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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 \( \mathrm{9 \, V} \).

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