Kirchhoff's Second Law

Spec mapping: OCR H556 Module 4.3 — Electrical circuits (Kirchhoff's second law as conservation of energy per unit charge around any closed loop; $\sum \varepsilon = \sum IR$ ; sign convention for batteries and resistors; multi-loop and multi-source circuit analysis; Wheatstone bridge balance condition). Refer to the official OCR H556 specification document for exact wording.

Kirchhoff's first law (lesson 3) handles junctions: charge entering equals charge leaving — conservation of charge. Kirchhoff's second law handles loops: total EMF around a closed path equals total $IR$ -drop around that path — conservation of energy per unit charge. Together they let you solve any linear DC circuit, no matter how tangled — including networks with multiple batteries or non-series-parallel topology (Wheatstone bridges) where the lesson-10 reduction techniques cannot reach.

This lesson states the second law, derives it from energy conservation, articulates a bullet-proof sign convention, and walks through single-loop and multi-loop examples up to the Wheatstone bridge. By the end you should be able to label currents, write equations, and solve any A-Level DC network — including circuits that look impossible.

Statement of the Law

Kirchhoff's Second Law. Around any closed loop in a circuit, the algebraic sum of the EMFs equals the algebraic sum of the potential differences across the resistive components.

In algebraic form:

$\boxed{\; \sum \varepsilon = \sum IR \quad \text{(around any closed loop)} \;}$

or equivalently:

$\sum \varepsilon - \sum IR = 0$

A closed loop is any path through the circuit that starts and ends at the same point. The signs of $\varepsilon$ and $IR$ depend on the direction of traversal and the assumed current direction — covered in the sign convention below.

Why It Must Be True — Conservation of Energy

Kirchhoff-2 is a direct consequence of conservation of energy applied to a unit charge.

An EMF source $\varepsilon$ does work $\varepsilon Q$ on a charge $Q$ passing through it (from $-$ to $+$ internally). Per coulomb, the work is $\varepsilon$ joules — that's the definition of EMF.
A resistor $R$ removes energy $IRQ$ from charge $Q$ that flows through it. Per coulomb, the energy taken is $IR$ joules — that's the definition of pd.

If a charge $Q$ traverses a closed loop and returns to its starting point, it must end with the same electrical potential energy it started with (electrical PE is a function of position only — see lesson 4). Therefore:

$\text{Energy in from EMFs} = \text{Energy out to resistors}$

$\sum_{\text{loop}} \varepsilon Q = \sum_{\text{loop}} IR \, Q$

Dividing both sides by $Q$ :

$\sum_{\text{loop}} \varepsilon = \sum_{\text{loop}} IR$

This is Kirchhoff's second law. Without it a perpetual motion machine would be possible — a charge could circulate around a loop and emerge with more energy than it started with.

Sign Conventions — The Hard Part

Kirchhoff-2's algebra is straightforward; the signs are where students lose marks. Here is a reliable procedure.

Choose a direction to traverse the loop (clockwise or anticlockwise — it does not matter, but be consistent).
Assume a direction for each branch current. Draw arrows.
Crossing a battery from $-$ to $+$ (in the direction the EMF pushes positive charge): add $+\varepsilon$ .
Crossing a battery from $+$ to $-$ (against the EMF push): add $-\varepsilon$ .
Crossing a resistor in the direction of assumed current: add $-IR$ (voltage drops).
Crossing a resistor against the assumed current: add $+IR$ .
Sum to zero around the loop.
Solve the resulting linear equations.

A negative current in the solution means the actual direction is opposite to your assumed direction. This is not an error; just interpret the sign at the end. Do not "fix" it by re-doing the algebra.

flowchart TD
  A["Pick traverse direction (CW or ACW)"] --> B["Assign assumed current direction per branch"]
  B --> C{"Crossing a battery"}
  C -- "- to +" --> D["Add +ε"]
  C -- "+ to -" --> E["Add -ε"]
  B --> F{"Crossing a resistor"}
  F -- "with assumed I" --> G["Add -IR"]
  F -- "against I" --> H["Add +IR"]
  D --> Z["Sum to zero around the loop"]
  E --> Z
  G --> Z
  H --> Z
  Z --> Y["Solve simultaneous equations"]
  Y --> X{"Any I < 0?"}
  X -- "Yes" --> W["Real direction is opposite to assumed"]
  X -- "No" --> V["Assumed direction was correct"]

Worked Example 1 — Single Loop, One EMF

A $12$ V ideal battery is connected in a simple loop with a $4$ $\Omega$ resistor and an $8$ $\Omega$ resistor in series. Find the current.

Traverse clockwise starting from the battery's negative terminal. Assume $I$ flows clockwise.

Cross battery $-$ to $+$ : $+12$ V.
Cross $4$ $\Omega$ in direction of $I$ : $-4I$ .
Cross $8$ $\Omega$ in direction of $I$ : $-8I$ .
Return to start: sum $= 0$ .

$12 - 4I - 8I = 0 \quad\Rightarrow\quad 12 = 12I \quad\Rightarrow\quad I = \mathbf{1.0 \text{ A}}$

Sanity check: $R_{\text{total}} = 4 + 8 = 12$ $\Omega$ , so $I = V/R = 12/12 = 1.0$ A. ✓ The single-loop, single-EMF case reduces to Ohm's law applied to the series combination.

Worked Example 2 — Single Loop, Two EMFs Aiding

Two cells of EMF $6$ V and $2$ V are in a loop, both pushing current in the same direction, with a single $5$ $\Omega$ resistor.

Traverse in the push direction. Assume $I$ in the push direction.

$+6$ V (first cell, $-$ to $+$ ).
$+2$ V (second cell, $-$ to $+$ ).
$-5I$ (resistor, in direction of $I$ ).
$= 0$ .

$8 - 5I = 0 \quad\Rightarrow\quad I = \mathbf{1.6 \text{ A}}$

Now reverse the small cell — it now opposes the larger:

$+6 - 2 - 5I = 0 \quad\Rightarrow\quad 4 - 5I = 0 \quad\Rightarrow\quad I = \mathbf{0.8 \text{ A}}$

The current is halved when the EMFs oppose. This is exactly what happens during battery charging: an external EMF (e.g. car alternator) is driven against a smaller cell EMF (battery) to recharge the cell — current flows backwards through the cell.

Worked Example 3 — Single Loop With Internal Resistance

A $1.5$ V cell with internal resistance $r = 0.5$ $\Omega$ drives a $4.5$ $\Omega$ external resistor. Find the current and terminal pd.

Traverse clockwise:

$+1.5$ V (EMF).
$-0.5 \, I$ (internal resistance, in direction of $I$ ).
$-4.5 \, I$ (external resistor).
$= 0$ .

$1.5 - 0.5 I - 4.5 I = 0 \quad\Rightarrow\quad 1.5 = 5.0 I \quad\Rightarrow\quad I = \mathbf{0.30 \text{ A}}$

Terminal pd:

$V_{\text{terminal}} = I R_{\text{ext}} = 0.30 \times 4.5 = \mathbf{1.35 \text{ V}}$

Or equivalently, $V = \varepsilon - Ir = 1.5 - 0.30 \times 0.5 = 1.35$ V ✓ (lesson 9). Kirchhoff-2 reproduces the internal-resistance formula $\varepsilon = I(R+r)$ rigorously.

Multi-Loop Circuits — The General Recipe

When a network has more than one loop, the procedure generalises:

Label each branch current with a symbol ( $I_1, I_2, \ldots$ ) and an assumed direction.
Apply Kirchhoff's first law at each junction to relate currents (one equation per independent junction).
Apply Kirchhoff's second law around each independent loop (one equation per independent loop).
Solve the simultaneous equations.
Interpret negative results as currents flowing opposite to your assumed direction.

For a network with $b$ branches and $n$ nodes (junctions), the number of independent loop equations is $b - n + 1$ (the circuit rank). At A-Level you rarely encounter more than 2-3 loops, but the counting argument tells you exactly how many you need: as many equations as unknowns.

Worked Example 4 — Two-Cell, Two-Loop Network

Cell A (EMF $12$ V, $r_A \approx 0$ ) drives current $I_1$ through $R_1 = 2$ $\Omega$ into a junction. Cell B (EMF $6$ V, $r_B \approx 0$ ) drives current $I_2$ through $R_2 = 3$ $\Omega$ into the same junction. From the junction, a shared resistor $R_3 = 4$ $\Omega$ carries $I_3$ to ground.

flowchart LR
  CA["Cell A: 12 V"] --> R1["R_1 = 2 Ω"]
  R1 --> J((Junction))
  J --> R3["R_3 = 4 Ω"]
  R3 --> GND["Ground"]
  J --> R2["R_2 = 3 Ω"]
  R2 --> CB["Cell B: 6 V"]
  CB --> GND
  CA --> GND

Step 1 — Junction equation (Kirchhoff-1). Assume $I_1$ and $I_2$ both flow into the junction; $I_3$ flows out:

$I_1 + I_2 = I_3 \quad (\star)$

Step 2 — Loop equation 1 (cell A, $R_1$ , $R_3$ , back to A):

$12 - 2 I_1 - 4 I_3 = 0$

Step 3 — Loop equation 2 (cell B, $R_2$ , $R_3$ , back to B):

$6 - 3 I_2 - 4 I_3 = 0$

Step 4 — Substitute $I_3 = I_1 + I_2$ from ( $\star$ ):

Loop 1: $12 - 2 I_1 - 4(I_1 + I_2) = 0 \Rightarrow 6 I_1 + 4 I_2 = 12 \Rightarrow 3 I_1 + 2 I_2 = 6$ .

Loop 2: $6 - 3 I_2 - 4(I_1 + I_2) = 0 \Rightarrow 4 I_1 + 7 I_2 = 6$ .

Step 5 — Solve the pair. Multiply equation 1 by $7/2$ :

$10.5 \, I_1 + 7 \, I_2 = 21$

Subtract equation 2: $6.5 \, I_1 = 15 \Rightarrow I_1 = \mathbf{2.31 \text{ A}}$ .

Substitute back: $3(2.31) + 2 I_2 = 6 \Rightarrow 6.92 + 2 I_2 = 6 \Rightarrow I_2 = \mathbf{-0.46 \text{ A}}$ .

$I_3 = I_1 + I_2 = 2.31 - 0.46 = \mathbf{1.85 \text{ A}}$ .

Interpretation. $I_2 < 0$ means the actual current in branch 2 flows opposite to the assumed direction — cell B is being charged by cell A, not discharging. Cell A's larger EMF dominates the network, driving current through the shared $R_3$ to ground and pushing current backwards through cell B. This is exactly how a car alternator charges the car battery while the engine is running.

Wheatstone Bridge — The Classic Non-Series-Parallel Network

The Wheatstone bridge has four resistors $R_1, R_2, R_3, R_4$ in a diamond, with a sensitive galvanometer (or modern voltmeter) between the top and bottom nodes and a battery driving the left and right nodes. It cannot be reduced by series-parallel rules — the galvanometer branch is neither in series nor in parallel with anything obvious. But Kirchhoff's laws crack it immediately.

The bridge is balanced when no current flows through the galvanometer ( $I_G = 0$ ). By Kirchhoff-1 at the top and bottom nodes:

The current through $R_1$ equals the current through $R_2$ (call it $I_{\text{top}}$ ).
The current through $R_3$ equals the current through $R_4$ (call it $I_{\text{bot}}$ ).

Apply Kirchhoff-2 to the upper-left half-loop (going through $R_1$ then galvanometer then $R_3$ , back to the battery node). Since $I_G = 0$ , there is no $IR$ across the galvanometer:

$I_{\text{top}} R_1 = I_{\text{bot}} R_3$

Similarly, upper-right half-loop:

$I_{\text{top}} R_2 = I_{\text{bot}} R_4$

Dividing the two equations:

$\boxed{\; \frac{R_1}{R_2} = \frac{R_3}{R_4} \;}$

This is the Wheatstone bridge balance condition. Before digital multimeters, this is how precision resistance measurements were made — you placed an unknown resistor in one position, then varied a known precision resistor in another position until the galvanometer read zero. From the balance condition, the unknown resistance is then read off from the known three.

Modern bridge applications: strain gauges (Wheatstone bridge with one or four strain-gauge arms subtracts baseline and amplifies the resistance change — see lesson 12 worked example 6), platinum resistance thermometers, capacitance bridges in AC variants, and high-precision instrumentation.

When to Use Kirchhoff vs Series-Parallel

Kirchhoff's laws are the most general method, but also the most algebra. For any circuit reducible by series-parallel rules (lesson 10) — single battery, no cross-links — start with the reduction, find the total current, and work outward.

Reach for Kirchhoff's laws when:

Multiple EMF sources are present (lesson 10 reduction cannot handle them).
Bridge or ladder topology (some resistor is neither in series nor in parallel with its neighbour).
Mixed EMF/internal resistance with multiple cells in different positions.

Exam Tip: If the OCR question has only one battery and the resistor network is reducible (e.g. $R_1$ in series with $R_2 \| R_3$ ), use lesson-10 series-parallel. If the question has multiple batteries, internal resistances on different cells, or a "bridge" topology, switch directly to Kirchhoff's laws.

Synoptic Links

Kirchhoff's first law (lesson 3, Module 4.3) — together with the second law forms the complete pair for any linear DC network. First law = charge conservation; second law = energy conservation per unit charge.
Internal resistance (lesson 9, Module 4.2) — Kirchhoff-2 reproduces $\varepsilon = I(R+r)$ rigorously and extends to multi-cell circuits where the lesson-9 single-cell formula no longer suffices.
Series and parallel (lesson 10, Module 4.3) — series-parallel reduction is a consequence of Kirchhoff's laws for simple topologies; the laws themselves remain valid for the topologies where reduction fails.
Potential dividers (lesson 12, Module 4.3) — the divider equation $V_{\text{out}} = V_{\text{in}} R_2/(R_1+R_2)$ follows from Kirchhoff-2 around the series loop under the open-circuit assumption.
Mesh and nodal analysis (undergraduate EE) — modern circuit-analysis tools are systematic procedures built on Kirchhoff's laws; nodal analysis uses Kirchhoff-1 + Ohm, mesh analysis uses Kirchhoff-2 + Ohm.

Specimen question modelled on the OCR H556 paper format

Question (9 marks): A circuit contains two cells in parallel, both connected (with correct polarities) across an external load $R = 6.0$ $\Omega$ . Cell A has EMF $\varepsilon_A = 4.0$ V and internal resistance $r_A = 1.0$ $\Omega$ . Cell B has EMF $\varepsilon_B = 3.0$ V and internal resistance $r_B = 0.5$ $\Omega$ .

(a) Define Kirchhoff's second law and explain why it follows from conservation of energy. [3]

Kirchhoff's Second Law

Kirchhoff's Second Law

Statement of the Law

Why It Must Be True — Conservation of Energy

Sign Conventions — The Hard Part

Worked Example 1 — Single Loop, One EMF

Worked Example 2 — Single Loop, Two EMFs Aiding

Worked Example 3 — Single Loop With Internal Resistance

Multi-Loop Circuits — The General Recipe

Worked Example 4 — Two-Cell, Two-Loop Network

Wheatstone Bridge — The Classic Non-Series-Parallel Network

When to Use Kirchhoff vs Series-Parallel

Synoptic Links

Specimen question modelled on the OCR H556 paper format

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