Topic 6 — Kirchhoff's Rules

Part 1 · The idea, told simply

Series and parallel formulas run out. Put two batteries in different branches of a network and there is no way to "reduce" anything — the circuit simply is not a series-parallel arrangement. Kirchhoff's two rules handle every case.

Story track · the junction rule

Imagine a road junction where three roads meet. In one hour, 500 cars drive in along the first road and 300 along the second. How many drive out along the third?

800, obviously. Cars do not pile up at the junction and they do not vanish. Whatever goes in must come out.

Charge behaves exactly the same way. Charge is never created or destroyed, and in a steady circuit it never accumulates anywhere. So at any junction:

Σ I(entering) = Σ I(leaving) ← conservation of CHARGE

Story track · the loop rule

Now imagine walking around a hilly park and returning to the exact bench you started from. You went up some slopes and down others. Whatever your route, the total height change must be zero — because you are back where you began, and the bench has not moved.

Electric potential works the same way. Potential is a property of a place. Walk around any closed loop of a circuit and come back to your starting point, and the total change in potential must be zero:

Σ ΔV around any closed loop = 0 ← conservation of ENERGY

The battery is the uphill climb; the resistors are the downhill slides. Around a full loop they must exactly cancel.

Animation 1 · The junction rule

Set the incoming currents. The outgoing one has no choice in the matter.

I₁ = I₂ =

Part 2 · The sign convention — the whole game

The rules themselves are easy. Applying them costs marks because every element you cross contributes a term whose sign depends on which way you are walking. Here is the complete system.

The four cases — memorise these, there are no others

What you cross	Which way you walk	Term
Resistor R carrying current I	ALONG the current	− I R (a drop)
Resistor R carrying current I	AGAINST the current	+ I R (a rise)
Cell of emf ε	from − terminal to + terminal	+ ε (a rise)
Cell of emf ε	from + terminal to − terminal	− ε (a drop)

Notice what the cell rule does NOT depend on: the direction of the current. A cell's sign is decided purely by which terminal you enter from. A resistor's sign is decided by the current direction. Mixing these two up is the single most common error in this topic.

Animation 2 · The sign machine

Pick an element and a walking direction. The sign appears. Drill it until you stop having to think.

Walking a loop, one step at a time

The safest method is mechanical. Fix a direction for the loop — clockwise, say — and keep it for the whole loop. Then walk, writing down one term per element, and set the total to zero.

Animation 3 · The potential walk

Step around the loop and watch the running total of potential. It must return to exactly zero.

When the answer comes out negative

Story track

You have to guess the current directions before you can write the equations. What if you guess wrong?

Nothing bad happens. The algebra tells you. If a current comes out negative, it means the real current is the same size but flows opposite to the arrow you drew. You do not need to redo anything — just report the direction as reversed.

This is a genuine convenience and NCERT states it explicitly: "If I turns out to be negative, the current actually flows in a direction opposite to the arrow." Guess freely.

Animation 4 · A wrong guess, and what it costs you

A two-battery circuit where one branch current comes out negative. Watch what that actually means.

Part 3 · Working the method

Maths track · the procedure, in order

Label every branch current with a symbol and an arrow. Guess the directions; wrong guesses are self-correcting.
Use the junction rule immediately to reduce the number of unknowns. If a junction has currents I₁ and I₂ entering, label the third branch (I₁ + I₂) rather than introducing I₃. This is what NCERT means by "to reduce the number of unknowns at the outset".
Count what you need. With n unknowns you need n independent equations.
Pick loops and walk them, fixing one direction per loop and applying the four sign cases.
Solve, then sanity-check by substituting into a loop you did not use. If it balances, the answer is right.

How many equations are independent?

A common exam question. For a network with n junctions and b branches:

Independent junction equations = n − 1 Independent loop equations = b − (n − 1) Total independent equations = b (one per unknown branch current)

The last junction gives you nothing new — it is the sum of all the others. Similarly, extra loops give equations that are combinations of the ones you already have. NCERT notes this in Example 3.6: applying the loop rule to the remaining loops "does not provide any additional independent equation".

Maths track · NCERT Example 3.6 worked through

The three loop equations for that network are

7I₁ − 6I₂ − 2I₃ = 10 (loop ADCA) I₁ + 6I₂ + 2I₃ = 10 (loop ABCA) 2I₁ − 4I₂ − 4I₃ = −5 (loop BCDEB)

Solving the three simultaneously:

I₁ = 2.5 A I₂ = 5/8 A = 0.625 A I₃ = 15/8 A = 1.875 A

The check NCERT then performs is the one worth copying: take a loop you did not use — BADEB — and add up its potential changes. They come to zero, confirming the solution. In an exam this check costs thirty seconds and catches sign errors that are otherwise invisible.

Symmetry can save you the whole calculation

NCERT Example 3.5 (the cube of twelve resistors) makes the point honestly: because of the symmetry, "the great power of Kirchhoff's rules has not been very apparent". When branches are indistinguishable, equal currents can be written down by inspection and the equations shrink dramatically. Always look for symmetry before grinding through simultaneous equations.

Animation 5 · Counting the equations you actually need

Add branches and junctions and watch the bookkeeping. The last junction equation is always redundant.

junctions n = branches b =

Trap · the six that cost marks

Changing the traversal direction halfway around a loop. Fix it once, keep it.
Giving a cell a sign based on the current direction. A cell's sign depends only on which terminal you enter from.
Forgetting the internal resistance of a cell, which is a resistor like any other and follows the resistor rule.
Using the junction rule at every junction, including the last. Only n − 1 of them are independent.
Panicking at a negative answer. It simply means the arrow was drawn the wrong way.
Not checking the solution in an unused loop. Thirty seconds that catches most sign errors.

Part 4 · Formula sheet

The two rules

JUNCTION RULE (charge conservation) Σ I(in) = Σ I(out) or Σ I = 0 with signs Valid because charge does not accumulate in a steady circuit Bending or reorienting the wire changes nothing LOOP RULE (energy conservation) Σ ΔV = 0 around any closed loop Valid because potential is a property of position

Sign convention — the four cases

Resistor, walking ALONG the current → − I R Resistor, walking AGAINST the current → + I R Cell, walking from − to + → + ε Cell, walking from + to − → − ε Cell sign depends ONLY on the terminals, never on the current Internal resistance r follows the RESISTOR rule

Terminal voltage inside a loop

Crossing a real cell from N to P (with current I from N to P inside): V = ε − I r Crossing the same cell from P to N: V = ε + I r

Counting equations

n junctions, b branches: independent junction equations = n − 1 independent loop equations = b − n + 1 total = b unknown branch currents Extra equations are always combinations of these

Standard results reached with these rules

Balanced Wheatstone bridge: P/Q = R/S ⇒ I_g = 0 Cube of 12 equal R, body diagonal: 5R/6 (symmetry + loop rule) Two cells in series: ε_eq = ε₁ ± ε₂, r_eq = r₁ + r₂ Two cells in parallel: ε_eq/r_eq = ε₁/r₁ + ε₂/r₂

Method and sanity checks

1. Label all branch currents with arrows 2. Apply the junction rule FIRST to cut unknowns 3. Fix one traversal direction per loop and keep it 4. Solve, then verify in a loop you did not use Negative current → real direction is opposite the arrow A negative answer is never an error by itself

Part 5 · 50 NEET-pattern questions with full solutions

Includes 4 graph-based questions, 3 assertion–reason questions, and 10 questions tagged Sign — deliberately heavy, because sign errors account for most lost marks here. Year labels are not attached: the patterns are authentic, the wording is mine.