# Kirchhoff's laws

> A-Level Physics · CIE 9702
> Source: https://www.owlsprep.com/study/cie-9702-u10-kirchhoff-s-laws/

This module covers Kirchhoff's two fundamental laws for DC circuit analysis, which let you solve complex circuits that cannot be reduced using series or parallel rules alone. You will learn to apply them to find unknown currents and potential differences.

**Prerequisites:** [Series and parallel resistor combinations](https://www.owlsprep.com/study/cie-9702-u10-resistors-in-series-and-parallel/); Basic conservation of charge and energy

## Learning objectives

- State Kirchhoff's first and second laws
- Link Kirchhoff's laws to conservation of charge and energy
- Apply Kirchhoff's laws to solve complex DC circuit problems
- Interpret negative current values from calculations

## Kirchhoff's First Law (Junction Rule)

**Kirchhoff's First Law** — The sum of currents entering a junction equals the sum of currents leaving the junction. This is a direct consequence of conservation of charge: charge cannot be created or destroyed at a junction.

*Notation:* \sum I_{in} = \sum I_{out}

*Example:* If 3 A and 2 A enter a junction, 5 A must leave it.

When analysing circuits, first label all currents at every junction with an assumed direction. If you pick the wrong direction, the calculated current will be negative, indicating the actual direction is opposite to your assumption.

**Worked example:** At a junction, current $I_1 = 2.0$ A enters, and $I_2 = 1.5$ A leaves. Find the magnitude and direction of the third current $I_3$.

1. Apply KCL, taking currents entering the junction as positive:
2. $$I_1 = I_2 + I_3$$
3. Rearrange to solve for $I_3$:
4. $$I_3 = I_1 - I_2 = 2.0 - 1.5 = 0.5 \ \text{A}$$
5. Since $I_3$ is positive, our assumption that it leaves the junction is correct. The third current is 0.5 A leaving the junction.

> **Exam tip:** Always state the direction of the final current to earn full marks in structured questions.

*Calculator:* allowed

## Kirchhoff's Second Law (Loop Rule)

**Kirchhoff's Second Law** — Around any closed loop in a circuit, the sum of electromotive forces (emfs) equals the sum of potential differences across all resistors in the loop. This is a consequence of conservation of energy.

*Notation:* \sum \varepsilon = \sum IR

*Example:* For a single loop with one battery and one resistor, this simplifies to $\varepsilon = IR$.

Sign convention is critical here: when traversing the loop, assign a positive value to emf if you move from the negative to positive terminal of a battery. Assign a negative potential difference $-IR$ if you move across a resistor in the same direction as the current.

**Worked example:** A single loop contains a 6.0 V battery and two series resistors of $2.0 \ \Omega$ and $4.0 \ \Omega$. Find the current in the loop using KVL.

1. Traverse the loop clockwise starting at the negative terminal of the battery.
2. Apply KVL: sum of emfs equals sum of IR drops:
3. $$\varepsilon = I R_1 + I R_2$$
4. Substitute the given values:
5. $$6.0 = I (2.0 + 4.0) = 6I$$
6. Solve for I:
7. $$I = 1.0 \ \text{A}$$

> **tip**
>
> Stick to the same sign convention every time you solve a problem to avoid avoidable sign errors.

*Calculator:* allowed

## Solving Complex Two-Loop Circuits

Most CIE exam questions requiring Kirchhoff's laws involve two connected loops with multiple junctions. To solve these, you write a system of simultaneous equations from KCL and KVL, then solve for unknown currents.

1. Label all currents and choose their assumed directions at every junction
2. Write one independent KCL equation for each junction
3. Write one independent KVL equation for each closed loop
4. Solve the system of simultaneous equations
5. Interpret negative values to find actual current directions

**Worked example:** Two loops share a $2 \ \Omega$ resistor. Loop 1 has a 12 V battery and $1 \ \Omega$ resistor. Loop 2 has a 6 V battery and $3 \ \Omega$ resistor. Find the current in the shared resistor.

1. Label currents: $I_1$ in loop 1, $I_2$ in loop 2. By KCL: total current through shared resistor is $I = I_1 + I_2$:
2. Write KVL for loop 1:
3. $$12 = 1I_1 + 2(I_1 + I_2) = 3I_1 + 2I_2$$
4. Write KVL for loop 2:
5. $$6 = 3I_2 + 2(I_1 + I_2) = 2I_1 + 5I_2$$
6. Solve simultaneous equations: multiply first by 2, second by 3, subtract to get $I_2 = -6/11 \approx -0.55$ A, then $I_1 \approx 4.36$ A:
7. Total current in shared resistor: $I = I_1 + I_2 \approx 3.8$ A. Negative $I_2$ means its direction is opposite to our assumption.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Mixing up the physical basis of the two laws
  - Why it fails: This common mix-up costs easy marks when examiners ask for an explanation of the laws
  - Correct: Remember: KCL = conservation of charge, KVL = conservation of energy
- **Wrong:** Ignoring negative current values
  - Why it fails: Candidates often discard negative values instead of interpreting them correctly
  - Correct: A negative value just means the actual current direction is opposite to your assumed direction
- **Wrong:** Inconsistent sign conventions for loop traversal
  - Why it fails: Randomly assigning signs leads to incorrect final values
  - Correct: Always use the same convention: +emf for negative to positive, -IR for same direction as current
- **Wrong:** Writing more independent equations than needed
  - Why it fails: Extra equations are redundant and can introduce errors when solving
  - Correct: For n independent loops, you need n total equations (n-1 KCL, n KVL)

## Cheatsheet

| Law | Mathematical Form | Physical Basis | Key Tip |
| --- | --- | --- | --- |
| First (Junction) | $\sum I_{in} = \sum I_{out}$ | Conservation of charge | Negative = opposite direction |
| Second (Loop) | $\sum \varepsilon = \sum IR$ | Conservation of energy | Stick to consistent signs |

## What's next

Kirchhoff's laws are the foundation of all circuit analysis, and they underpin every topic you will cover in DC and later AC circuits. Now that you understand how to apply these rules, you can move on to more practical circuit concepts including internal resistance of batteries and potential dividers, which are extremely common exam topics that build directly on this foundation. Mastery of Kirchhoff's laws also prepares you for more advanced circuit analysis in A2 Physics, including AC impedance and complex network problems.

- [Internal Resistance of Cells](https://www.owlsprep.com/study/cie-9702-u10-internal-resistance/)
- [Series and parallel resistor combinations](https://www.owlsprep.com/study/cie-9702-u10-series-and-parallel-resistor-combinations/)
- [Potential divider](https://www.owlsprep.com/study/cie-9702-u10-potential-divider/)

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