 through R1, from the source towards A.
1
Look at the circuit
The bottom wire is ground. The top of V1 is 15 V above ground, so that node is known: 15 V.
Only node A is unknown, so you need one equation.
2
Write KCL at A
Two resistors leave A: R1 goes to the 15 V node and R2 goes to ground. The current source pushes 1 A into A.
(VA − 15)/3 + VA/6 = 1
3
Solve for VA
Multiply both sides by 6 to remove the fractions.
2·(VA − 15) + VA = 6
3·VA − 30 = 6 ⇒ VA = 12 V
4
Find the current in R₁
IR1 = (15 − 12)/3 = 1 A
The result is positive, so 1 A really flows from the source towards A.
5
Check the answer
R2 carries 12/6 = 2 A down to ground. Into A flow 1 A through R1 and 1 A from the current source.
That is 2 A in and 2 A out, so KCL holds at A.
Answer
VA = 12 V and IR1 = 1 A.
5Two unknown nodes
With two unknown nodes you write one KCL equation at each node and solve the two equations together.
At node A, three resistors leave: R1, R2 and R3. At node B, two resistors leave and the current source pushes 2 A in.
A: (VA − 6)/2 + VA/3 + (VA − VB)/3 = 0
B: (VB − VA)/3 + VB/3 = 2
Multiply the first equation by 6 and the second by 3:
7·VA − 2·VB = 18
−VA + 2·VB = 6
Add the two equations. The VB terms cancel, so 6·VA = 24.
VA = 4 V, VB = 5 V
Notice that R3 appears in both equations, as (VA − VB) at A and as (VB − VA) at B. This is a quick check that your equations are right.
6Common mistakes
Mixing up the sign of a shared resistor: it is (VA − VB) at A, but (VB − VA) at B.
Forgetting a current source that enters or leaves the node.
Writing a KCL equation for the reference node, or treating a known node as unknown.
Dropping the minus sign of a negative current. A negative result means the current flows against the arrow.
Self-check At node A, one 4 Ω resistor goes to ground and another 4 Ω resistor goes to a 12 V node. A 2 A source pushes current into A. What is VA?
10 V
12 V
2 V
6 V
Answer: 10 V. (VA − 12)/4 + VA/4 = 2 gives 2·VA − 12 = 8, so VA = 10 V. Forgetting the source gives 6 V; treating it as leaving A gives 2 V.
7How exam questions look
Easy: one unknown node A, with a voltage source and either a current source or a second voltage source. One KCL equation is enough.
Medium: two unknown nodes A and B, with a current source at B. Write two equations and solve them together.
Hard: two unknown nodes and three sources: two voltage sources and a current source. The method is the same, with more terms.
The question asks for a node voltage, or for the current through one resistor in the direction of the arrow.
For a branch current, first find the node voltages. Then use I = (Vstart − Vend)/R along the arrow.
Only node A is unknown. Write one KCL equation: the sum of the branch currents leaving A is zero.
Each branch current is (VA − Vk)/Rk. Collect the VA terms on one side.
Σ1/R = 0.9583 S.
Show the answer and the step-by-step solution
Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.
KCL at node A
Take the bottom wire as ground. Each branch carries (VA − Vk)/Rk away from A, where Vk is the voltage of its source (zero for a plain resistor). The currents leaving A add up to zero.
(VA − 20)/8 + (VA − 18)/3 + (VA − 8)/2 = 0
Solve for VA
VA = (20/8 + 18/3 + 8/2) / (1/8 + 1/3 + 1/2) = 12.5/0.9583 = 13.04 V
Answer
VA = 13.04 V
Example 2Medium
The voltage source Vₛ between A and B has its + terminal at B, so it makes VB − VA = Vₛ. Use nodal analysis with a supernode to find the current Ix through the voltage source, from A to B as shown by the arrow.
Is = 1.5 AVs = 3 VR1 = 6 ΩR2 = 3 Ω
Find Ix(A)
Show hints
A voltage source between two non-ground nodes fixes their difference: VB − VA = Vs.
Write one KCL equation for the supernode around A, B and the source: VA/R₁ + VB/R₂ = Is.
VA = 1 V.
Show the answer and the step-by-step solution
Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.
Form a supernode
The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.
VB = VA + Vs = VA + 3
VA/R1 + VB/R2 = Is
Solve
VA/6 + (VA + 3)/3 = 1.5
VA = (1.5 − 3/3)/(1/6 + 1/3) = 1 V
VB = 1 + 3 = 4 V
Current through the voltage source
Everything that passes through the source from A to B leaves node B through R₂.
Ix = VB/R2 = 4/3 = 1.333 A
Answer
Ix = 1.333 A
A negative result means the real current flows against the arrow.
Example 3Hard
Use nodal analysis to find the current IR4 through R4 in the direction of the arrow.
Take the bottom wire as ground (0 V). Nodes tied straight to a voltage source are already known, so the unknowns are VA and VB.
Write KCL at A: for every resistor, (VA − Vother end)/R, and set their sum equal to any source current flowing into A.
Solving gives VA = 2.455 V. Then IR4 = (voltage at the tail of the arrow − voltage at the head)/R4.
Show the answer and the step-by-step solution
Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.
Set up the nodes
Ground is the bottom wire. Known node voltages come straight from the sources (V1 = 6 V, V2 = 9 V). Unknowns: VA, VB.
KCL at node A
Currents leaving A through the resistors equal the current pushed into A by sources.
VA/R3 + (VA − V1)/R1 + (VA − VB)/R2 = 0
VA/30 + (VA − 6)/20 + (VA − VB)/20 = 0
0.1333·VA − 0.05·VB = 0.3
KCL at node B
Currents leaving B through the resistors equal the current pushed into B by sources.
VB/R5 + (VB − VA)/R2 + (VB − V2)/R4 = Is
VB/30 + (VB − VA)/20 + (VB − 9)/20 = (−0.5)
−0.05·VA + 0.1333·VB = −0.05
Solve
Δ = (0.1333)(0.1333) − (−0.05)(−0.05) = 0.01528
VA = Δ1/Δ = 0.0375/0.01528 = 2.455 V
VB = Δ2/Δ = 0.008333/0.01528 = 0.5455 V
Answer
IR4 = (VB − V2)/R4 = (0.5455 − 9)/20 = −0.4227 A
Questions and answers
What is Nodal analysis?
Nodal analysis is a standard method to find every voltage in a circuit. It works even when series and parallel rules are not enough.
A node is a point where two or more parts of a circuit meet. A whole stretch of plain wire counts as one node, because a wire has no voltage drop along it.
How do I solve Nodal analysis problems step by step?
Choose a reference node (ground) and label the other node voltages.
Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
Express every branch current as (Va − Vb)/R.
Solve the system of equations, then find any current you need.
What are common mistakes in Nodal analysis?
Mixing up the sign of a branch current between the two node equations.
Writing an equation for the reference node.
Forgetting a current source that enters or leaves a node.
What do the Nodal analysis exercises cover?
Easy: 1 unknown node
Medium: 2 unknown nodes
Hard: 2 nodes, 3 sources
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.