Kirchhoff's Laws and Circuit Analysis

Kirchhoff's two laws are the foundation of all circuit analysis. They are based on two fundamental conservation laws — conservation of charge and conservation of energy. Together, they allow us to solve any circuit, no matter how complex.

Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.7 — Kirchhoff's laws and circuit analysis, including the statement and conservation-law foundations of the current law (KCL) and voltage law (KVL), their application to single- and multi-loop circuits, and the systematic procedure for solving simultaneous equations that arise from non-trivial networks. The Wheatstone bridge as a special-case application is also implicit in the spec coverage. (Refer to the official AQA specification document for exact wording.)

Synoptic links:

• 3.5.1.5 — Series and parallel circuits: the series voltage-sum and parallel current-sum rules are direct consequences of KVL and KCL respectively. Kirchhoff's laws generalise to any topology — series and parallel are the two simplest cases.
• 3.5.1.6 — EMF and internal resistance: the equation ε = I(R + r) is a one-loop application of KVL with one EMF source and two resistive elements. The general framework here subsumes the special case.
• 3.4 — Conservation laws in mechanics: KCL is conservation of charge applied at a junction; KVL is conservation of energy applied around a loop. The structural parallel with momentum and energy conservation in mechanics is exact — these are the same conservation principles in different physical contexts.

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Kirchhoff's First Law (KCL — Current Law)

Statement: The sum of the currents entering any junction equals the sum of the currents leaving that junction.

Equivalently: the algebraic sum of currents at any junction is zero.

$\sum I_{in} = \sum I_{out}$∑Iin​=∑Iout​

or

$\sum I = 0 \text{ (using sign convention: positive for currents entering, negative for leaving)}$∑I=0 (using sign convention: positive for currents entering, negative for leaving)

Physical basis: Conservation of charge. Charge cannot accumulate at a junction — what flows in must flow out.

Worked Example 1 — KCL at a Junction

At a junction, three wires carry currents of 3.0 A, 5.0 A, and 2.0 A. The first two are entering the junction. What is the current in the third wire, and in which direction?

Solution:

Sum entering = 3.0 + 5.0 = 8.0 A

Current leaving through wire 3 = 8.0 − 2.0 = 6.0 A (leaving)

Wait — let me re-read. If the third wire carries 2.0 A, and the first two carry 3.0 A and 5.0 A entering, then:

If the third wire is leaving: we need I₃ such that 3.0 + 5.0 = I₃

Actually, there must be more wires. Let me restate:

At a junction, four wires meet. I₁ = 3.0 A (entering), I₂ = 5.0 A (entering), I₃ = 2.0 A (leaving). Find I₄.

Currents in = Currents out: 3.0 + 5.0 = 2.0 + I₄

I₄ = 8.0 − 2.0 = 6.0 A (leaving the junction)

Worked Example 2 — Multiple Junctions

In a circuit, 2.0 A flows from A to B through a 10 Ω resistor. At junction B, the current splits: 1.2 A goes through a 15 Ω resistor. What current flows through the other branch?

Solution:

By KCL at junction B: 2.0 = 1.2 + I₂

I₂ = 2.0 − 1.2 = 0.8 A

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Kirchhoff's Second Law (KVL — Voltage Law)

Statement: The sum of the EMFs around any closed loop in a circuit equals the sum of the potential differences (IR drops) around that loop.

$\sum \varepsilon = \sum IR$∑ε=∑IR

Or equivalently: the algebraic sum of all voltages (EMFs and p.d.s) around any closed loop is zero.

Physical basis: Conservation of energy. The total energy supplied per coulomb (EMFs) equals the total energy dissipated per coulomb (p.d.s) around any closed path.

Sign Convention for KVL

When traversing a loop in a chosen direction:

• EMF is positive if you pass through the source from − to + (in the direction of the EMF)
• EMF is negative if you pass from + to − (against the EMF)
• IR drop is positive if you traverse in the direction of current flow
• IR drop is negative if you traverse against the current flow

Worked Example 3 — Single Loop

A circuit contains a 12 V battery with internal resistance 0.5 Ω, connected to two resistors in series: R₁ = 4.5 Ω and R₂ = 7.0 Ω. Find the current.

Solution:

Applying KVL around the loop:

ε = IR₁ + IR₂ + Ir

12 = I(4.5 + 7.0 + 0.5) = I × 12.0

I = 12 / 12.0 = 1.0 A

p.d. across R₁ = 1.0 × 4.5 = 4.5 V p.d. across R₂ = 1.0 × 7.0 = 7.0 V p.d. across r = 1.0 × 0.5 = 0.5 V

Check: 4.5 + 7.0 + 0.5 = 12.0 V = ε ✓

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Solving Multi-Loop Circuits

For circuits with multiple loops and/or multiple sources, use the following systematic approach:

Step-by-Step Method

1. Label all currents with assumed directions (if you guess wrong, the answer will be negative — that's fine)
2. Apply KCL at each junction to relate the currents
3. Apply KVL around enough independent loops to generate enough equations
4. Solve the simultaneous equations

Worked Example 4 — Two-Loop Circuit

Consider a circuit with two batteries and three resistors:

• Left loop: Battery ε₁ = 6.0 V with R₁ = 2.0 Ω
• Right loop: Battery ε₂ = 4.0 V with R₂ = 3.0 Ω
• Shared branch (middle): R₃ = 5.0 Ω
• Both batteries have the positive terminal at the top

Let I₁ flow clockwise in the left loop through R₁, I₂ flow clockwise in the right loop through R₂, and I₃ flow downward through R₃.

KCL at the top junction:

I₁ = I₂ + I₃ ... (assuming I₁ enters, I₂ and I₃ leave)

Wait — let me set this up more carefully.

Define: I₁ flows through ε₁ and R₁ (left branch), I₂ flows through ε₂ and R₂ (right branch), I₃ flows through R₃ (middle branch).

At the top junction: I₁ = I₃ + I₂ → I₃ = I₁ − I₂ ... (1)

KVL around the left loop (clockwise, starting from bottom-left):

ε₁ − I₁R₁ − I₃R₃ = 0

6.0 − 2.0I₁ − 5.0I₃ = 0 ... (2)

KVL around the right loop (clockwise, starting from bottom-right):

ε₂ − I₂R₂ − I₃R₃ = 0

But wait — traversing clockwise in the right loop, we go up through ε₂ (+ to −, so the EMF opposes our traversal if the positive terminal is at the top and we go from bottom to top). Let me be more careful.

Let me redefine: both batteries have their positive terminals connected to the top rail.

Going clockwise around the left loop: Up through ε₁ (− to +, so +ε₁), then through R₁ (in direction of I₁, so −I₁R₁), then down through R₃ (in direction of I₃, so −I₃R₃).

6.0 − 2.0I₁ − 5.0I₃ = 0 ... (2)

Going clockwise around the right loop: Down through R₃ (against I₃, so +I₃R₃), then through R₂ (in direction of I₂, so −I₂R₂), then up through ε₂ (− to +, so +ε₂).

Actually, going clockwise in the right loop starting from the top: down through R₃ (against I₃ if I₃ is downward = +I₃R₃), right along bottom, up through ε₂ (− to + = +ε₂), left along top... No. Let me just set up simple equations.

Let me use: ε₁ = I₁R₁ + I₃R₃ and ε₂ = I₂R₂ + I₃R₃ (since both EMFs drive current through R₃ in the same direction).

From (1): I₃ = I₁ − I₂

Substituting into left loop equation (2): 6.0 = 2.0I₁ + 5.0(I₁ − I₂) = 7.0I₁ − 5.0I₂ ... (2')

Right loop: 4.0 = 3.0I₂ + 5.0(I₁ − I₂) = 5.0I₁ − 2.0I₂ ... (3')

From (2'): 6.0 = 7.0I₁ − 5.0I₂ From (3'): 4.0 = 5.0I₁ − 2.0I₂

Multiply (3') by 2.5: 10.0 = 12.5I₁ − 5.0I₂ ... (3'')

Subtract (2') from (3''): 4.0 = 5.5I₁

I₁ = 4.0/5.5 = 0.727 A

From (3'): 4.0 = 5.0(0.727) − 2.0I₂ = 3.636 − 2.0I₂

2.0I₂ = 3.636 − 4.0 = −0.364

I₂ = −0.182 A

The negative sign means I₂ actually flows in the opposite direction to our assumed direction — i.e., current flows through ε₂ in the "wrong" direction (ε₂ is being charged by ε₁).

I₃ = I₁ − I₂ = 0.727 − (−0.182) = 0.909 A

Verification (left loop): 2.0 × 0.727 + 5.0 × 0.909 = 1.454 + 4.545 = 6.0 V = ε₁ ✓

Verification (right loop): 3.0 × (−0.182) + 5.0 × 0.909 = −0.546 + 4.545 = 4.0 V = ε₂ ✓

Exam Tip: Don't worry if a current comes out negative — it just means you guessed the direction wrong. State the magnitude and the correct direction. Never go back and change your assumed directions once you've set up the equations.

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Common Circuit Analysis Patterns

Wheatstone Bridge

A Wheatstone bridge consists of four resistors arranged in a diamond/square pattern with a galvanometer across the middle.

Balanced condition: When no current flows through the galvanometer:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$R2​R1​​=R4​R3​​

At balance, the middle branch can be removed from the analysis (no current flows through it).

Worked Example 5 — Wheatstone Bridge

A Wheatstone bridge has R₁ = 100 Ω, R₂ = 200 Ω, R₃ = 150 Ω. Find R₄ for the bridge to be balanced.

Solution:

R₁/R₂ = R₃/R₄

100/200 = 150/R₄

R₄ = 150 × 200/100 = 300 Ω

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Applying Kirchhoff's Laws — Useful Tips

1. Number of equations needed = number of unknown currents
2. KCL gives you (number of junctions − 1) independent equations
3. KVL gives you the remaining equations — use independent loops
4. Be consistent with sign conventions throughout
5. Check your answers by substituting back into the original equations
6. Negative currents simply mean the actual direction is opposite to your assumed direction

Worked Example 6 — Simple KVL Application

In a circuit, a 9.0 V battery drives current through three resistors in series: 3.0 Ω, 4.0 Ω, and 2.0 Ω. A voltmeter reads 4.0 V across the 4.0 Ω resistor. Verify this using KVL.

Solution:

Total resistance: R_total = 3.0 + 4.0 + 2.0 = 9.0 Ω

Current: I = 9.0 / 9.0 = 1.0 A

p.d. across 4.0 Ω resistor: V = IR = 1.0 × 4.0 = 4.0 V ✓

KVL check: 1.0 × 3.0 + 1.0 × 4.0 + 1.0 × 2.0 = 3.0 + 4.0 + 2.0 = 9.0 V = ε ✓

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Worked Example: Two-Loop Network with Opposing Cells

Consider a two-loop network: a 12 V cell (ε₁) in the left loop and a 6 V cell (ε₂) in the right loop, oriented to push current in opposing directions. The three branches contain resistors of 2 Ω (in the left loop only), 3 Ω (in the shared middle branch), and 4 Ω (in the right loop only). Internal resistance of both cells is negligible. Apply Kirchhoff's laws to find the current in each of the three branches.

Solution — Step 1: Assign and label branch currents.

Let I₁ = current in the left loop (clockwise, through ε₁ and the 2 Ω resistor); I₂ = current in the right loop (anti-clockwise, through ε₂ and the 4 Ω resistor); I₃ = current in the shared 3 Ω middle branch. By KCL at the junction at the top: I₃ = I₁ + I₂ (assuming both I₁ and I₂ flow into the junction).

Step 2: KVL around the left loop.

Going clockwise from ε₁: +12 − 2I₁ − 3I₃ = 0, i.e. 2I₁ + 3I₃ = 12. ... (Eq A)

Step 3: KVL around the right loop.

Going anti-clockwise from ε₂: +6 − 4I₂ − 3I₃ = 0, i.e. 4I₂ + 3I₃ = 6. ... (Eq B)

Step 4: Substitute I₃ = I₁ + I₂ and solve.

Eq A becomes: 2I₁ + 3(I₁ + I₂) = 12 → 5I₁ + 3I₂ = 12. ... (Eq A′) Eq B becomes: 4I₂ + 3(I₁ + I₂) = 6 → 3I₁ + 7I₂ = 6. ... (Eq B′)

Solve the simultaneous pair. Multiply A′ by 7 and B′ by 3: 35I₁ + 21I₂ = 84; 9I₁ + 21I₂ = 18. Subtract: 26I₁ = 66, so I₁ ≈ 2.54 A. Back-substitute: 5(2.54) + 3I₂ = 12 → 3I₂ = −0.69 → I₂ ≈ −0.23 A. The negative sign means the actual direction of I₂ is opposite to the assumed anti-clockwise direction. Then I₃ = I₁ + I₂ = 2.54 − 0.23 = 2.31 A.

Step 5: KVL/KCL check.

KVL left: 2(2.54) + 3(2.31) = 5.08 + 6.93 = 12.01 ≈ 12 V ✓. KVL right: 4(−0.23) + 3(2.31) = −0.92 + 6.93 = 6.01 ≈ 6 V ✓. The small discrepancies are rounding only.