recall Kirchhoff’s second law and understand that it is a consequence of conservation of energy

Kirchhoff’s Second Law (Loop Rule)

Statement of the law

In any closed conducting loop, the algebraic sum of the potential differences (voltage drops and rises) is zero:

\$\sum{k=1}^{n} \Delta Vk = 0\$

where each \$\Delta V_k\$ is taken as positive for a rise in potential (e.g., moving from the negative to the positive terminal of a battery) and negative for a drop (e.g., across a resistor in the direction of current).

Why the law holds – Conservation of Energy

The loop rule is a direct consequence of the principle that energy cannot be created or destroyed. As a test charge \$q\$ moves once around a closed loop, the net work done on it must be zero because it returns to its starting point with the same kinetic energy. The work done by the electric field is \$q\Delta V\$, so:

\$\sum{k} q\,\Delta Vk = 0 \;\;\Longrightarrow\;\; \sum{k} \Delta Vk = 0\$

This is exactly the statement of Kirchhoff’s second law. It ensures that the electrical energy supplied by sources (batteries, generators) equals the energy dissipated in the circuit elements (resistors, lamps, etc.) for each complete traversal of the loop.

Applying the loop rule

1. Identify a closed loop in the circuit.
2. Choose a direction to travel around the loop (clockwise or anticlockwise).
3. For each element encountered, write the voltage change:

   • Battery: \$+V\$ when moving from – to + terminal, \$-V\$ in the opposite sense.
   • Resistor: \$-IR\$ when moving in the direction of the current (Ohm’s law \$V=IR\$).
   • Other components (capacitor, inductor) can be treated similarly with their appropriate voltage expressions.

4. Sum all the expressions and set the total equal to zero.
5. Solve the resulting equation(s) for the unknown quantity (current, voltage, resistance, etc.).

Worked example

Consider the circuit below (see suggested diagram). The loop contains a 12 V battery, a resistor \$R1 = 4\;\Omega\$, and a resistor \$R2 = 6\;\Omega\$ in series.

Applying the loop rule (clockwise direction):

\$+12\;\text{V} - I R1 - I R2 = 0\$

Substituting the resistances:

\$12 - I(4) - I(6) = 0 \;\;\Longrightarrow\;\; 12 - 10I = 0\$

Hence the current is:

\$I = \frac{12}{10} = 1.2\;\text{A}\$

Key symbols and their meanings

Common pitfalls

• Forgetting to assign a consistent sign convention throughout the loop.
• Mixing directions: the sign of \$IR\$ depends on whether you move with or against the assumed current direction.
• Neglecting voltage contributions from non‑resistive elements (e.g., internal resistance of a battery).
• Applying the loop rule to an open circuit – the rule only holds for closed paths.

Summary

Kirchhoff’s second law states that the sum of all potential differences around any closed loop is zero. It follows directly from the conservation of energy: a charge that returns to its starting point cannot have a net gain or loss of electrical energy. By translating this principle into algebraic equations, we can analyse complex circuits and determine unknown currents and voltages.

Self‑check questions

1. Write the loop equation for a circuit containing a 9 V battery, a \$2\;\Omega\$ resistor, and an unknown resistor \$Rx\$ in series. Solve for \$Rx\$ if the measured current is \$1.5\;\text{A}\$.
2. Explain why the loop rule would be violated if a battery were assumed to create energy without any corresponding voltage drop elsewhere.
3. In a circuit with two loops sharing a common resistor, describe how you would apply Kirchhoff’s second law to solve for the currents in each loop.