Kirchhoff laws — solve a circuit with two loops

Question

In the circuit shown, $E_1 = 12$ V (internal resistance $r_1 = 1\,\Omega$ ), $E_2 = 6$ V (internal resistance $r_2 = 2\,\Omega$ ), $R_1 = 4\,\Omega$ , $R_2 = 3\,\Omega$ . Find the currents $I_1$ , $I_2$ , and $I_3$ using Kirchhoff’s laws.

The circuit has two loops: Loop 1 (left) contains $E_1$ , $r_1$ , and $R_1$ ; Loop 2 (right) contains $E_2$ , $r_2$ , and $R_2$ . The two loops share a common branch with current $I_3$ .

    A----[R1=4Ω]----B----[R2=3Ω]----C
    |               |               |
  [E1,r1]        [I3]             [E2,r2]
   12V,1Ω          |               6V,2Ω
    |               |               |
    D---------------E---------------F

Let $I_1$ flow in the left loop (clockwise), $I_2$ in the right loop (clockwise), and $I_3 = I_1 - I_2$ in the middle branch.

Solution — Step by Step

KCL (Junction Rule): The sum of currents entering a junction equals the sum leaving it. Based on conservation of charge.

KVL (Loop Rule): The algebraic sum of EMFs equals the algebraic sum of potential drops ( $IR$ ) around any closed loop. Based on conservation of energy.

At junction B (or E), currents $I_1$ , $I_2$ , and $I_3$ meet:

I_1 = I_2 + I_3 \implies I_3 = I_1 - I_2

This is our KCL equation.

Going clockwise around the left loop: A → R₁ → B → E → D → A

Starting from A, moving clockwise:

Across $R_1 = 4\,\Omega$ (drop in direction of $I_1$ ): $-4I_1$
Across $I_3$ through middle branch: This is a shared branch. Going from B to E with current $I_3 = I_1 - I_2$ downward: $-R_3 \cdot I_3$ — but wait, let’s simplify with a concrete topology.

For a standard two-loop problem without a middle resistor, let’s assume the common branch has negligible resistance (wire). Then:

Loop 1 (left): EMF $E_1$ drives current through $r_1$ and $R_1$ :

E_1 - I_1 r_1 - I_1 R_1 = 0

Wait — the loops share a common wire, so current $I_3$ flows through the common branch. Let’s be explicit.

Corrected setup with the two loops sharing only a wire (junction):

Loop 1 (clockwise, left): $E_1 = I_1(r_1 + R_1)$

12 = I_1(1 + 4) = 5I_1 \implies I_1 = 2.4 \text{ A}

Loop 2 (clockwise, right): $E_2 = I_2(r_2 + R_2)$

6 = I_2(2 + 3) = 5I_2 \implies I_2 = 1.2 \text{ A}

By KCL: $I_3 = I_1 - I_2 = 2.4 - 1.2 = 1.2$ A

These are the mesh currents — in a real shared-branch circuit, we proceed with KVL for each loop including the shared branch current.

For a standard circuit where a resistor $R_m$ is shared between two loops:

Let mesh current $I_1$ (loop 1, clockwise), mesh current $I_2$ (loop 2, clockwise).

Loop 1 (applying KVL, signs for each element):

E_1 - I_1 r_1 - (I_1 - I_2)R_m - I_1 R_1 = 0

Loop 2:

-E_2 - I_2 r_2 - (I_2 - I_1)R_m - I_2 R_2 = 0

Here the signs depend on relative directions of EMF and current in each loop. Rearrange and solve simultaneously.

With numbers (say $R_m = 2\,\Omega$ , other values as before):

Loop 1: $12 - I_1(1) - (I_1 - I_2)(2) - I_1(4) = 0$ $12 = I_1(1 + 2 + 4) - 2I_2 = 7I_1 - 2I_2$ …(1)

Loop 2: $-6 - I_2(2) - (I_2 - I_1)(2) - I_2(3) = 0$ $-6 = -2I_1 + I_2(2 + 2 + 3) = -2I_1 + 7I_2$ …(2)

From (2): $2I_1 - 7I_2 = 6$ → $I_1 = (6 + 7I_2)/2$

Substitute in (1): $12 = 7 \cdot \frac{6 + 7I_2}{2} - 2I_2 = \frac{42 + 49I_2}{2} - 2I_2$

$24 = 42 + 49I_2 - 4I_2 = 42 + 45I_2$

$45I_2 = -18 \implies I_2 = -0.4$ A (negative sign means actual direction is opposite to assumed clockwise)

$I_1 = (6 + 7(-0.4))/2 = (6 - 2.8)/2 = 3.2/2 = 1.6$ A

$I_3 = I_1 - I_2 = 1.6 - (-0.4) = 2.0$ A

Why This Works

KVL and KCL together provide enough independent equations to solve any resistive circuit. For a circuit with $n$ junctions and $b$ branches:

KCL gives $n - 1$ independent equations
KVL (mesh analysis) gives $b - n + 1$ independent equations
Total: $b$ equations for $b$ unknown branch currents

Mesh analysis (KVL for independent loops) is the most systematic approach. Always define mesh current directions consistently (all clockwise or all counterclockwise), then apply KVL carefully with sign conventions.

Alternative Method

Nodal analysis (KCL at each node) is an alternative to KVL. For circuits with many parallel branches, nodal analysis is often faster.

Common Mistake

The most frequent error: forgetting the sign of shared branch current. In loop 1, the shared branch carries current $(I_1 - I_2)$ in the direction of $I_1$ . In loop 2, the same branch carries current $(I_2 - I_1)$ in the direction of $I_2$ . Students often write the same sign for both loops, which gives wrong equations. When you go around loop 2 and encounter the shared branch, the current in it is opposing your loop direction — account for this with the correct sign.

If your final current comes out negative, don’t redo the problem — a negative value simply means the actual current direction is opposite to what you assumed. The magnitude is still correct. Just note “current flows in the opposite direction to assumed” in your answer.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next