Series and parallel circuits

1. Overview

Electrical components can be connected in two ways: series or parallel. Understanding how current, voltage (p.d.), and resistance behave in these arrangements is essential for designing circuits, from simple flashlights to complex household wiring.

Key Definitions

• Series Circuit: A circuit where components are connected one after another in a single loop.
• Parallel Circuit: A circuit where components are connected on separate branches, providing multiple paths for the current.
• Junction: A point in a parallel circuit where the circuit divides or combines.
• e.m.f. (Electromotive Force): The total energy supplied by a source (like a battery) to the electrical charge.
• Potential Difference (p.d.): The energy transferred per unit charge between two points in a circuit.

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Core Content

Current in Series

In a series circuit, there is only one path for the electrons to flow. Therefore, the current is the same at every point in the circuit.

• If an ammeter reads 2A near the battery, it will read 2A anywhere else in that loop.

Constructing Circuits

• Series: Connect components end-to-end. If one component breaks, the whole circuit stops working.
• Parallel: Connect components across each other. If one branch breaks, the others continue to function.
• 📊A simple series circuit with one battery and two lamps in a single loop vs. a parallel circuit with one battery and two lamps on separate branches.

Combined e.m.f. of Sources in Series

When multiple cells are connected in series (facing the same direction), their total e.m.f. is the sum of their individual e.m.f.s.

• Worked Example: Three 1.5V cells are connected in series.
  • Total e.m.f. = $1.5V + 1.5V + 1.5V = 4.5V$.

Combined Resistance in Series

The total resistance ($R_{Total}$) is the sum of all individual resistors.

• Formula: $R_{Total} = R_1 + R_2 + ...$
• Worked Example: A $5\Omega$ resistor and a $10\Omega$ resistor are in series.
  • $R_{Total} = 5 + 10 = 15\Omega$.

Current in Parallel

• The current from the source (the "main" part of the circuit) is always larger than the current in each individual branch.
• The total current is shared between the branches.

Combined Resistance in Parallel

• When resistors are connected in parallel, the total resistance is less than the resistance of the smallest individual resistor.
• This happens because you are providing more "paths" for the current to flow through.

Advantages of Parallel Circuits for Lighting

1. Independent Control: Each lamp can be switched on or off independently.
2. Constant Voltage: Every lamp receives the full voltage of the source, so they all shine brightly.
3. Reliability: If one lamp blows, the rest of the circuit stays closed and the other lamps remain lit.

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Extended Content (Extended Only)

Conservation of Charge at Junctions

Current is the flow of electrons. Because charge cannot be created or destroyed, the sum of currents entering a junction must equal the sum of currents leaving it.

• If 5A enters a junction and splits into two branches, and one branch has 2A, the other must have 3A ($5 - 2 = 3$).

Potential Difference in Series

The total p.d. supplied by the source is shared between the components.

• $V_{Source} = V_1 + V_2 + ...$
• Worked Example: A 12V battery is connected to two identical resistors in series. Each resistor will have a p.d. of 6V.

Potential Difference in Parallel

The p.d. across each branch in a parallel arrangement is the same as the p.d. across the whole arrangement.

• If a 9V battery is connected to three parallel branches, each branch has exactly 9V across it.

Calculating Parallel Resistance

For two resistors in parallel, use the reciprocal formula: $$\frac{1}{R_{Total}} = \frac{1}{R_1} + \frac{1}{R_2}$$

• Worked Example: Calculate the resistance of a $4\Omega$ and $6\Omega$ resistor in parallel.
  1. $\frac{1}{R_{Total}} = \frac{1}{4} + \frac{1}{6}$
  2. $\frac{1}{R_{Total}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$
  3. $R_{Total} = \frac{12}{5} = 2.4\Omega$

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Key Equations

Equation	Symbols	Units
$R_s = R_1 + R_2$	$R$: Resistance	Ohms ($\Omega$)
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$	$R$: Resistance	Ohms ($\Omega$)
$I_{in} = I_{out}$	$I$: Current	Amperes (A)
$V_{total} = V_1 + V_2$ (Series)	$V$: Potential Difference	Volts (V)

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Common Mistakes to Avoid

• ❌ Wrong: Thinking current is "used up" as it goes around a circuit.
• ✓ Right: Current is the same at all points in a series circuit; energy (p.d.) is what is transferred.
• ❌ Wrong: Thinking adding a resistor in parallel increases the total resistance.
• ✓ Right: Adding a parallel resistor decreases total resistance because you are adding a new path for the current.
• ❌ Wrong: Forgetting to "flip" the fraction at the end of a parallel resistance calculation.
• ✓ Right: Always calculate $1 \div (\text{your answer})$ to get $R_{Total}$.
• ❌ Wrong: Assuming the current in a branch can be higher than the total supply current.
• ✓ Right: The supply current is always the sum of all branch currents.

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Exam Tips

1. The "Smaller than Smallest" Check: After calculating parallel resistance, always check if your answer is smaller than the lowest value resistor in that arrangement. If it isn't, you've made a calculation error.
2. Identify Junctions: When looking at a complex diagram, circle the junctions. This helps you identify which components are in parallel and where the current will split.
3. Follow the Path: To check if components are in series, trace the path from the battery. If your finger never has to choose between two paths, those components are in series.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student sets up a circuit to investigate the current through two resistors connected in parallel.

(a) State two advantages of connecting lamps in parallel rather than in series in a lighting circuit. [2]

(b) The student measures the current through resistor R1 as 0.5 A and the current through resistor R2 as 0.3 A. Calculate the total current supplied by the battery. [1]

(c) Explain why the total current supplied by the battery is larger than the current in each branch. [2]

Worked Solution:

(a)

1. If one lamp fails in a parallel circuit, the other lamps will continue to work. This is because each lamp has its own independent path for current.
2. Lamps in parallel have the same brightness/potential difference across them. This is because the potential difference across each branch in a parallel circuit is the same.

How to earn full marks:

• State that other lamps remain lit if one fails.
• State that all lamps have the same brightness.

(b)

1. The total current is the sum of the currents in each branch. $I_{total} = I_1 + I_2 = 0.5 A + 0.3 A = 0.8 A$ $I_{total} = \boxed{0.8 A}$

How to earn full marks:

• Correctly add the two currents.
• Correct unit given.

(c)

1. In a parallel circuit, the total current splits between the branches. Each branch provides a path for current to flow.
2. The total resistance of the parallel combination is less than the resistance of any individual branch. Therefore, for a given voltage, the total current is larger than the current in each branch because the total resistance is less.

How to earn full marks:

• State that the total current splits between the branches.
• State that the total resistance is less, leading to a larger total current.

Common Pitfall: Students sometimes think current is "used up" as it flows through a circuit. Remember that current is the flow of charge, and charge is conserved. The total current entering a junction must equal the total current leaving it.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A potential divider circuit is set up as shown in the diagram. The battery has an e.m.f. of 9.0 V. Resistor R1 has a resistance of 2000 $\Omega$ and resistor R2 has a resistance of 1000 $\Omega$.

(a) Calculate the total resistance of the circuit. [2]

(b) Calculate the current flowing through the circuit. [2]

(c) Calculate the potential difference across resistor R2. [2]

(d) A light-dependent resistor (LDR) is used to replace R2. Describe how the potential difference across the LDR changes as the light intensity increases. Explain your answer. [2]

Worked Solution:

(a)

1. The total resistance in a series circuit is the sum of individual resistances. $R_{total} = R_1 + R_2 = 2000 \Omega + 1000 \Omega = 3000 \Omega$ $R_{total} = \boxed{3000 \Omega}$

How to earn full marks:

• Correctly add the two resistances.
• Correct unit given.

(b)

1. Using Ohm's Law: V = IR, so I = V/R $I = \frac{V}{R} = \frac{9.0 V}{3000 \Omega} = 0.003 A$ $I = \boxed{0.003 A}$

How to earn full marks:

• Correctly substitute values into Ohm's Law.
• Correct unit given.

(c)

1. Using Ohm's Law: V = IR $V_2 = I \times R_2 = 0.003 A \times 1000 \Omega = 3.0 V$ $V_2 = \boxed{3.0 V}$

How to earn full marks:

• Correctly substitute values into Ohm's Law, using the current from part (b) and the resistance of R2.
• Correct unit given. ECF from (b) is allowed.

(d)

1. As light intensity increases, the resistance of the LDR decreases. This is a property of light-dependent resistors.
2. Since the resistance of the LDR decreases, the potential difference across it also decreases. The potential difference is divided proportionally to the resistances. As the LDR's resistance drops, it takes a smaller share of the total potential difference.

How to earn full marks:

• State that the resistance of the LDR decreases with increasing light intensity.
• State that the potential difference across the LDR decreases.

Common Pitfall: Students often forget the relationship between light intensity and LDR resistance. Also, remember that in a potential divider, the voltage is divided proportionally to the resistances.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

Two resistors are connected in parallel to a 6.0 V battery. Resistor R1 has a resistance of 12 $\Omega$ and resistor R2 has a resistance of 6 $\Omega$.

(a) State that the potential difference across an arrangement of parallel resistances is the same as the potential difference across one branch in the arrangement of the parallel resistances. [1]

(b) Calculate the current through resistor R1. [2]

(c) Calculate the current through resistor R2. [2]

(d) Calculate the total current supplied by the battery. [1]

Worked Solution:

(a)

1. The potential difference across each branch in a parallel circuit is the same.

How to earn full marks:

• States that the potential difference across each branch in a parallel circuit is the same.

(b)

1. Using Ohm's Law: V = IR, so I = V/R $I_1 = \frac{V}{R_1} = \frac{6.0 V}{12 \Omega} = 0.5 A$ $I_1 = \boxed{0.5 A}$

How to earn full marks:

• Correctly substitute values into Ohm's Law.
• Correct unit given.

(c)

1. Using Ohm's Law: V = IR, so I = V/R $I_2 = \frac{V}{R_2} = \frac{6.0 V}{6 \Omega} = 1.0 A$ $I_2 = \boxed{1.0 A}$

How to earn full marks:

• Correctly substitute values into Ohm's Law.
• Correct unit given.

(d)

1. The total current is the sum of the currents in each branch. $I_{total} = I_1 + I_2 = 0.5 A + 1.0 A = 1.5 A$ $I_{total} = \boxed{1.5 A}$

How to earn full marks:

• Correctly add the two currents.
• Correct unit given. ECF from (b) and (c) is allowed.

Common Pitfall: A common mistake is to forget that the voltage is the same across all components in parallel. Also, be careful with units and make sure to include them in your final answer.

Exam-Style Question 4 — Extended Response [10 marks]

Question:

Three resistors, R1, R2, and R3, are connected in a circuit as shown. The battery has an e.m.f. of 12.0 V and negligible internal resistance. R1 = 4.0 $\Omega$, R2 = 6.0 $\Omega$, and R3 = 12.0 $\Omega$. R2 and R3 are connected in parallel, and that combination is in series with R1.

(a) Calculate the combined resistance of R2 and R3. [3]

(b) Calculate the total resistance of the entire circuit. [2]

(c) Calculate the current supplied by the battery. [2]

(d) Calculate the potential difference across resistor R1. [1]

(e) Determine the current flowing through resistor R2. [2]

Worked Solution:

(a)

1. For resistors in parallel: $\frac{1}{R_{parallel}} = \frac{1}{R_2} + \frac{1}{R_3}$ This is the formula for calculating the reciprocal of the combined resistance of parallel resistors.
2. Substituting the values: $\frac{1}{R_{parallel}} = \frac{1}{6.0 \Omega} + \frac{1}{12.0 \Omega} = \frac{2}{12.0 \Omega} + \frac{1}{12.0 \Omega} = \frac{3}{12.0 \Omega}$ Finding the sum of the reciprocals.
3. Therefore, $R_{parallel} = \frac{12.0 \Omega}{3} = 4.0 \Omega$ $R_{parallel} = \boxed{4.0 \Omega}$ Taking the reciprocal to find the combined resistance.

How to earn full marks:

• Correctly use the parallel resistance formula.
• Correctly substitute the values.
• Correct final answer with the correct unit.

(b)

1. The combined resistance of the parallel section is in series with R1, so we add the resistances. $R_{total} = R_1 + R_{parallel} = 4.0 \Omega + 4.0 \Omega = 8.0 \Omega$ $R_{total} = \boxed{8.0 \Omega}$

How to earn full marks:

• Correctly add R1 to the combined parallel resistance from part (a). ECF from (a) is allowed.
• Correct final answer with the correct unit.

(c)

1. Using Ohm's Law: V = IR, so I = V/R $I = \frac{V}{R_{total}} = \frac{12.0 V}{8.0 \Omega} = 1.5 A$ $I = \boxed{1.5 A}$

How to earn full marks:

• Correctly substitute values into Ohm's Law, using the total voltage and total resistance.
• Correct final answer with the correct unit. ECF from (b) is allowed.

(d)

1. Using Ohm's Law: V = IR $V_1 = I \times R_1 = 1.5 A \times 4.0 \Omega = 6.0 V$ $V_1 = \boxed{6.0 V}$

How to earn full marks:

• Correctly substitute values into Ohm's Law, using the current from part (c) and the resistance of R1.
• Correct final answer with the correct unit. ECF from (c) is allowed.

(e)

1. The potential difference across the parallel combination of R2 and R3 is the same. The voltage across R1 is 6V, so the voltage across the parallel section must be 12-6 = 6V. $V_{parallel} = 12.0 V - 6.0 V = 6.0 V$ This is because the total voltage across the series circuit is the sum of the voltage across each component.
2. Using Ohm's Law: I = V/R $I_2 = \frac{V_{parallel}}{R_2} = \frac{6.0 V}{6.0 \Omega} = 1.0 A$ $I_2 = \boxed{1.0 A}$

How to earn full marks:

• Correctly calculate the potential difference across the parallel section.
• Correctly substitute values into Ohm's Law, using the voltage across the parallel section and the resistance of R2.
• Correct final answer with the correct unit.

Common Pitfall: When dealing with combined series-parallel circuits, it's crucial to break down the problem step-by-step. First, find the equivalent resistance of the parallel section, then treat that as a single resistor in series with the other components. Also, remember that the current through the series resistor (R1) is the total current supplied by the battery.