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Kirchhoff's Second Law

Key idea: Apply Kirchhoff’s second law (sum of potential changes around a closed loop is zero) using a clear sign convention for e.m.f. and resistors (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Recall circuit symbols and draw or interpret circuit diagrams.
  • Draw circuit diagrams containing sources, switches, resistors, meters, lamps, thermistors, light-dependent resistors and diodes.
  • Apply resistance and resistivity, interpret I–V characteristics and explain temperature effects.
  • Analyse e.m.f., terminal potential difference and internal resistance in real sources.
  • Analyse series, parallel and potential-divider resistor networks.
  • Combine capacitors in series and parallel.
  • Analyse charging and discharging in RC circuits using the time constant.
Syllabus note (9478)

Kirchhoff’s laws are not explicitly listed in the 9478 syllabus, but they are a common enrichment tool for solving non-trivial d.c. circuit networks.

1. Definitions (Must Know)

A. Kirchhoff’s second law (loop rule / voltage law)

Kirchhoff’s second law states that for any closed loop:

the algebraic sum of potential changes is zero

∑ Δ V = 0

This is a statement of conservation of energy: a charge that returns to the same point has no net change in electrical potential energy.

2. Key Ideas (What Earns Marks)

  • Choose a loop direction, then apply one consistent sign convention.
  • Typical sign convention while traversing a loop:
    • across a source from − to +: + ε
    • across a source from + to −: -ε
    • across a resistor in the direction of conventional current: -IR
    • across a resistor opposite to current: + IR
  • Combine with Kirchhoff’s first law (junction rule) to solve multi-loop circuits.
Fast workflow
  1. Label currents. 2) Write junction equations. 3) Write loop equations. 4) Solve simultaneous equations. 5) Negative current means your assumed direction is opposite.

3. Detailed Explanations

A. Why the loop rule works (energy per unit charge)

Potential difference is energy per unit charge. Around a closed loop, a charge returns to the same point, so net energy change per unit charge is zero:

∑ Δ V = 0

This is why you can add e.m.f. rises and subtract resistive drops in one equation.

B. Sign convention diagram

Potential-change signs when traversing a circuit loopAcross a cell, travel from negative to positive gives a positive emf change. Across a resistor, travel opposite to conventional current gives a positive potential change; reversing either traversal reverses the sign.Crossing a sourceCrossing a resistor−+εloop traversal− to + gives +εRconventional current Iloop traversalopposite I gives +IR
Use one consistent sign convention for sources and resistors when traversing a loop; reversing the traversal reverses the sign.

4. Common Mistakes

  • Flipping the sign convention mid-loop.
  • Using IR drops for resistors but forgetting internal resistance terms when stated.
  • Writing too many loop equations (some are not independent).
  • Forgetting that a negative answer means the direction is opposite to your assumption.

5. Exam Tips

  • Pick loop directions that match your assumed current directions to reduce sign errors.
  • If the circuit has many loops, start with the smallest independent loops.
  • After solving, do a quick check by substituting back into one loop equation.

6. Worked Examples

Modelled example 1

Two sources feeding a shared resistor (multi-loop)

Core

Problem

Use Kirchhoff’s laws to find the branch currents in the circuit shown.

Two sources feeding a shared resistorJunction A is at the top centre and junction B at the bottom centre. The left branch has a 10-volt cell, positive terminal up, and a 10-ohm resistor R1 carrying current I1 to A. The right branch has a 20-volt cell, positive terminal up, and a 20-ohm resistor R2 carrying current I2 to A. The middle branch from A down to B is a 40-ohm resistor R3 carrying current I3, where I3 equals I1 plus I2.ABR₁ = 10 ΩR₂ = 20 ΩR₃ = 40 Ω10 V20 V++I₁I₂I₃
Example circuit for setting up junction and loop equations.

Assume I₁ and I₂ flow towards the junction and combine as I₃. Given R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 40 Ω, and sources of 10 V and 20 V.

Study the worked solution

  1. Write the junction relation

    Method

    I₁ + I₂ = I₃.

    Reason

    The two assumed incoming branch currents combine in the shared resistor.

    Working

    I₃ = I₁ + I₂

  2. Write one equation for each independent loop

    Method

    The source rises equal the resistor drops in each loop.

    Reason

    Each closed path returns to the same potential.

    Working

    10 = 10I₁ + 40I₃, 20 = 20I₂ + 40I₃

  3. Eliminate the shared current

    Method

    Substitute I₃ = I₁ + I₂ into both loop equations.

    Reason

    This leaves two simultaneous equations in I₁ and I₂.

    Working

    1 = 5I₁ + 4I₂, 1 = 2I₁ + 3I₂

  4. Solve the branch currents

    Method

    I₂ = 0.429 A and I₁ = -0.143 A.

    Reason

    Solving the two independent equations determines both signed currents.

    Working

    I₂ = 3/7 A, I₁ = -1/7 A

  5. Recover and interpret the shared current

    Method

    I₃ = 0.286 A, while I₁ actually flows opposite to its assumed arrow.

    Reason

    I₃ = I₁ + I₂ and a negative signed current reverses the chosen reference direction.

    Working

    I₃ = -1/7 + 3/7 = 2/7 = 0.286 A

Guided practice 2

Single loop with internal resistance

About 4 min

Problem

A cell has e.m.f. ε = 1.5 V and internal resistance r = 0.50 Ω, connected to R = 2.0 Ω. Find the current.

Try this before viewing the solution

Unit: A

Hints

Hint 1: write every loop change
Traversing with the current gives + ε-Ir-IR = 0.
View solution step by step

  1. Account for both resistances

    Method

    ε-Ir-IR = 0.

    Reason

    The source supplies energy per unit charge and both resistances dissipate it.

    Working

    ε = I(R + r)

  2. Calculate

    Method

    I = 0.60 A.

    Reason

    The total series resistance is 2.50 Ω.

    Working

    I = 1.5/(2.0 + 0.50) = 0.60 A

Common misconception 3

Loop equation sign check (resistor)

Find and correct the mistake

Learner claim

You traverse a resistor in the same direction as conventional current. A learner writes + IR because the loop direction follows the current. Diagnose the sign.

Try this before viewing the solution

Potential-change term

View solution step by step

  1. Identify the physical change

    Method

    Potential decreases through a passive resistor in the current direction.

    Reason

    Electrical energy per unit charge is transferred to thermal energy.

    Working

    Δ V = V_after-V_before < 0

  2. Write the loop term

    Method

    The term is -IR.

    Reason

    IR is the magnitude of the resistive potential drop.

    Working

    Δ V_R = -IR

Examiner practice 4

Two resistors in series (single loop)

3 marks

Examination question

A 12 V source is connected in series with R₁ = 4.0 Ω and R₂ = 8.0 Ω. Use Kirchhoff’s second law to find the current. [3 marks]

Try this before viewing the solution

Unit: A

View solution step by step

  1. Write the loop equation

    1 mark

    Method

    12-IR₁-IR₂ = 0.

    Reason

    The source rise balances both series-resistor drops.

    Working

    12-I(4.0)-I(8.0) = 0

  2. Collect the drops

    1 mark

    Method

    12 = 12I.

    Reason

    The same current passes through both series resistors.

    Working

    I(4.0 + 8.0) = 12

  3. Calculate

    1 mark

    Method

    I = 1.0 A.

    Reason

    A 12 V rise is balanced by the drop across 12 Ω total resistance.

    Working

    I = 12/12.0 = 1.0 A

Challenge 5

Two opposing sources in series

Minimal support

Independent transfer

Two sources of 9.0 V and 3.0 V oppose each other in a single loop with total resistance R = 6.0 Ω. Find the current magnitude and identify which source sets its direction.

Try this before viewing the solution

Unit: A
Source setting current direction

Hints

Hint 1: combine signed source changes
Choose the 9.0 V source direction as positive: εₙₑₜ = 9.0-3.0.
View solution step by step

  1. Find the signed net e.m.f.

    Method

    εₙₑₜ = 6.0 V in the direction of the 9.0 V source.

    Reason

    The two sources oppose, so their e.m.f.s subtract.

    Working

    9.0-3.0 = 6.0 V

  2. Calculate current magnitude

    Method

    I = 1.0 A.

    Reason

    The net source rise is balanced by the drop across 6.0 Ω.

    Working

    I = 6.0/6.0 = 1.0 A

  3. State the direction

    Method

    Current flows in the direction driven by the 9.0 V source.

    Reason

    The larger source determines the sign of the net e.m.f.

    Working

    εₙₑₜ > 0 in the 9.0 V reference direction.

7. Mind Stretchers

Mind stretcher 1: Reversing loop directionExtension

Show that reversing your loop direction does not change the final current you calculate (it only multiplies the loop equation by −1).

Show Answer

If you traverse the same loop in the opposite direction, every potential change flips sign:

  • rises become drops (+ ε ↔ -ε),
  • drops become rises (-IR ↔ + IR).

So the entire equation is multiplied by -1, e.g. ε-IR = 0 ⇒ -ε + IR = 0

Both equations are equivalent and give the same current when you solve them.

Mind stretcher 2: Negative current interpretationExtension

In a multi-loop solution, you obtain I₁ = -0.20 A for a branch current you assumed pointed towards a junction. Explain what this means physically.

Show Answer

It means the true current direction is opposite to your assumed arrow. The branch current has magnitude 0.20 A and flows away from the junction (or in the opposite direction along that branch).

Mind stretcher 3: Optional (Enrichment)Extension

A. More Kirchhoff practice circuits

  • /a-level/h2-physics/dc-circuits/using-kirchhoffs-laws-on-a-single-loop-circuit
  • /a-level/h2-physics/dc-circuits/using-kirchhoffs-laws-on-a-charging-circuit
  • /a-level/h2-physics/dc-circuits/using-kirchhoffs-laws-on-a-complex-circuit

B. When the loop rule can fail (induction, beyond d.c. circuits)

Kirchhoff’s second law assumes the electric field is conservative around the loop.

If the magnetic flux through a loop changes with time, an induced e.m.f. appears and ∑ Δ V around the loop is no longer zero in the simple “resistive drops + source rises” sense. This is handled under electromagnetic induction (Faraday’s law).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027