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Kirchhoff's laws take one page in the textbook and cost students marks for years. The rules are simple. What goes wrong is the bookkeeping: a battery counted with the wrong sign, a resistor drop added instead of subtracted, a current direction "corrected" halfway through. This post gives you one sign convention, a fully worked two-loop circuit and the checks that catch errors before the examiner does.
What do Kirchhoff's two laws actually say?
The NCERT Class 12 chapter on Current Electricity states them this way. Junction rule: at any junction, the sum of currents entering equals the sum of currents leaving. Loop rule: the algebraic sum of changes in potential around any closed loop is zero. The first is conservation of charge. The second holds because potential depends only on position, so if you walk round a loop and return to the start, the net change must be zero.
Both rules are in the JEE (Main) 2026 syllabus under Current Electricity ("Kirchhoff's laws and their applications, Wheatstone bridge, Metre Bridge"), and the JEE (Advanced) 2026 information brochure lists "Kirchhoff's laws and simple applications". Check the syllabus for your own exam year.
Why do students get the signs wrong?
Part of the problem is a weak picture of what a circuit does. In their study of a 29-question diagnostic test on DC circuits, Engelhardt and Beichner (2004) found that high school and university students often held several misconceptions even after instruction. The idea used most often in interviews was that a battery is a constant source of current, and students tended to confuse current with voltage and resistance.
If you think a battery pushes out a fixed current, you will be surprised when a loop equation gives a negative current through it. That surprise leads to the most common mistake of all: changing the assumed direction mid-solution.
One sign convention to use every time
Pick a direction to walk around each loop. Then, for every element you cross:
Battery, from − to +: potential rises, write +E. From + to −, write −E. This depends only on the direction you walk, not on which way current flows.
Resistor, walking with your assumed current: potential drops, write −IR. Walking against it, write +IR.
Guess every current direction at the start, mark it on the diagram with an arrow and never change it. If a current comes out negative, it flows opposite to your arrow. That is information, not an error.
Worked example: a two-loop circuit
Three branches join a top node A to a bottom node B.
Left branch: a 12 V battery (positive terminal towards A) in series with 2 Ω.
Middle branch: a 4 Ω resistor.
Right branch: a 2 V battery (positive terminal towards A) in series with 2 Ω.
Find the current in each branch.
Step 1: Assume directions. Let I₁ flow up the left branch, I₂ up the right branch. By the junction rule at A, the current down the middle branch is I₁ + I₂.
Step 2: Left loop (up the left branch, down the middle). Cross the 12 V battery from − to +: +12. Cross 2 Ω with I₁: −2I₁. Cross 4 Ω with the middle current: −4(I₁ + I₂).
12 − 2I₁ − 4(I₁ + I₂) = 0, which gives 3I₁ + 2I₂ = 6.
Step 3: Right loop (up the right branch, down the middle).
2 − 2I₂ − 4(I₁ + I₂) = 0, which gives 2I₁ + 3I₂ = 1.
Step 4: Solve. Multiply the first equation by 3 and the second by 2: 9I₁ + 6I₂ = 18 and 4I₁ + 6I₂ = 2. Subtracting, 5I₁ = 16, so I₁ = 3.2 A. Then 2I₂ = 6 − 9.6, so I₂ = −1.8 A. The middle current is 3.2 − 1.8 = 1.4 A.
Step 5: Read the signs. I₂ is negative, so 1.8 A flows down the right branch, into the positive terminal of the 2 V battery. That battery is being charged.
Step 6: Check. Work out the potential of A above B along each branch. Left: 12 − 2(3.2) = 5.6 V. Middle: 4(1.4) = 5.6 V. Right: 2 − 2(−1.8) = 5.6 V. All three agree. As a second check, the 12 V battery supplies 12 × 3.2 = 38.4 W, the 2 V battery absorbs 2 × 1.8 = 3.6 W, and the resistors dissipate 20.48 + 6.48 + 7.84 = 34.8 W. 38.4 − 3.6 = 34.8 W, so energy balances.
Which other mistakes cost marks?
Writing the battery sign from the current direction. The sign of E depends on the direction you cross it, nothing else.
Too many unknowns. Use the junction rule first, as in Step 1, so you need fewer loop equations.
Choosing a loop that repeats information. With two independent loops, a third loop equation adds nothing new. The outer loop here is a useful check, not a third equation.
Forgetting internal resistance. If a cell has internal resistance r, it sits in series with the cell and gets its own −Ir term.
Skipping the check. The "same potential along every branch" test takes thirty seconds. Our post on neat working and fewer careless errors explains why a tidy layout makes checks like this quicker.
If you keep setting up the right equations but get the wrong answer, compare your working line by line with a full solution. The Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see exactly which sign went astray.
A weekly practice routine
Day 1: Redraw three circuits from your book and mark assumed currents and loop directions before writing any equation. Our guide on drawing the diagram first shows why this step matters.
Day 2: Solve two two-loop circuits and do both checks (branch potential and power balance).
Day 3: One Wheatstone bridge and one circuit with internal resistance.
Day 4: A mixed set of five circuit questions, timed. Log every error as sign, set-up or algebra.
Tonight, take one circuit you got wrong and solve it again with the six steps above, ending with the potential check.
