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

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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):
$$\displaystyle \sum_{k=1}^{n} I_k = 0$$

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:
$$\displaystyle \sum_{k=1}^{n} V_k = 0$$
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 $R_1,R_2,\dots,R_n$ 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: $V_k = I R_k$.
  3. Applying KVL round the loop, $$V = \sum_{k=1}^{n} V_k = I\sum_{k=1}^{n} R_k.$$
  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 $R_1,R_2,\dots,R_n$ in parallel across a source of voltage $V$.
  2. Each branch experiences the same voltage $V$, so the current in branch $k$ is $I_k = \dfrac{V}{R_k}$ (Ohm’s law).
  3. Applying KCL at the node where the branches join, $$I_{\text{total}} = \sum_{k=1}^{n} I_k = \sum_{k=1}^{n}\frac{V}{R_k}.$$
  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

Configuration Equivalent Resistance
Two resistors in series $R_{\text{eq}} = R_1 + R_2$
Three resistors in series $R_{\text{eq}} = R_1 + R_2 + R_3$
Two resistors in parallel $R_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2}$
Three resistors in parallel $\displaystyle \frac{1}{R_{\text{eq}}}= \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
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 $R_1$ and $R_2$ across a source $V_{\text{in}}$, the voltage across $R_2$ (the “output” voltage) is obtained from KVL: $$V_{\text{in}} = I(R_1+R_2) \quad\Longrightarrow\quad I = \frac{V_{\text{in}}}{R_1+R_2}.$$ Hence $$V_{\text{out}} = I R_2 = V_{\text{in}}\;\frac{R_2}{R_1+R_2}.$$

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 $R_1 = 4\;\Omega$, $R_2 = 6\;\Omega$, $R_3 = 12\;\Omega$ 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 $R_s$ in series with a parallel block of $R_1 = 4\;\Omega$ 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: $$R_{\text{total}} = R_s + R_{p}= 2\;\Omega + 3\;\Omega = 5\;\Omega.$$
  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): $$I_1 = \frac{V_{\text{across }p}}{R_1}= \frac{I_{\text{total}}R_{p}}{R_1}= \frac{2.4\times3}{4}=1.8\;\text{A},$$ $$I_2 = I_{\text{total}}-I_1 = 0.6\;\text{A}.$$

Example 3 – Potential Divider

Problem: A $12\;\text{V}$ source feeds a series pair $R_1 = 3\;\Omega$ and $R_2 = 9\;\Omega$. Determine the voltage across $R_2$.

  1. Use the divider formula: $$V_{R_2}= V_{\text{in}}\frac{R_2}{R_1+R_2}=12\;\text{V}\times\frac{9}{3+9}=12\times\frac{9}{12}=9\;\text{V}.$$

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: $R_{\text{eq}} = \sum R_k$.
  • Parallel: $\displaystyle \frac{1}{R_{\text{eq}}}= \sum \frac{1}{R_k}$.
  • Potential divider: $V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1+R_2}$.
  • 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 $R_1 = 3\;\Omega$ and $R_2 = 6\;\Omega$. 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 $R_1 = 2\;\Omega$ and $R_2 = 8\;\Omega$. 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 $R_s$ (2 Ω) which then splits into two parallel branches $R_1$ (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.

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