5.2 Kirchhoff's Voltage & Current Laws

5.2 Kirchhoff's Voltage & Current Laws

Quick Answer: KCL — at any node, ΣI = 0 (what enters equals what leaves). KVL — around any closed loop, ΣV = 0 (rises cancel drops). Combine both with V = IR on each resistor to solve branched and multi-loop DC circuits.

After Ohm’s law, Kirchhoff’s laws are the second pillar of topic 3.6 DC Circuits. Real aircraft and shop circuits are rarely a single resistor on an ideal battery. Currents split at junctions; voltages share around loops. Gustav Kirchhoff’s two conservation statements turn those sketches into solvable equations.

Kirchhoff’s Current Law (KCL)

Statement: The algebraic sum of currents meeting at a node is zero:

Σ I = 0

Equivalently: total current entering a node = total current leaving the node. Charge does not pile up at an ordinary circuit junction in steady DC.

Sign convention (pick one and stick to it)

Common classroom choice: currents into the node are positive; currents out are negative (or the reverse). For maintenance-exam speed, many candidates simply write:

I₁ = I₂ + I₃

when I₁ enters and I₂, I₃ leave. Both styles are fine if consistent.

Worked KCL Example — Parallel Branch Split

A 28 V source feeds two parallel resistors: R₁ = 14 Ω and R₂ = 28 Ω. Find each branch current and the total source current.

1. Voltage across each parallel branch = 28 V.
2. I₁ = 28/14 = 2 A.
3. I₂ = 28/28 = 1 A.
4. At the positive node, KCL: I_total = I₁ + I₂ = 3 A.

Check: equivalent parallel resistance R_eq = (14 × 28)/(14 + 28) = 392/42 ≈ 9.33 Ω, then I_total = 28/9.33 ≈ 3 A. KCL and the parallel formula agree.

Worked KCL Example — Missing Branch Current

Three wires meet at node N. Measured: I_A = 4.5 A into N, I_B = 1.2 A out of N. Find I_C.

Using “in = out”: 4.5 = 1.2 + I_C ⇒ I_C = 3.3 A out of N.

If a stem says I_C is also into N, then 4.5 + I_C = 1.2 would force a negative value—meaning your assumed direction for I_C was wrong. Flip the arrow and take the absolute value as the physical current.

Kirchhoff’s Voltage Law (KVL)

Statement: The algebraic sum of potential differences around any closed loop is zero:

Σ V = 0

Energy gained through sources equals energy lost through drops in a conservative electrostatic field around a closed path (steady DC idealisation used in Module 3).

Walking a loop

1. Choose a start node and a travel direction (clockwise or counterclockwise).
2. When you traverse a source from − to +, count a voltage rise (+E).
3. When you traverse a resistor in the direction of assumed current, count a drop (−IR).
4. Sum to zero; solve.

Worked KVL Example — Single Loop

Series loop: battery E = 24 V, resistors R₁ = 5 Ω, R₂ = 7 Ω. Find current and each drop.

1. KVL: E − I R₁ − I R₂ = 0 ⇒ 24 − I(5 + 7) = 0.
2. I = 24/12 = 2 A.
3. V₁ = 2 × 5 = 10 V, V₂ = 2 × 7 = 14 V.
4. Check: 10 + 14 = 24 V.

Multi-Loop Worked Example

Consider two loops sharing a middle resistor (a classic Module 3 topology):

• Left loop source E₁ = 12 V in series with R₁ = 4 Ω.
• Right loop source E₂ = 6 V (opposing sense relative to the shared branch as drawn below) in series with R₃ = 2 Ω.
• Shared middle resistor R₂ = 4 Ω.

Define mesh currents:

• I₁ clockwise in the left loop.
• I₂ clockwise in the right loop.

On R₂, the net current is (I₁ − I₂) if both meshes are clockwise.

Left-loop KVL (clockwise from bottom-left):

12 − 4 I₁ − 4(I₁ − I₂) = 0

Simplify: 12 − 4 I₁ − 4 I₁ + 4 I₂ = 0 ⇒ 12 − 8 I₁ + 4 I₂ = 0

Divide by 4: 3 − 2 I₁ + I₂ = 0 ⇒ I₂ = 2 I₁ − 3 … (1)

Right-loop KVL (clockwise): treat E₂ = 6 V as a drop when traversing against its rise in this orientation (common textbook opposing-source layout):

−6 − 2 I₂ − 4(I₂ − I₁) = 0

Simplify: −6 − 2 I₂ − 4 I₂ + 4 I₁ = 0 ⇒ −6 + 4 I₁ − 6 I₂ = 0

Divide by 2: −3 + 2 I₁ − 3 I₂ = 0 ⇒ 2 I₁ − 3 I₂ = 3 … (2)

Substitute (1) into (2):

2 I₁ − 3(2 I₁ − 3) = 3 ⇒ 2 I₁ − 6 I₁ + 9 = 3 ⇒ −4 I₁ = −6 ⇒ I₁ = 1.5 A

Then I₂ = 2(1.5) − 3 = 0 A.

Shared branch current I₁ − I₂ = 1.5 A (left to right if that was the I₁ sense through R₂).

Element checks with Ohm’s law:

• Drop on R₁: 1.5 × 4 = 6 V.
• Drop on R₂: 1.5 × 4 = 6 V.
• Left loop: 12 − 6 − 6 = 0 ✓.
• Right loop with I₂ = 0: no drop on R₃; shared drop 6 V balances the 6 V source ✓.

This example shows why Module 3 loves multi-loop items: one mesh current can be zero while energy still circulates in the other loop, and KCL at the top node of R₂ still holds (1.5 A in, 1.5 A out along the shared path with zero right-mesh contribution).

Compact Second Multi-Loop (Numbers Only)

Series–parallel hybrid often reduced by KCL first:

Supply E = 30 V, series R_s = 5 Ω, then a parallel pair R_a = 10 Ω and R_b = 15 Ω to return.

1. Parallel equivalent: R_p = (10 × 15)/(10 + 15) = 6 Ω.
2. Total R = 5 + 6 = 11 Ω.
3. Supply current I_s = 30/11 ≈ 2.727 A (KCL total into the parallel node).
4. Voltage across parallel node: V_p = I_s × 6 ≈ 16.36 V.
5. I_a = 16.36/10 ≈ 1.636 A, I_b = 16.36/15 ≈ 1.091 A.
6. KCL check: 1.636 + 1.091 ≈ 2.727 A.

You used Ohm for every resistor, KVL implicitly (30 = drop on R_s + V_p), and KCL at the parallel junction.

Exam Habit Table

Common Sign and Topology Errors

• Writing ΣV = E + IR + IR (adding drops with the same sign as the source) instead of E − IR − IR = 0.
• Applying supply voltage to a resistor that is not directly across the supply.
• Forgetting that parallel branches share voltage while series elements share current.
• Changing the positive-current direction mid-solution without rewriting equations.

Draw the circuit, label I and V, write KCL/KVL, substitute Ohm, then solve. That sequence is what CAAS Level 2 application expects on topic 3.6.