4.3.2 Series and Parallel Circuits
Objective: Know how to construct and use series and parallel circuits. 🚀
Series Circuits – The “One‑Way Street” Analogy
Imagine a single lane road where all cars must travel one after another.
In a series circuit, the current flows through each component one after the other – just like cars on a single lane.
The total resistance is the sum of all individual resistances:
\$R{\text{total}} = R1 + R2 + R3 + \dots\$
The voltage supplied by the battery is divided across each component:
\$V{\text{total}} = V1 + V2 + V3 + \dots\$
The current is the same through every component.
Parallel Circuits – The “Multi‑Lane Highway” Analogy
Picture a highway with several lanes, each lane carrying its own cars.
In a parallel circuit, the current splits into different paths, each path containing a component.
The voltage across each component is the same:
\$V{\text{total}} = V1 = V2 = V3 = \dots\$
The total resistance is found using:
\$\frac{1}{R{\text{total}}} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} + \dots\$
The total current is the sum of the currents in each branch.
Key Differences at a Glance
- Current: same in series, splits in parallel.
- Voltage: splits in series, same across each in parallel.
- Resistance: additive in series, decreases in parallel.
- Failure: one component fails → whole series circuit stops; parallel circuit still works.
Calculating with Series Circuits
- List all resistances.
- Sum them to get \(R_{\text{total}}\).
- Use Ohm’s law \(I = \frac{V}{R_{\text{total}}}\) to find current.
- Find voltage drop across each resistor: \(Vi = I \times Ri\).
Calculating with Parallel Circuits
- List all resistances.
- Compute reciprocal sum: \(\frac{1}{R{\text{total}}} = \sum \frac{1}{Ri}\).
- Find \(R_{\text{total}}\) by taking reciprocal.
- Use Ohm’s law \(I = \frac{V}{R_{\text{total}}}\) for total current.
- Find current in each branch: \(Ii = \frac{V}{Ri}\).
Example 1 – Series Circuit
Three resistors: \(R1 = 4\,\Omega\), \(R2 = 6\,\Omega\), \(R_3 = 10\,\Omega\).
Battery voltage \(V = 12\,\text{V}\).
- \(R_{\text{total}} = 4 + 6 + 10 = 20\,\Omega\)
- Current: \(I = \frac{12}{20} = 0.60\,\text{A}\)
- Voltage drops:
\(V_1 = 0.60 \times 4 = 2.4\,\text{V}\)
\(V_2 = 0.60 \times 6 = 3.6\,\text{V}\)
\(V_3 = 0.60 \times 10 = 6.0\,\text{V}\)
Example 2 – Parallel Circuit
Two resistors: \(R1 = 8\,\Omega\), \(R2 = 12\,\Omega\).
Battery voltage \(V = 9\,\text{V}\).
- \(\frac{1}{R_{\text{total}}} = \frac{1}{8} + \frac{1}{12} = 0.125 + 0.0833 = 0.2083\)
- \(R_{\text{total}} = \frac{1}{0.2083} \approx 4.80\,\Omega\)
- Total current: \(I = \frac{9}{4.80} \approx 1.88\,\text{A}\)
- Branch currents:
\(I_1 = \frac{9}{8} = 1.125\,\text{A}\)
\(I_2 = \frac{9}{12} = 0.75\,\text{A}\)
Exam Tips 📚
- Always check whether the circuit is series or parallel before calculating.
- For series, remember “add resistances”. For parallel, remember “add reciprocals”.
- Use Ohm’s law in the form that matches the known values (e.g., \(I = V/R\) or \(V = IR\)).
- When a component fails, the whole series circuit stops; a parallel circuit keeps running.
- Practice converting between total resistance and individual resistances to build confidence.
Quick Reference Table
| Circuit Type | Total Resistance | Voltage Distribution | Current Distribution |
|---|
| Series | \(R{\text{total}} = \sum Ri\) | \(V{\text{total}} = \sum Vi\) | \(I\) same through all components |
| Parallel | \(\displaystyle \frac{1}{R{\text{total}}} = \sum \frac{1}{Ri}\) | \(V{\text{total}} = Vi\) for all branches | \(I{\text{total}} = \sum Ii\) |