CIE iGCSE Co-ordinated Sciences-P4.3.2 Series and parallel circuits- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P4.3.2 Series and parallel circuits – Study Notes
CIE iGCSE Co-ordinated Sciences-P4.3.2 Series and parallel circuits – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
Core
- Know that the current at every point in a series circuit is the same
- Know how to construct and use series and parallel circuits
- Calculate the combined resistance of two or more resistors in series
- Know the advantages of connecting lamps in parallel in a circuit
- Know that, for a parallel circuit, the current from the source is larger than the current in each branch
- Know that the combined resistance of two resistors in parallel is less than that of either resistor by itself
Supplement
- Recall and use in calculations, the fact that:
(a) the sum of the currents entering a junction in a parallel circuit is equal to the sum of the currents that leave the junction
(b) the total p.d. across the components in a series circuit is equal to the sum of the individual p.d.s across each component
(c) the p.d. across each branch of a parallel arrangement of components is the p.d. across the whole arrangement
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Current in a Series Circuit
In a series circuit, the current is the same at every point in the circuit. This happens because there is only one path for charges to flow.
Key Points:
- Same number of electrons pass through each component per second.
- No matter where you measure, the current is identical.
- Current is measured in amperes (A).
Formula for Current:
Current is defined as:
$\mathrm{I = \dfrac{Q}{t}}$
Where:
- \(\mathrm{I}\) = Current (A)
- \(\mathrm{Q}\) = Charge (Coulombs)
- \(\mathrm{t}\) = Time (s)
Important Understanding:
- If current at the battery is \( \mathrm{0.3 \, A} \), then current everywhere in the loop is also \( \mathrm{0.3 \, A} \).
- Unlike voltage, current does not split in series circuits.
Example :
A battery provides a current of \( \mathrm{0.5 \, A} \) in a series circuit with two resistors. What is the current through each resistor?
▶️ Answer/Explanation
Step 1: In series, current is the same everywhere.
Step 2: Current through resistor 1 = \( \mathrm{0.5 \, A} \).
Step 3: Current through resistor 2 = \( \mathrm{0.5 \, A} \).
Final Answer: Current = \( \mathrm{0.5 \, A} \) through each resistor.
Series Circuits
In a series circuit, components are connected end-to-end, so there is only one path for the current to flow.
Key Points:
- The current is the same</strong at every point in the circuit.
- The total voltage supplied by the battery is shared across all components.
- If one component breaks, the whole circuit stops working.
Constructing and Using Series Circuits:
- Connect components end-to-end (in a single loop).
- Use an ammeter in series to measure current.
- Use a voltmeter in parallel across a component to measure voltage across it.
Combined Resistance in Series:
When resistors are connected in series, their resistances add together:
\(\mathrm{R_{total} = R_1 + R_2 + R_3 + \dots}\)
Where:
- \(\mathrm{R_{total}}\) = Total resistance (\(\Omega\))
- \(\mathrm{R_1, R_2, R_3}\) = Individual resistances (\(\Omega\))
Example :
Three resistors of \( \mathrm{4 \, \Omega} \), \( \mathrm{6 \, \Omega} \), and \( \mathrm{10 \, \Omega} \) are connected in series. Calculate the total resistance.
▶️ Answer/Explanation
Step 1: Use the series formula: \(\mathrm{R_{total} = R_1 + R_2 + R_3}\).
Step 2: \(\mathrm{R_{total} = 4 + 6 + 10 = 20 \, \Omega}\).
Final Answer: Total resistance = \( \mathrm{20 \, \Omega} \).
Parallel Circuits
In a parallel circuit, components are connected across the same two points, so the current has multiple paths to flow.
Key Points:
- The total current is shared between the branches.
- The voltage across each branch is the same as the supply voltage.
- If one component breaks, current can still flow through other branches.
Constructing and Using Parallel Circuits:
- Connect components so that each one has its own separate branch.
- Use an ammeter in series with each branch to measure the current in that branch.
- Use a voltmeter across a branch to measure the voltage (same as supply).
Combined Resistance in Parallel:
When resistors are connected in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances:
\(\mathrm{\dfrac{1}{R_{total}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots}\)
Where:
- \(\mathrm{R_{total}}\) = Total resistance (\(\Omega\))
- \(\mathrm{R_1, R_2, R_3}\) = Individual resistances (\(\Omega\))
Example :
Two resistors of \( \mathrm{6 \, \Omega} \) and \( \mathrm{12 \, \Omega} \) are connected in parallel. Calculate the total resistance.
▶️ Answer/Explanation
Step 1: Use the parallel formula: \(\mathrm{\dfrac{1}{R_{total}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}\).
Step 2: \(\mathrm{\dfrac{1}{R_{total}} = \dfrac{1}{6} + \dfrac{1}{12}}\).
Step 3: \(\mathrm{\dfrac{1}{R_{total}} = \dfrac{2}{12} + \dfrac{1}{12} = \dfrac{3}{12}}\).
Step 4: \(\mathrm{R_{total} = \dfrac{12}{3} = 4 \, \Omega}\).
Final Answer: Total resistance = \( \mathrm{4 \, \Omega} \).
Advantages of Connecting Lamps in Parallel
Each lamp gets the full supply voltage, so they all shine equally bright.
- If one lamp breaks, the others continue to work (independent branches).
- Switches can be added in each branch to control lamps separately.
Current in a Parallel Circuit:
The total current from the source is the sum of the currents in the separate branches.
- Therefore, the source current is always larger than the current in each branch.
\(\mathrm{I_{total} = I_1 + I_2 + I_3 + \dots}\)
Combined Resistance in Parallel:
The combined resistance of two resistors in parallel is less than the resistance of either resistor on its own.
- This is because adding a parallel branch provides an extra path for current.
\(\mathrm{\dfrac{1}{R_{total}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots}\)
Example :
Two lamps are connected in parallel. The current in lamp 1 is \( \mathrm{0.4 \, A} \), and in lamp 2 it is \( \mathrm{0.6 \, A} \). Calculate the total current from the source.
▶️ Answer/Explanation
Step 1: Use the rule: \(\mathrm{I_{total} = I_1 + I_2}\).
Step 2: \(\mathrm{I_{total} = 0.4 + 0.6 = 1.0 \, A}\).
Final Answer: Total current = \( \mathrm{1.0 \, A} \).
Division of Current in Parallel:
When two resistors are connected in parallel, the current divides inversely proportional to their resistances.
The branch with the smaller resistance carries the larger current.
\(\mathrm{I_1 = \dfrac{R_2}{R_1 + R_2} \times I_{total}}\)
\(\mathrm{I_2 = \dfrac{R_1}{R_1 + R_2} \times I_{total}}\)
Example :
Two resistors, \( \mathrm{R_1 = 4 \, \Omega} \) and \( \mathrm{R_2 = 6 \, \Omega} \), are connected in parallel. The total current supplied is \( \mathrm{5.0 \, A} \). Calculate the current through each resistor.
▶️ Answer/Explanation
Step 1: Use current division formulas:
\(\mathrm{I_1 = \dfrac{R_2}{R_1 + R_2} \times I_{total}}\)
\(\mathrm{I_2 = \dfrac{R_1}{R_1 + R_2} \times I_{total}}\)
Step 2: For \( \mathrm{I_1} \): \(\mathrm{I_1 = \dfrac{6}{4 + 6} \times 5 = \dfrac{6}{10} \times 5 = 3.0 \, A}\).
Step 3: For \( \mathrm{I_2} \): \(\mathrm{I_2 = \dfrac{4}{4 + 6} \times 5 = \dfrac{4}{10} \times 5 = 2.0 \, A}\).
Step 4: Check: \(\mathrm{I_{total} = I_1 + I_2 = 3.0 + 2.0 = 5.0 \, A}\) ✅
Final Answer: Current in \( \mathrm{R_1} = 3.0 \, A\), Current in \( \mathrm{R_2} = 2.0 \, A\).
(a) Currents at a Junction in a Parallel Circuit
Rule: At any junction in a parallel circuit, the total current entering the junction is equal to the total current leaving it.
\(\mathrm{I_{in} = I_{out}}\)
Explanation:
- Charge is conserved, so no current is lost at a junction.
- The source current splits between the branches and then recombines.
Example :
A current of \( \mathrm{2.0 \, A} \) flows into a junction. It splits into two branches: \( \mathrm{1.2 \, A} \) in branch 1 and \( \mathrm{0.8 \, A} \) in branch 2. Show that current is conserved.
▶️ Answer/Explanation
Step 1: Current entering = \( \mathrm{2.0 \, A} \).
Step 2: Current leaving = \( \mathrm{1.2 + 0.8 = 2.0 \, A} \).
Final Answer: \( \mathrm{I_{in} = I_{out}} \), so current is conserved.
(b) Potential Difference in a Series Circuit
Rule: In a series circuit, the total potential difference (p.d.) across all components is equal to the sum of the individual p.d.s across each component.
\(\mathrm{V_{total} = V_1 + V_2 + V_3 + \dots}\)
Explanation:
- The supply voltage is shared between the components in proportion to their resistances.
- The more resistance a component has, the bigger the share of the p.d. it receives.
Example :
A battery supplies \( \mathrm{12 \, V} \) across two resistors in series. The p.d. across resistor 1 is \( \mathrm{7 \, V} \). Calculate the p.d. across resistor 2.
▶️ Answer/Explanation
Step 1: Use the rule: \(\mathrm{V_{total} = V_1 + V_2}\).
Step 2: \(\mathrm{12 = 7 + V_2}\).
Step 3: \(\mathrm{V_2 = 12 – 7 = 5 \, V}\).
Final Answer: Potential difference across resistor 2 = \( \mathrm{5 \, V} \).
(c) Potential Difference in a Parallel Circuit
Rule: In a parallel circuit, the p.d. across each branch is the same as the p.d. across the supply (the whole arrangement).
\(\mathrm{V_{supply} = V_{branch1} = V_{branch2} = \dots}\)
Explanation:
- Each branch is directly connected across the battery terminals.
- Therefore, every branch receives the full supply voltage, regardless of its resistance.
Example :
A battery provides \( \mathrm{9 \, V} \) to a parallel circuit with three lamps. What is the p.d. across each lamp?
▶️ Answer/Explanation
Step 1: Rule: In parallel, \(\mathrm{V_{supply} = V_{branch}}\).
Step 2: Each branch has the full battery voltage.
Final Answer: Each lamp has a p.d. of \( \mathrm{9 \, V} \).