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Loop equations work, but with several loops they multiply quickly, and one sign slip spoils them all. The node potential method names the voltage at each junction instead of the currents. Every current then follows from Ohm's law, and the junction rule gives one equation per unknown voltage. This post sets out the method and works two circuits with checks.
Where does this sit in the syllabus?
The NCERT Class 12 chapter Current Electricity states Kirchhoff's rules in section 3.12. The junction rule says that at any junction the sum of currents entering equals the sum leaving; the loop rule says the changes in potential around any closed loop add to zero. The node method is not a new law: it is the junction rule written with potentials, with the loop rule built in because every point has one potential.
The Syllabus for JEE (Main) 2026 lists "Kirchhoff's laws and their applications, Wheatstone bridge, Metre Bridge" under Unit 12, and the NEET (UG) 2026 syllabus has the same line. Check the syllabus for your own year.
What are the steps?
Pick a reference node. Usually the negative terminal of the main battery. Call it 0 V.
Label the other nodes. Where an ideal battery joins two nodes, their potentials differ by its emf, so one label covers both.
Write one equation per unknown node: the currents leaving it add to zero. Through a resistor R to a node at potential V_B, the current leaving node X is (V_X − V_B)/R.
For a battery in a branch, subtract its emf. If a branch from X to B holds a battery of emf E with its positive terminal facing X and a resistor R, the current leaving X is (V_X − V_B − E)/R.
Solve, then find every current, and check.
Worked example 1: an unbalanced Wheatstone bridge
A 10 V battery with no internal resistance connects node A (positive) to node G (negative, 0 V). From A, a 2 Ω resistor goes to node C and a 4 Ω resistor to node D. From C, a 4 Ω resistor goes to G; from D, a 2 Ω resistor goes to G. A 2 Ω resistor joins C and D.
The battery fixes V_A = 10 V, so only V_C and V_D are unknown.
Node C: (V_C − 10)/2 + V_C/4 + (V_C − V_D)/2 = 0. Multiply by 4: 5V_C − 2V_D = 20.
Node D: (V_D − 10)/4 + V_D/2 + (V_D − V_C)/2 = 0. Multiply by 4: −2V_C + 5V_D = 10.
Solve: multiply the first by 5 and the second by 2 and add, giving 21V_C = 120. So V_C = 40/7 ≈ 5.71 V, and V_D = (10 + 2V_C)/5 = 30/7 ≈ 4.29 V.
Now every current:
A to C: (10 − 40/7)/2 = 15/7 ≈ 2.14 A.
A to D: (10 − 30/7)/4 = 10/7 ≈ 1.43 A.
C to D: (40/7 − 30/7)/2 = 5/7 ≈ 0.71 A.
C to G: (40/7)/4 = 10/7 ≈ 1.43 A.
D to G: (30/7)/2 = 15/7 ≈ 2.14 A.
Battery current: 15/7 + 10/7 = 25/7 ≈ 3.57 A, so the equivalent resistance is 10 ÷ (25/7) = 2.8 Ω.
Checks: at C, 15/7 in equals 5/7 + 10/7 out. At D, 10/7 + 5/7 in equals 15/7 out. The battery supplies 10 × 25/7 ≈ 35.7 W, and the five resistors dissipate the same total. The bridge is unbalanced because 2/4 is not equal to 4/2, which is why current flows from C to D. A loop-current solution of the same bridge needs three unknowns.
Worked example 2: three branches between two nodes
Three branches join node P to node G (0 V). Branch 1: a 12 V battery, positive towards P, in series with 3 Ω. Branch 2: a 6 V battery, positive towards P, in series with 6 Ω. Branch 3: a 2 Ω resistor.
One unknown, V_P. Currents leaving P add to zero:
(V_P − 12)/3 + (V_P − 6)/6 + V_P/2 = 0
Multiply by 6: 2V_P − 24 + V_P − 6 + 3V_P = 0, so V_P = 5 V.
Branch 1 carries (12 − 5)/3 = 7/3 ≈ 2.33 A into P.
Branch 2 carries (6 − 5)/6 = 1/6 ≈ 0.17 A into P.
Branch 3 carries 5/2 = 2.5 A from P to G.
Check: 7/3 + 1/6 = 15/6 = 2.5 A.
The pattern generalises: V_P = Σ(E/R) ÷ Σ(1/R). If V_P had come out above 6 V, the 6 V battery would be charging, and the sign of its current would tell you.
Why does thinking in voltages help?
Smith and van Kampen (2011) studied preservice science teachers working on circuits with several batteries. Most could not explain the effect of adding batteries, because they reasoned with current and resistance where reasoning about voltage was needed, and many treated a battery as a source of constant current. Labelling every node with a potential forces the voltage reasoning those circuits need. Our post on Kirchhoff's laws and their common errors covers the loop method if your teacher expects that instead.
Which mistakes cost marks?
Mixing "leaving" and "entering" in one equation. Write every term as leaving.
Battery signs. Decide which terminal faces the node before writing the term.
Forgetting internal resistance. Treat it as a resistor in series with the emf.
A poor reference node. Choose the node with the most connections or the battery's negative terminal.
If you solve the equations but cannot tell whether the answer makes sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, which you can compare with your own checks.
A practice routine
Day 1: Redo example 1 with the C–D resistor changed to 4 Ω and confirm the currents still balance at each node.
Day 2: Two circuits with batteries in more than one branch.
Day 3: One metre-bridge or balanced-bridge problem, where the node method shows V_C = V_D at once.
Day 4: A mixed set of four, each finished with a power check.
Today, replace the 2 Ω in example 2 with 4 Ω and show that V_P becomes 20/3 ≈ 6.67 V, so the 6 V battery now charges.
