P6: Series and Parallel Circuits
Rules for current, voltage and resistance in series and parallel circuits — and why parallel circuits are used in the home.
P6 Series And Parallel Circuits
📋 Key Definitions
Series circuit: A circuit where components are connected one after the other in a single loop. There is only one path for the current to flow.
Parallel circuit: A circuit where components are connected in separate branches. Each branch provides a different path for the current.
📊 Series vs Parallel — The Rules
| Property | Series Circuit | Parallel Circuit |
|---|---|---|
| Current | Same everywhere: I_total = I_1 = I_2 = I_3 | Shared between branches: I_total = I_1 + I_2 + I_3 |
| Potential difference | Shared across components: V_total = V_1 + V_2 + V_3 | Same across each branch: V_total = V_1 = V_2 = V_3 |
| Resistance | Total resistance = sum of individual resistances: R_total = R_1 + R_2 | Adding resistors in parallel reduces total resistance |
Adding resistors in parallel: When you add a resistor in parallel, you provide an extra path for current. This means the total resistance decreases because more current can flow overall.
Adding resistors in series: When you add a resistor in series, you make it harder for current to flow through the single loop. The total resistance increases.
🏠 Why Parallel Circuits Are Used in Homes
Domestic circuits are wired in parallel because: (1) each appliance gets the full mains voltage (230 V); (2) each appliance can be switched on or off independently; (3) if one appliance breaks, the others still work.
- In a series circuit, if one component breaks the whole circuit stops working
- In a parallel circuit, each branch is independent — one lamp blowing does not affect others
- Each appliance in a parallel circuit receives the same voltage (230 V from the mains)
📐 Series Resistance Formula
R_total = R_1 + R_2 + R_3 (for series circuits)
Parallel resistance: The total resistance of a parallel circuit is always less than the smallest individual resistance. For two resistors in parallel: 1/R_total = 1/R_1 + 1/R_2.
📝 Worked Examples
Worked Example 1 — Series Circuit (Current and Voltage)
Question: Two resistors are connected in series to a 12 V battery. The current through the circuit is 2 A. What is the current through each resistor? The voltage across R_1 is 4 V. What is the voltage across R_2?
Solution:
In series, current is the same everywhere: I = 2 A through both resistors.
In series, V_total = V_1 + V_2. So V_2 = 12 - 4 = 8 V.
Worked Example 2 — Series Circuit (Resistance)
Question: A 4 Ω resistor and a 6 Ω resistor are connected in series to a 10 V battery. Calculate the total resistance, the current in the circuit, and the voltage across each resistor.
Solution:
R_total = R_1 + R_2 = 4 + 6 = 10 Ω
I = V / R_total = 10 / 10 = 1 A
V across 4 Ω: V = I x R = 1 x 4 = 4 V
V across 6 Ω: V = I x R = 1 x 6 = 6 V (check: 4 + 6 = 10 V = V_total)
Worked Example 3 — Parallel Circuit (Current)
Question: Two resistors are connected in parallel to a 12 V supply. The current through R_1 is 3 A and the current through R_2 is 1 A. Calculate the total current from the supply.
Solution:
In parallel: I_total = I_1 + I_2 = 3 + 1 = 4 A
Note: each branch has the full 12 V across it (voltage is the same in parallel).
Worked Example 4 — Parallel Circuit (Voltage and Resistance)
Question: A 20 Ω and a 30 Ω resistor are connected in parallel to a 12 V supply. Calculate the current through each branch, the total current, and the total resistance.
Solution:
In parallel, each branch gets the full 12 V.
I_1 = V / R_1 = 12 / 20 = 0.6 A
I_2 = V / R_2 = 12 / 30 = 0.4 A
I_total = I_1 + I_2 = 0.6 + 0.4 = 1.0 A
R_total = V / I_total = 12 / 1.0 = 12 Ω (less than either individual resistance)
Worked Example 5 — Adding Resistors
Question: A circuit has a single 10 Ω resistor connected to a battery. A second 10 Ω resistor is added (a) in series and (b) in parallel. Calculate the total resistance in each case and explain the difference.
Solution:
(a) In series: R_total = 10 + 10 = 20 Ω — resistance doubles (harder for current to flow through one path with more resistance).
(b) In parallel: 1/R_total = 1/10 + 1/10 = 2/10 = 1/5. So R_total = 5 Ω — resistance halves (two paths for current, so more current flows overall).
Worked Example 6 — Mixed Problem
Question: Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series to a 15 V battery. Calculate the total resistance, the current, and the voltage across the 5 Ω resistor.
Solution:
R_total = 2 + 3 + 5 = 10 Ω
I = V / R_total = 15 / 10 = 1.5 A (same through all resistors in series)
V across 5 Ω = I x R = 1.5 x 5 = 7.5 V
Worked Example 7 — Domestic Parallel Circuit
Question: A house circuit has three lamps connected in parallel to a 230 V supply. Lamp 1 draws 0.5 A, Lamp 2 draws 0.25 A, and Lamp 3 draws 0.75 A. Calculate the total current drawn and explain what happens if Lamp 2 is switched off.
Solution:
I_total = 0.5 + 0.25 + 0.75 = 1.5 A
If Lamp 2 is switched off, its branch is broken but the other branches still work. I_total becomes 0.5 + 0.75 = 1.25 A. Each remaining lamp still receives 230 V. This is why parallel circuits are used in homes.
❓ Practice Questions
Q1: Foundation State the rules for current and voltage in a series circuit.
Q2: Foundation Two 6 Ω resistors are connected in series to a 12 V battery. Calculate the total resistance and the current in the circuit.
Q3: Foundation Explain why domestic circuits use parallel rather than series wiring. Give two reasons.
Q4: Higher A 10 Ω and a 15 Ω resistor are connected in parallel to a 6 V supply. Calculate the current through each branch and the total current from the supply.
Q5: Higher A circuit has a 12 Ω resistor and a 4 Ω resistor in series. The supply voltage is 24 V. Calculate: (a) the total resistance, (b) the current, (c) the voltage across each resistor.
Q6: Higher Explain why adding a resistor in parallel decreases the total resistance of the circuit.
✅ Answers
- In a series circuit: current is the same through all components (I_total = I_1 = I_2), and the total voltage is shared across the components (V_total = V_1 + V_2).
- R_total = 6 + 6 = 12 Ω. I = V / R = 12 / 12 = 1 A.
- 1) Each appliance receives the full mains voltage (230 V). 2) Each appliance can be switched on or off independently without affecting the others. If one breaks, the rest still work.
- I through 10 Ω = 6 / 10 = 0.6 A. I through 15 Ω = 6 / 15 = 0.4 A. I_total = 0.6 + 0.4 = 1.0 A.
- (a) R_total = 12 + 4 = 16 Ω. (b) I = 24 / 16 = 1.5 A. (c) V across 12 Ω = 1.5 x 12 = 18 V. V across 4 Ω = 1.5 x 4 = 6 V. Check: 18 + 6 = 24 V.
- Adding a resistor in parallel provides an additional path for current to flow. Even though the new path has its own resistance, the total current from the supply increases because current can now flow through both paths. Since I_total increases for the same voltage, R_total = V / I_total must decrease.
🎯 Exam Tips
- Remember the key difference: series = one loop, current same, voltage shared; parallel = multiple branches, current shared, voltage same.
- In series, adding a resistor increases total resistance. In parallel, adding a resistor decreases total resistance.
- When a question asks about domestic circuits, always mention parallel because each appliance gets full voltage and works independently.
- For series calculations: find total R first, then find I, then find V across each component.
- For parallel calculations: each branch gets full voltage, so calculate I through each branch separately, then add currents for the total.
- The total resistance in parallel is always less than the smallest individual resistance — use this to check your answer.
🔬 Required Practical
Required Practical: Investigating Resistance in Series and Parallel
Aim: To investigate how the total resistance of a circuit changes when resistors are added in series and in parallel.
Method: Connect two identical resistors in series with a battery and an ammeter. Measure the current and calculate the total resistance using R = V / I. Repeat with the same two resistors connected in parallel. Compare the total resistance in each arrangement. Vary the number of resistors and repeat.
Variables: IV: Circuit arrangement (series or parallel) and number of resistors, DV: Total resistance, Control: Supply voltage, same resistors used throughout
🔢 Maths Skills
Mathematical Skills
For series circuits: R_total = R₁ + R₂ + R₃. For parallel circuits with two resistors: 1/R_total = 1/R₁ + 1/R₂, then invert to find R_total. Always use V = IR to find unknowns in each branch. In series, find total R first, then I, then V across each component. In parallel, each branch has the full supply voltage.
Maths Example
A 12 Ω and a 6 Ω resistor are connected in parallel to a 24 V supply. Calculate the total resistance and the total current.
1/R_total = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4
R_total = 4 Ω
I_total = V / R_total = 24 / 4 = 6 A
⚠️ Common Misconceptions
Watch Out!
1. Wrong: Current is "used up" by components in a series circuit Correct: Current is the same through all components in series — it is energy that is transferred, not the current itself
2. Wrong: Parallel circuits have the same current everywhere Correct: Only the voltage is the same across each branch; the current splits at junctions — I_total = I₁ + I₂ + I₃
3. Wrong: Adding resistors in parallel increases the total resistance Correct: Adding resistors in parallel provides extra paths for current, so total resistance decreases — it is always less than the smallest individual resistance
✍️ 6-Mark Question
Extended Answer
6 marks: Explain why domestic circuits are wired in parallel rather than in series. Refer to current, voltage and what happens if one appliance fails.
In a parallel circuit, each appliance is connected in its own separate branch across the supply. This means each appliance receives the full mains voltage of 230 V, so they all operate at their correct power rating. Each appliance can be switched on or off independently without affecting the others. If one appliance breaks or a lamp blows, the circuit in that branch is broken but current still flows through the other branches, so the other appliances continue to work. In a series circuit, if one component fails the entire circuit is broken and nothing works. Also in series, the voltage is shared between components, so appliances would not receive the correct voltage and would not work properly. The current in series is the same through all components, meaning all appliances would have to be on at the same time.
Mark scheme: 1 mark — each appliance gets full 230 V, 1 mark — can be switched independently, 1 mark — if one fails others still work, 1 mark — series would share voltage, 1 mark — series: one failure stops all, 1 mark — series: all must be on together
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student sets up a circuit with two 10 Ω resistors and a 12 V battery. They measure the following:
| Arrangement | Total current (A) | V across R₁ (V) | V across R₂ (V) |
|---|---|---|---|
| Series | 0.6 | 6.0 | 6.0 |
| Parallel | 2.4 | 12.0 | 12.0 |
(a) Show that the series current measurement is consistent with R_total = R₁ + R₂.
(b) One of the resistors is replaced with a faulty 10 Ω resistor that has a much higher actual resistance. In which arrangement (series or parallel) would this fault be easier to detect from current readings alone? Justify your answer.
(c) The student's ammeter has a zero error of +0.05 A. Explain how this affects the calculated total resistance in both arrangements and how to correct for it.
Answers: (a) R_total = 10 + 10 = 20 Ω. I = V/R_total = 12/20 = 0.6 A — consistent. (b) Series — in series, total resistance changes significantly (R_total changes from 20 Ω), so current changes noticeably. In parallel, R_total is dominated by the good resistor (~10 Ω) so the change is smaller and harder to detect. (c) All current readings are 0.05 A too high, so calculated R_total is too low. Subtract 0.05 A from each reading before calculating resistance to correct the zero error.
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