use the formula for the combined resistance of two or more resistors in parallel

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

Learning Objectives

  • Recall the first (current) and second (voltage) Kirchhoff’s laws, including the sign conventions required by the syllabus.

  • Derive the combined‑resistance formulas for series and parallel networks using KCL and KVL.

  • Apply Kirchhoff’s laws to circuits that contain both series and parallel elements.

  • Use the potential‑divider relation and the concept of internal resistance (EMF vs terminal voltage).

  • Identify and avoid common mistakes when analysing resistive circuits.

1. Recall – Kirchhoff’s Laws (Syllabus 10.2)

First law – Kirchhoff’s Current Law (KCL)

The algebraic sum of currents at a junction is zero. Currents entering the node are taken as positive, currents leaving as negative (or vice‑versa, provided the same convention is kept consistently):

\$\displaystyle \sum{k=1}^{n} Ik = 0\$



Second law – Kirchhoff’s Voltage Law (KVL)

The algebraic sum of potential differences round any closed loop is zero. Traversing a loop, a rise in potential (e.g. moving from – to + of a source) is taken as +V, a drop across a resistor (current direction same as traversal) is taken as –V:

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


Example: For a loop that goes from the negative terminal of a 12 V battery, through a resistor \$R\$, and back to the positive terminal,

\$+12\;\text{V} \;-\; I R \;=\;0 \;\;\Longrightarrow\;\; I=\frac{12}{R}.\$

2. Key Concepts for Resistive Networks

  • Series connection: the same current flows through each resistor; the total voltage is the sum of the individual voltage drops.

  • Parallel connection: the same voltage appears across each branch; the total current is the sum of the branch currents.

  • Both ideas are direct consequences of KCL (for currents) and KVL (for voltages).

3. Derivation of Equivalent‑Resistance Formulas

3.1 Series‑resistance

  1. Consider \$n\$ resistors \$R1,R2,\dots,R_n\$ in series across a source of voltage \$V\$.

  2. Because the current \$I\$ is the same through each resistor, Ohm’s law gives the voltage drop on each branch: \$Vk = I Rk\$.

  3. Applying KVL round the loop,

    \$V = \sum{k=1}^{n} Vk = I\sum{k=1}^{n} Rk.\$

  4. Define the equivalent resistance \$R{\text{eq}}\$ by \$V = I R{\text{eq}}\$.

  5. Equating the two expressions for \$V\$ yields

    \$\boxed{R{\text{eq}} = \sum{k=1}^{n} R_k}.\$

3.2 Parallel‑resistance

  1. Consider \$n\$ resistors \$R1,R2,\dots,R_n\$ in parallel across a source of voltage \$V\$.

  2. Each branch experiences the same voltage \$V\$, so the current in branch \$k\$ is \$Ik = \dfrac{V}{Rk}\$ (Ohm’s law).

  3. Applying KCL at the node where the branches join,

    \$I{\text{total}} = \sum{k=1}^{n} Ik = \sum{k=1}^{n}\frac{V}{R_k}.\$

  4. Define \$R{\text{eq}}\$ by \$I{\text{total}} = \dfrac{V}{R_{\text{eq}}}\$.

  5. Eliminate \$V\$ to obtain the fundamental relation

    \$\boxed{\frac{1}{R{\text{eq}}}= \sum{k=1}^{n}\frac{1}{R_k}}.\$

4. Formula Summary

ConfigurationEquivalent Resistance
Two resistors in series\$R{\text{eq}} = R1 + R_2\$
Three resistors in series\$R{\text{eq}} = R1 + R2 + R3\$
Two resistors in parallel\$R{\text{eq}} = \dfrac{R1 R2}{R1 + R_2}\$
Three resistors in parallel\$\displaystyle \frac{1}{R{\text{eq}}}= \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}\$
General case ( \$n\$ resistors )\$\displaystyle \frac{1}{R{\text{eq}}}= \sum{k=1}^{n}\frac{1}{R_k}\$

5. Potential Divider (Syllabus 10.2 5)

For two series resistors \$R1\$ and \$R2\$ across a source \$V{\text{in}}\$, the voltage across \$R2\$ (the “output” voltage) is obtained from KVL:

\$V{\text{in}} = I(R1+R2) \quad\Longrightarrow\quad I = \frac{V{\text{in}}}{R1+R2}.\$

Hence

\$V{\text{out}} = I R2 = V{\text{in}}\;\frac{R2}{R1+R2}.\$

6. Internal Resistance of a Source (EMF vs Terminal Voltage)

A real battery is modelled as an ideal emf \$E\$ in series with an internal resistance \$r\$.

  • Terminal voltage (the voltage you measure across the external terminals) is

    \$V_{\text{term}} = E - I r,\$

    where \$I\$ is the current drawn from the battery.

  • When \$I=0\$ (open circuit) \$V_{\text{term}} = E\$ (the emf).

  • When the battery supplies a load \$R_{\text{L}}\$, the total current is

    \$I = \frac{E}{r + R_{\text{L}}}.\$

7. Worked Examples

Example 1 – Pure Parallel Network

Problem: Find \$R{\text{eq}}\$ for \$R1 = 4\;\Omega\$, \$R2 = 6\;\Omega\$, \$R3 = 12\;\Omega\$ in parallel.

  1. Write the reciprocal sum:

    \$\frac{1}{R_{\text{eq}}}= \frac{1}{4} + \frac{1}{6} + \frac{1}{12}.\$

  2. Common denominator \$12\$: \$\displaystyle \frac{1}{4}= \frac{3}{12},\; \frac{1}{6}= \frac{2}{12},\; \frac{1}{12}= \frac{1}{12}.\$

  3. Add: \$\displaystyle \frac{1}{R_{\text{eq}}}= \frac{3+2+1}{12}= \frac{6}{12}= \frac{1}{2}.\$

  4. Invert: \$\displaystyle R_{\text{eq}} = 2\;\Omega.\$

Example 2 – Series + Parallel Combination

Problem: A circuit consists of a \$2\;\Omega\$ resistor \$Rs\$ in series with a parallel block of \$R1 = 4\;\Omega\$ and \$R_2 = 12\;\Omega\$. The whole network is connected across a \$12\;\text{V}\$ battery (ideal source). Find the total resistance and the current supplied by the battery.

  1. First find the equivalent resistance of the parallel part:

    \$\frac{1}{R{p}} = \frac{1}{4} + \frac{1}{12}= \frac{3+1}{12}= \frac{4}{12}= \frac{1}{3}\;\;\Longrightarrow\;\;R{p}=3\;\Omega.\$

  2. Combine with the series resistor:

    \$R{\text{total}} = Rs + R_{p}= 2\;\Omega + 3\;\Omega = 5\;\Omega.\$

  3. Use Ohm’s law for the whole circuit:

    \$I{\text{total}} = \frac{V}{R{\text{total}}}= \frac{12\;\text{V}}{5\;\Omega}= 2.4\;\text{A}.\$

  4. Currents in the parallel branches (optional):

    \$I1 = \frac{V{\text{across }p}}{R1}= \frac{I{\text{total}}R{p}}{R1}= \frac{2.4\times3}{4}=1.8\;\text{A},\$

    \$I2 = I{\text{total}}-I_1 = 0.6\;\text{A}.\$

Example 3 – Potential Divider

Problem: A \$12\;\text{V}\$ source feeds a series pair \$R1 = 3\;\Omega\$ and \$R2 = 9\;\Omega\$. Determine the voltage across \$R_2\$.

  1. Use the divider formula:

    \$V{R2}= V{\text{in}}\frac{R2}{R1+R2}=12\;\text{V}\times\frac{9}{3+9}=12\times\frac{9}{12}=9\;\text{V}.\$

Example 4 – Internal Resistance

Problem: A \$1.5\;\text{V}\$ AA cell has an internal resistance \$r = 0.2\;\Omega\$. It supplies a lamp of \$R_{\text{L}} = 4.8\;\Omega\$. Find the terminal voltage and the current through the lamp.

  1. Total resistance: \$r+R_{\text{L}} = 0.2+4.8 = 5.0\;\Omega\$.

  2. Current: \$I = \dfrac{E}{r+R_{\text{L}}}= \dfrac{1.5}{5.0}=0.30\;\text{A}\$.

  3. Terminal voltage: \$V_{\text{term}} = E - I r = 1.5 - (0.30)(0.2)=1.44\;\text{V}.\$

8. Common Pitfalls

  • Adding resistances directly for parallel branches – only valid for series.

  • Assuming the voltage differs between parallel resistors; the voltage is identical across each branch (KVL).

  • Neglecting to count all currents when applying KCL, which leads to an incorrect total current.

  • Mixing up sign conventions in KVL (treating a rise as a drop or vice‑versa). Write a clear loop direction and stick to the +/– rule.

  • Rounding intermediate fractions too early – keep exact fractions until the final answer to avoid cumulative error.

9. Summary

KCL guarantees that the algebraic sum of currents at a node is zero, while KVL ensures the algebraic sum of potential differences around any closed loop is zero. Using these laws together with Ohm’s law leads directly to:

  • Series: \$R{\text{eq}} = \sum Rk\$.

  • Parallel: \$\displaystyle \frac{1}{R{\text{eq}}}= \sum \frac{1}{Rk}\$.

  • Potential divider: \$V{\text{out}} = V{\text{in}}\dfrac{R2}{R1+R_2}\$.

  • Internal resistance: \$V_{\text{term}} = E - I r\$.

Mastering these relations enables systematic analysis of any resistive network required by the Cambridge A‑Level syllabus.

10. Practice Questions

  1. Three resistors \$5\;\Omega\$, \$10\;\Omega\$ and \$20\;\Omega\$ are connected in parallel. Calculate \$R_{\text{eq}}\$.

  2. A \$12\;\text{V}\$ battery supplies a parallel network of \$R1 = 3\;\Omega\$ and \$R2 = 6\;\Omega\$. Determine the total current drawn from the battery.

  3. In a circuit, a \$2\;\Omega\$ resistor is in series with a parallel combination of \$4\;\Omega\$ and \$12\;\Omega\$. Find the total resistance of the circuit.

  4. A \$9\;\text{V}\$ source is connected to a series pair \$R1 = 2\;\Omega\$ and \$R2 = 8\;\Omega\$. What is the voltage across \$R_2\$?

  5. A \$6\;\text{V}\$ battery with internal resistance \$r = 0.5\;\Omega\$ drives a load \$R_{\text{L}} = 3\;\Omega\$. Find the terminal voltage and the power dissipated in the load.

Suggested diagram: a 12 V source feeding a series resistor \$Rs\$ (2 Ω) which then splits into two parallel branches \$R1\$ (4 Ω) and \$R_2\$ (12 Ω). Nodes A (input) and B (common return) are labelled to illustrate KCL at A and KVL around each loop.