Kirchhoff's Second Law

Kirchhoff's first law (Lesson 3) handled junctions. Kirchhoff's second law handles loops. Together they let us solve any linear DC circuit, no matter how tangled, including circuits with several batteries that series-parallel simplification cannot reduce.

This lesson states the second law, explains why it must be true, works through single-loop and multi-loop examples, and gives you a systematic procedure you can apply to any OCR H556 circuit question. It is a core topic of Module 4.3.1.

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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 components.

Equivalently:

Σε = ΣIR (around a closed loop)

or as an algebraic sum equal to zero:

Σε − ΣIR = 0

A "closed loop" is any path through the circuit that starts and finishes at the same point.

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Why It Must Be True — Conservation of Energy

Kirchhoff's second law is a direct consequence of the conservation of energy. Recall:

• An EMF source supplies energy εQ to each coulomb of charge that passes through it (from − to + internally).
• A resistor dissipates energy IRQ per coulomb (= VQ).

If a charge Q travels all the way around a closed loop and returns to its starting point, it must end up with the same electrical potential energy it started with (because potential energy is a function of position only). All the energy the EMFs put in along the loop must be exactly matched by all the energy the components take out:

Total energy in = Total energy out → Σε = ΣIR

This is the rigorous justification. Without it, a perpetual motion machine would be possible.

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Sign Conventions — The Hard Part

To apply the law you must be careful with signs. Here is a reliable procedure:

1. Pick a direction to traverse the loop (clockwise or anticlockwise — it does not matter, so long as you are consistent).
2. When you cross a battery from − to + (i.e. with the EMF "pushing you along"), add +ε.
3. When you cross a battery from + to −, add −ε.
4. When you cross a resistor in the direction of the assumed current, add −IR (a voltage drop).
5. When you cross a resistor against the assumed current, add +IR.
6. Sum to zero around the loop.

If a current turns out negative after you solve, its real direction is opposite to the one you assumed. That is not an error — just invert the sign at the end.

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Worked Example 1 — Single Loop, One EMF

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

Traverse clockwise, starting from the battery's negative terminal:

• Cross battery from − to +: +12 V
• Cross 4 Ω in direction of I: −4I
• Cross 8 Ω in direction of I: −8I
• Return to start: sum = 0

12 − 4I − 8I = 0 12 = 12I I = 1.0 A

Same answer as R_total = 12 Ω, I = V/R_total = 12/12 = 1 A. Good — the single-loop, single-source case reduces to what we already know.

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Worked Example 2 — Single Loop, Two EMFs

Two cells of EMF 6 V and 2 V are connected in a loop (correct polarities — both pushing the current in the same direction) with a 5 Ω resistor.

Traverse in the direction both cells push:

• +6 V (first cell)
• +2 V (second cell)
• −5I (resistor)
• = 0

8 − 5I = 0 → I = 1.6 A

Now imagine the smaller cell is reversed — it now opposes the bigger one.

• +6 V
• −2 V
• −5I
• = 0

4 − 5I = 0 → I = 0.8 A

The current is halved because the two EMFs partially cancel. In practice this is how a battery charges: an external, larger EMF drives current backwards through a smaller cell to recharge it.

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Worked Example 3 — Single Loop with Internal Resistance

A 1.5 V cell with internal resistance 0.5 Ω is connected to a 4.5 Ω external resistor. Find the current and terminal pd.

Traverse:

• +1.5 V (EMF)
• −0.5 × I (internal resistance)
• −4.5 × I (external resistor)
• = 0

1.5 = 0.5I + 4.5I = 5.0I → I = 0.3 A

Terminal pd = I × R_ext = 0.3 × 4.5 = 1.35 V. Or: ε − Ir = 1.5 − 0.3 × 0.5 = 1.35 V ✓ (Lesson 9).

You can see Kirchhoff's second law simply gives you the internal-resistance formula ε = I(R + r) rigorously.

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Multi-Loop Circuits

When there are multiple loops you apply Kirchhoff's second law to each independent loop and Kirchhoff's first law at each junction. You need as many equations as unknowns.

General procedure:

1. Label each branch current with a symbol (I₁, I₂, …) and assume a direction.
2. Use Kirchhoff's first law at each junction to relate the currents.
3. Use Kirchhoff's second law around each independent loop.
4. Solve the simultaneous equations.

Worked Example 4 — Two-Loop Circuit