ΣI = 0 at junction. ΣV = 0 around loop. Worked circuit example.

Kirchhoff's Laws — Junction and Loop Rules

Question

A circuit has two batteries and three resistors arranged as shown. Using Kirchhoff’s Laws, find the current through each branch:

Battery $E_1 = 10\text{ V}$ , internal resistance $r_1 = 1\text{ }\Omega$ in branch AB
Battery $E_2 = 6\text{ V}$ , internal resistance $r_2 = 2\text{ }\Omega$ in branch BC
External resistor $R = 4\text{ }\Omega$ in branch AC
Junction A connects all three branches

Find currents $I_1$ , $I_2$ , and $I_3$ .

Solution — Step by Step

Label the three branch currents $I_1$ , $I_2$ , $I_3$ with assumed directions — say, $I_1$ flows from A to B, $I_2$ from B to C, and $I_3$ from A to C. If we get a negative answer, the actual direction is opposite. This is perfectly fine and expected.

At any junction, charge can’t pile up — whatever flows in must flow out. This is Kirchhoff’s Current Law (KCL):

\sum I = 0 \quad \text{at a junction}

At junction A, currents $I_1$ and $I_3$ leave, while $I_2$ arrives (based on our assumed directions):

I_2 = I_1 + I_3

This gives us Equation 1.

We go around the loop in one fixed direction — clockwise here. The sign convention: EMF is positive if we traverse the battery from − to +; a resistor drop is negative if current flows in our traversal direction.

Loop 1 (through $E_1$ , $r_1$ , $r_2$ , $E_2$ ):

E_1 - I_1 r_1 - I_2 r_2 - E_2 = 0

10 - I_1(1) - I_2(2) - 6 = 0

I_1 + 2I_2 = 4 \quad \text{...(Equation 2)}

Loop 2 (through $R$ , $r_2$ , $E_2$ , going counterclockwise from A):

E_2 - I_2 r_2 + I_3 R = 0

Wait — we traverse $R$ with current $I_3$ , so the drop is $-I_3 R$ if we go against $I_3$ ‘s direction. Let’s be careful and traverse clockwise through the right loop (A → B → C → A via R):

-E_2 + I_2 r_2 - I_3 R = 0

-6 + 2I_2 - 4I_3 = 0

2I_2 - 4I_3 = 6 \quad \text{...(Equation 3)}

From Equation 1: $I_2 = I_1 + I_3$

Substitute into Equation 2:

I_1 + 2(I_1 + I_3) = 4 \implies 3I_1 + 2I_3 = 4 \quad \text{...(4)}

From Equation 3: $2I_2 - 4I_3 = 6 \implies I_2 = \frac{6 + 4I_3}{2} = 3 + 2I_3$

Substitute back: $I_1 = I_2 - I_3 = 3 + 2I_3 - I_3 = 3 + I_3$

Plug into (4): $3(3 + I_3) + 2I_3 = 4 \implies 9 + 5I_3 = 4 \implies I_3 = -1\text{ A}$

So: $I_1 = 3 + (-1) = 2\text{ A}$ , and $I_2 = 3 + 2(-1) = 1\text{ A}$

$I_1 = 2\text{ A}$ , $I_2 = 1\text{ A}$ , $I_3 = -1\text{ A}$

The negative sign on $I_3$ means current actually flows from C to A through $R$ , opposite to our initial guess.

Why This Works

KCL comes from conservation of charge — no charge accumulates at a junction in steady state. KVL comes from conservation of energy — the total work done on a charge going around any closed loop must be zero, because it returns to the same potential.

These two laws together give us exactly enough equations to solve any circuit. For $n$ junctions, KCL gives $(n-1)$ independent equations. For $m$ independent loops (mesh loops), KVL gives $m$ equations. Together they cover all unknowns.

The beauty is that it doesn’t matter which direction you assume for currents or which direction you traverse a loop. The algebra self-corrects — negative answers just flip the physical direction.

Alternative Method — Using Mesh Analysis

Instead of branch currents, assign mesh currents $I_a$ (clockwise in left loop) and $I_b$ (clockwise in right loop). Branch currents become combinations: the shared branch carries $I_a - I_b$ .

For left mesh: $10 - I_a(1) - (I_a - I_b)(0) - 6 - I_a(2) = ...$

Mesh analysis is faster for larger circuits because you automatically satisfy KCL — you only apply KVL. In JEE problems with 3+ loops, this saves significant time.

In JEE Main, circuit problems with two batteries almost always require exactly 2 loop equations + 1 junction equation. If you’re writing more than 3 equations for a 3-branch circuit, you’re overcomplicating it.

Common Mistake

Wrong sign convention for EMF in KVL. Students often add EMF regardless of traversal direction. The rule: if you walk through a battery from negative terminal to positive terminal, the EMF is +E. If you walk from positive to negative (against the current the battery drives), it’s −E. Getting this backwards flips the sign of your equation and gives completely wrong answers. Always draw the +/− terminals on the battery before writing the loop equation.

A second common slip: forgetting the internal resistance. In board exams and JEE, internal resistance is almost always included in multi-battery problems. If you treat $E_1 = 10\text{ V}$ as an ideal source and ignore $r_1 = 1\text{ }\Omega$ , your loop equation will be off by $I_1$ volts — and your answer won’t match any option.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Mesh Analysis

Common Mistake

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