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

Kirchhoff’s Laws – Cambridge IGCSE/A‑Level Physics (9702)

1. Syllabus context (DC circuits – Topic 10)

• 10.1 Practical circuits – symbols, emf, internal resistance
• 10.2 Kirchhoff’s first (junction) and second (loop/mesh) laws
• 10.3 Potential dividers (including potentiometers)
• 10.4 Practical applications – sensor circuits, Wheatstone bridge, bias networks
• 10.5 Paper 5 skills – planning, data analysis and evaluation of experiments
• Related A‑Level extensions – magnetic induction, AC circuits (preview in § 13)

2. Circuit‑symbol cheat‑sheet

3. Kirchhoff’s laws – statements

• First law (junction rule): The algebraic sum of currents at any junction is zero.

  \[

  \sum I{\text{in}}-\sum I{\text{out}}=0\qquad\text{or}\qquad\sum I=0

  \]

  This is a direct expression of charge conservation.

• Second law (loop/mesh rule): The algebraic sum of potential differences (voltage rises + and drops –) around a closed conducting loop is zero.

  \[

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

  \]

  It follows from the conservation of energy for a test charge that returns to its starting point.

4. Derivation of the loop rule from energy conservation

Take a test charge \(q\) that moves once round a closed loop. The work done by the electric field is

\[

W = q\sum{k}\Delta Vk .

\]

Because the charge finishes where it started, its kinetic energy is unchanged; therefore the net work must be zero:

\[

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

\]

Consequently, the total electrical energy supplied by sources equals the total energy dissipated (or stored) in the circuit elements.

5. Sign‑convention for voltages (rise + , drop –)

6. Including internal resistance of a source

A real battery is modelled as an ideal emf \(E\) in series with an internal resistance \(r\). In a loop equation it simply appears as an extra resistor:

\[

+E \;-\; I r \;-\; \text{(other drops)} =0 .

\]

The terminal voltage is therefore

\[

V_{\text{term}} = E - Ir .

\]

7. Quick‑fire DC‑circuit fundamentals (required for Topic 10)

• Series combination: \(R{\text{eq}} = R1+R_2+\dots\)
• Parallel combination: \(\displaystyle\frac{1}{R{\text{eq}}}= \frac{1}{R1}+ \frac{1}{R_2}+ \dots\)
• Power relations: \(P = VI = I^{2}R = \frac{V^{2}}{R}\)
• Open‑circuit test: Measure emf \(E\) with no load (current = 0).
• Short‑circuit test: Connect terminals directly; the current is \(I{\text{sc}} = \frac{E}{r}\). This gives the internal resistance via \(r = \frac{E}{I{\text{sc}}}\).

8. Potential dividers

For two series resistors \(R1\) and \(R2\) across a source \(V\):

\[

+V - I R1 - I R2 =0 \;\Longrightarrow\; I=\frac{V}{R1+R2}.

\]

Voltage across \(R_2\) (the “output”) is

\[

V{R2}= I R2 = V\frac{R2}{R1+R2}.

\]

8.1 Potentiometer (variable resistor) as a null‑method voltage divider

A potentiometer of total resistance \(R{\text{pot}}\) is supplied from a stable reference voltage \(V{\text{ref}}\). The sliding contact divides the total resistance into \(Ra\) and \(Rb\) (with \(Ra+Rb=R_{\text{pot}}\)). The voltage at the wiper is

\[

V{\text{wiper}} = V{\text{ref}}\frac{Ra}{R{\text{pot}}}.

\]

In a null‑balance experiment the wiper voltage is set equal to an unknown voltage \(Vx\); the balance point directly gives the ratio \(Ra/R{\text{pot}}\) and hence \(Vx\) without drawing current from the unknown source.

9. Real‑world application: Wheatstone bridge (sensor & bias network)

A Wheatstone bridge consists of four resistors forming a diamond. When the bridge is balanced, the potential difference between the two opposite nodes is zero, giving the relation

\[

\frac{R1}{R2} = \frac{R3}{R4}.

\]

Typical use: one arm contains a thermistor (temperature sensor). By measuring the bridge voltage (or by adjusting a known resistor until balance is achieved) the resistance – and therefore the temperature – can be determined.

10. Mesh‑analysis – step‑by‑step procedure

1. Draw a clear schematic and label all currents (choose a direction, usually clockwise).
2. Identify independent loops (meshes). The number of independent loops = \(b - n + 1\) (where \(b\) = number of branches, \(n\) = number of nodes).
3. Write a junction equation for every node (first law).
4. Write a loop equation for each independent mesh using the sign‑convention table.
5. For a resistor shared by two meshes, use the algebraic current difference (e.g. \(IA-IB\)) and give opposite signs in the two equations.
6. Solve the simultaneous equations (substitution, elimination or matrix method).
7. Check: total power supplied = total power dissipated; junction equations are satisfied.

11. Worked examples

11.1 Single‑loop series circuit (review)

Loop (clockwise):

\[

+12\;-\;I R1\;-\;I R2 =0\;\Longrightarrow\;12 - I(4+6)=0.

\]

\[

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

\]

11.2 Two‑mesh bridge circuit (mesh analysis)

Assume clockwise mesh currents \(IA\) and \(IB\). The shared resistor \(R3\) carries \(IA-I_B\).

• Mesh A: \(\; +E1 - IA R1 - (IA-IB)R3 =0\)
• Mesh B: \(\; +E2 - IB R2 - (IB-IA)R3 =0\)

Re‑arranged:

\[

\begin{cases}

E1 = IA(R1+R3) - IB R3\\[4pt]

E2 = IB(R2+R3) - IA R3

\end{cases}

\]

For \(E1=12\;\text{V},\;E2=6\;\text{V},\;R1=2\;\Omega,\;R2=3\;\Omega,\;R_3=4\;\Omega\):

\[

\begin{aligned}

12 &= IA(2+4)-4IB\\

6 &= IB(3+4)-4IA

\end{aligned}

\Longrightarrow\;

\begin{cases}

I_A = 1.0\;\text{A}\\

I_B = 0.5\;\text{A}

\end{cases}

\]

Current through \(R3\) is \(IA-I_B =0.5\;\text{A}\).

11.3 Potentiometer as a voltage divider (null method)

Reference voltage \(V{\text{ref}}=5.0\;\text{V}\) across a 10 kΩ potentiometer. The wiper is positioned so that the resistance from the left end to the wiper is \(Ra=3.0\;\text{kΩ}\).

\[

V{\text{wiper}} = V{\text{ref}}\frac{Ra}{R{\text{pot}}}=5.0\;\text{V}\times\frac{3.0}{10}=1.5\;\text{V}.

\]

If an unknown source is adjusted until its voltage exactly balances the wiper voltage (galvanometer reads zero), the unknown voltage is 1.5 V.

11.4 Wheatstone bridge – finding an unknown resistance

Bridge is balanced when the galvanometer reads zero. Given \(R1=100\;\Omega\), \(R2=200\;\Omega\), \(R3=150\;\Omega\), find the unknown \(Rx\) in the fourth arm.

\[

\frac{R1}{R2} = \frac{R3}{Rx}\;\Longrightarrow\;

Rx = R3\frac{R2}{R1}=150\;\Omega\times\frac{200}{100}=300\;\Omega.

\]

12. Practical activity – measuring internal resistance of a cell (Paper 5 skill)

1. Set up the circuit: a variable resistor \(R\) (or a set of known resistors) in series with a fresh AA cell and an ammeter (A) in series. Connect a voltmeter (V) across the cell terminals.
2. For each chosen \(R\) record the terminal voltage \(V\) and the current \(I\).
3. Plot \(V\) (vertical axis) against \(I\) (horizontal axis). The data should lie on a straight line described by

   \[

   V = E - Ir .

   \]

4. From the graph:

   • y‑intercept = emf \(E\) (open‑circuit voltage).
   • Negative slope = internal resistance \(r\).

5. Evaluation checklist (marks 1‑4 for Paper 5):

   • Identify independent variables (current) and dependent variable (voltage).
   • State how you will keep temperature, contact resistance and battery age constant.
   • Explain how you will minimise loading error of the voltmeter (use a high‑impedance V).
   • Describe how you will assess uncertainty (repeat measurements, error bars, linear‑fit residuals).

13. Preview of A‑Level extensions (Topics 20 & 21)

• Magnetic induction: Faraday’s law \(\displaystyle \mathcal{E} = -\frac{d\Phi_B}{dt}\) can be combined with the loop rule to analyse generators and inductive circuits.
• AC circuits: Impedance \(Z\) replaces resistance in the loop rule; the same algebraic method gives phase relationships for \(R\), \(L\) and \(C\) components.
• Energy considerations: Power in reactive elements (\(P = VI\cos\phi\)) and the concept of apparent power \(S = VI\) are natural extensions of the energy‑conservation idea behind Kirchhoff’s second law.

14. Summary

• Kirchhoff’s first law conserves charge; the second law conserves energy around a loop.
• Both laws are applied by writing one algebraic equation per independent junction and per independent mesh.
• Consistent sign conventions (rise +, drop –) are essential; internal resistance is treated as an ordinary resistor.
• Potential‑divider formulas, Wheatstone bridges and potentiometer null methods are direct applications of the loop rule.
• Mesh analysis provides a systematic way to solve multi‑loop circuits.
• Practical experiments (e.g. measuring internal resistance) link the theory to Paper 5 requirements.
• Understanding the loop rule as energy conservation prepares students for later A‑Level topics such as induction and AC analysis.

15. Self‑check questions

1. Write the loop equation for a series circuit containing a 9 V battery, a \(2\;\Omega\) resistor and an unknown resistor \(Rx\). If the measured current is \(1.5\;\text{A}\), calculate \(Rx\).
2. Explain why a circuit appears to “violate” the loop rule if the internal resistance of the battery is ignored.
3. In a two‑mesh circuit that shares a resistor \(R_3\), outline the steps you would take to find the two mesh currents using Kirchhoff’s laws.
4. Derive the potential‑divider expression \(V{\text{out}} = V{\text{in}}\frac{R2}{R1+R_2}\) from a single‑loop equation.
5. Plan an experiment to determine the internal resistance of a 1.5 V AA cell using the volt‑amp method. State the variables you would control, the quantities you would measure, and two possible sources of error.
6. For a Wheatstone bridge with \(R1=100\;\Omega\), \(R2=200\;\Omega\), \(R3=150\;\Omega\) and a galvanometer reading zero, calculate the unknown resistance \(Rx\).
7. Describe how the loop rule is modified when an inductor is present in a DC circuit that is switching on (i.e., when \(dI/dt\neq0\)).