4.3.2 Series and Parallel Circuits
Learning objectives (AO1 – AO3)
- Recall the definitions of voltage, current, resistance, power and e.m.f.
- State the qualitative relationships for current, voltage and resistance in series and in parallel (AO1).
- Derive and use the quantitative equations for series‑ and parallel‑combined resistance, e.m.f. and power (AO2).
- Apply Kirchhoff’s junction and loop rules to solve unfamiliar circuit problems (AO3).
- Predict the behaviour of a lighting circuit when lamps are wired in series or in parallel.
- Identify the practical advantages of parallel wiring for domestic lighting.
Reminder – standard circuit symbols (4.3.1)
| Component | Symbol |
|---|
| Cell / mains supply (e.m.f.) | χ (or a long‑short line pair) |
| Switch (open/closed) | □ / ■ |
| Resistor (or lamp) | ∥ (zig‑zag for resistor, circle with a filament for lamp) |
| Ammeter | A (connected in series) |
| Voltmeter | V (connected in parallel) |
Example schematic (cell, switch, lamp, ammeter and voltmeter) – students must be able to draw and interpret such diagrams.
Key definitions
- Voltage (V) – electrical potential difference between two points. Unit: volt (V).
- Current (I) – rate of flow of charge through a conductor. Unit: ampere (A).
- Resistance (R) – opposition to the flow of current. Unit: ohm (Ω).
- Power (P) – rate at which electrical energy is converted to other forms.
\[P = VI = I^{2}R = \frac{V^{2}}{R}\]
Unit: watt (W).
- Electromotive force (e.m.f.) – energy supplied per unit charge by a source. Measured in volts, symbol \(\mathcal{E}\) or \(E\).
Qualitative statements required by the syllabus (4.3.2)
- In a series circuit the same current flows through every component; the voltage divides in proportion to the resistances.
- In a parallel circuit the voltage across each branch is the same as the source voltage; the total current is the sum of the branch currents.
- The equivalent resistance of a series arrangement is larger than any individual resistance; for a parallel arrangement it is smaller than the smallest individual resistance.
- When ideal e.m.f. sources are connected in series their voltages add; when identical ideal sources are connected in parallel the terminal voltage equals the e.m.f. of each source.
Series circuits
- Current: \[I_{\text{total}} = I_{1}=I_{2}= \dots = I_{n}\]
- Voltage across each element: \[V_{k}=I_{\text{total}}R_{k}\qquad\text{and}\qquad V_{\text{total}}=\sum_{k=1}^{n}V_{k}\]
- Equivalent resistance: \[R_{\text{eq}} = \sum_{k=1}^{n}R_{k}\]
(for \(n\) identical resistors, \(R_{\text{eq}} = nR\)).
- Combined e.m.f.: \[E_{\text{eq}} = \sum_{k=1}^{n}E_{k}\]
- Potential‑divider use: In a series chain a known resistor can be used to obtain a required fraction of the source voltage:
\[V_{x}=V_{\text{source}}\frac{R_{x}}{R_{\text{total}}}\]
Parallel circuits
- Voltage across each branch: \[V_{1}=V_{2}= \dots =V_{n}=V_{\text{source}}\]
- Current in each branch: \[I_{k}= \frac{V_{\text{source}}}{R_{k}}\qquad\text{and}\qquad I_{\text{total}} = \sum_{k=1}^{n} I_{k}\]
- Equivalent resistance: \[\frac{1}{R_{\text{eq}}}= \sum_{k=1}^{n}\frac{1}{R_{k}}\]
(for \(n\) identical resistors, \(R_{\text{eq}} = \frac{R}{n}\)).
- Combined e.m.f.: When identical ideal sources are placed in parallel, the terminal voltage equals the e.m.f. of each source.
Kirchhoff’s rules (useful for unfamiliar contexts)
- Junction (current) rule: The algebraic sum of currents entering a junction equals the sum leaving it.
\[\sum I_{\text{in}} = \sum I_{\text{out}}\]
- Loop (voltage) rule: The algebraic sum of the potential differences around any closed loop is zero.
\[\sum V = 0\]
(take rises as positive, drops as negative).
Example: A 12 V battery supplies a series resistor \(R_{1}=4\;\Omega\) that branches into two parallel resistors \(R_{2}=6\;\Omega\) and \(R_{3}=12\;\Omega\). Using the junction rule at the node and the loop rule for each branch gives the same currents as obtained from the series/parallel formulas, confirming the rules.
Comparison of series and parallel formulas
| Quantity |
Series circuit |
Parallel circuit |
| Current through each element |
\(I_{\text{total}} = I_{1}=I_{2}=…\) |
\(I_{k}=V_{\text{source}}/R_{k}\) (different for each branch) |
| Voltage across each element |
\(V_{k}=I_{\text{total}}R_{k}\) (splits) |
\(V_{k}=V_{\text{source}}\) (same for all) |
| Equivalent resistance |
\(R_{\text{eq}}=\displaystyle\sum R_{k}\) |
\(\displaystyle\frac{1}{R_{\text{eq}}}= \sum\frac{1}{R_{k}}\) |
| Combined e.m.f. |
\(E_{\text{eq}}=\displaystyle\sum E_{k}\) |
\(E_{\text{eq}}=E_{k}\) (identical sources only) |
Worked example – three identical 60 W, 240 V lamps
For each lamp:
\[
R = \frac{V^{2}}{P}= \frac{240^{2}}{60}=9.60\times10^{2}\;\Omega\;(3\;\text{significant figures})
\]
1. Lamps in series
- Equivalent resistance: \(R_{\text{eq}} = 3R = 2.88\times10^{3}\;\Omega\).
- Total current: \[
I = \frac{V_{\text{source}}}{R_{\text{eq}}}= \frac{240}{2.88\times10^{3}} = 8.3\times10^{-2}\;\text{A}
\] (2 s.f.).
- Voltage across each lamp: \(V_{\text{lamp}} = IR = 0.083\times960 \approx 80\;\text{V}\).
- Power per lamp: \[
P_{\text{lamp}} = I^{2}R = (0.083)^{2}\times960 \approx 6.6\;\text{W}
\] – very dim.
2. Lamps in parallel
- Equivalent resistance: \[
\frac{1}{R_{\text{eq}}}= \frac{1}{960}+\frac{1}{960}+\frac{1}{960}= \frac{3}{960}
\;\;\Rightarrow\;\; R_{\text{eq}} = 3.20\times10^{2}\;\Omega
\] (2 s.f.).
- Total current: \[
I_{\text{total}} = \frac{240}{320}=7.5\times10^{-1}\;\text{A}
\] (2 s.f.).
- Current through each lamp: \[
I_{\text{lamp}} = \frac{240}{960}=2.5\times10^{-1}\;\text{A}
\] (2 s.f.).
- Power per lamp: \[
P_{\text{lamp}} = V I = 240 \times 0.25 = 60\;\text{W}
\] – full brightness.
Note on significant figures: All final answers are given to the same number of significant figures as the data supplied (here 2 s.f.).
Internal resistance of the supply: The calculations above assume an ideal source (\(r_{\text{int}}=0\)). In real mains circuits a small internal resistance exists, causing a tiny voltage drop when large currents flow; this is rarely required for IGCSE calculations but may appear in “unfamiliar context” questions.
Power loss in wiring
- For a conductor of resistance \(r\) carrying current \(I\), the heating loss is \(P_{\text{loss}} = I^{2}r\).
- In a parallel lighting circuit the total current is split among several branches, so each branch carries a smaller current and the associated \(I^{2}r\) loss in the branch conductors is reduced.
- This is one reason why domestic circuits use relatively thin (and cheaper) wiring for each lighting point.
Advantages of connecting lamps in parallel (lighting circuits)
- Uniform brightness: Every lamp receives the full mains voltage, so each operates at its rated power.
- Independent operation: Failure (open circuit) of one lamp does not affect the others.
- Control flexibility: Switches, dimmers or timers can be placed on individual branches without altering the operation of the remaining lamps.
- Reduced heating in conductors: The total current is divided, minimising I²R losses and allowing the use of thinner, cheaper wiring.
- Safety: Lower current in any single conductor reduces the risk of overheating and fire.
Practical skills linked to this topic (AO2 & AO3)
- Design & drawing: Produce a clear schematic of a lighting circuit showing cell, switch, lamp, ammeter and voltmeter (AO1).
- Construction: Build series and parallel arrangements of identical lamps (or resistors) on a breadboard.
- Measurement: Use a voltmeter to record the voltage across each lamp and an ammeter to record the total current. Record values in a table.
- Data analysis: Calculate the theoretical currents and voltages using the formulas, then compute percentage errors and discuss sources of discrepancy (wire resistance, internal resistance of the source, meter loading) (AO2).
- Experiment planning (AO3): Write a brief experimental plan before the activity, stating hypothesis, variables, safety precautions and method.
- Investigation of lamp failure: Remove one lamp from a parallel circuit, observe the effect on the remaining lamps, and explain the observation using the qualitative statements and Kirchhoff’s junction rule.
Summary
- Series: same current, voltage divides, \(R_{\text{eq}}\) larger, e.m.f. adds.
- Parallel: same voltage, current divides, \(R_{\text{eq}}\) smaller, terminal voltage equals each source (identical e.m.f.).
- Kirchhoff’s rules allow analysis of any combination of series and parallel elements.
- Parallel wiring is preferred for domestic lighting because it gives uniform brightness, independent operation, flexible control and reduced heating losses.
- Accurate prediction of circuit behaviour using the series‑ and parallel‑combination equations, together with careful measurement and evaluation, fulfills the requirements of the Cambridge IGCSE 0625 syllabus.