Grade 10 Physical Sciences
Term 2 · Weeks 1–2

Electric Circuits — Series and Parallel

Paper 1 — Physics · Grade 10

Building on your knowledge of voltage, current and Ohm's Law, this chapter examines how resistors behave when connected in series and in parallel. You will also investigate real batteries, which have an internal resistance that causes the terminal voltage to drop when current flows.

Week 1

8.1 Series Circuits

Define series circuitR_s = R₁ + R₂ + R₃Voltage divides in series: ε = V₁ + V₂Same current throughout series

Definition

Series circuit

A series circuit is a circuit in which the components are connected end-to-end in a single path so that the same current flows through each component.

Definition

Equivalent (total) resistance

The equivalent resistance of a combination of resistors is the single resistance that could replace the combination and carry the same current for the same potential difference.

Series CircuitEMF εR₁R₂Series: Rₜ = R₁ + R₂ · same current I through each resistorVoltage divides: ε = V₁ + V₂
Figure 8.1 — A series circuit with resistors R₁, R₂ and R₃ connected end-to-end. The same current I flows through every component. The total voltage of the source equals the sum of the individual voltage drops: ε = V₁ + V₂ + V₃.

Formula

Total resistance in series

Rs=R1+R2+R3R_s = R_1 + R_2 + R_3

R_s = total series resistance (Ω), R₁, R₂, R₃ = individual resistances (Ω)

SI unit: Ω

In a series circuit, there is only one path for current to follow. Because charge cannot accumulate anywhere in the circuit, the same current I passes through every resistor. The battery's voltage is 'shared out' (divided) among the resistors in proportion to their resistance: the larger the resistance, the bigger its share of the voltage. The voltage drops add up to the total EMF: ε = V₁ + V₂ + V₃. Each individual voltage drop is found using Ohm's Law: V = IR.

Worked Example

A 3 Ω and a 7 Ω resistor are connected in series to a 20 V battery. Calculate (a) the total resistance, (b) the current in the circuit, (c) the voltage across each resistor.

Given

  • R₁ = 3 Ω
  • R₂ = 7 Ω
  • ε = 20 V

Find

(a) R_total (b) I (c) V₁ and V₂

Solution

  1. 1(a) R_total = R₁ + R₂ = 3 + 7 = 10 Ω
  2. 2(b) I = ε / R_total = 20 / 10 = 2 A
  3. 3(c) V₁ = I × R₁ = 2 × 3 = 6 V
  4. 4 V₂ = I × R₂ = 2 × 7 = 14 V
  5. 5Check: V₁ + V₂ = 6 + 14 = 20 V = ε ✓
Answer: (a) R_total = 10 Ω (b) I = 2 A (c) V₁ = 6 V, V₂ = 14 V

Worked Example

Three resistors of 2 Ω, 4 Ω and 6 Ω are connected in series to a 24 V battery. Calculate (a) the total resistance, (b) the current, (c) the voltage across each resistor.

Given

  • R₁ = 2 Ω
  • R₂ = 4 Ω
  • R₃ = 6 Ω
  • ε = 24 V

Find

(a) R_total (b) I (c) V₁, V₂, V₃

Solution

  1. 1(a) R_total = R₁ + R₂ + R₃ = 2 + 4 + 6 = 12 Ω
  2. 2(b) I = ε / R_total = 24 / 12 = 2 A
  3. 3(c) V₁ = 2 × 2 = 4 V
  4. 4 V₂ = 2 × 4 = 8 V
  5. 5 V₃ = 2 × 6 = 12 V
  6. 6Check: 4 + 8 + 12 = 24 V = ε ✓
Answer: (a) R_total = 12 Ω (b) I = 2 A (c) V₁ = 4 V, V₂ = 8 V, V₃ = 12 V
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Practice Question

Two resistors, R₁ = 5 Ω and R₂ = 15 Ω, are connected in series to a battery with EMF ε = 40 V. Calculate (a) the total resistance, (b) the current in the circuit, (c) the voltage across each resistor.

(5 marks)

✓

Exam Tip

In a series circuit, if one component breaks (open-circuit), the single path for current is broken and ALL components stop working — like old-style Christmas lights where one broken bulb darkened the whole string.

Week 1

8.2 Parallel Circuits

Define parallel circuit1/R_p = 1/R₁ + 1/R₂Same voltage across parallel branchesCurrent divides in parallel: I = I₁ + I₂

Definition

Parallel circuit

A parallel circuit is a circuit in which the components are connected across the same two points, providing multiple paths for current flow. The same potential difference exists across each parallel branch.

Parallel CircuitR₁R₂Parallel: 1/Rₜ = 1/R₁ + 1/R₂ · same voltage V across each branchCurrent splits: I = I₁ + I₂
Figure 8.2 — A parallel circuit with resistors R₁ and R₂ connected between the same two nodes. The voltage V is the same across each branch. The total current from the battery equals the sum of the branch currents: I = I₁ + I₂.

Formula

Total resistance in parallel

1Rp=1R1+1R2+1R3\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

R_p = total parallel resistance (Ω), R₁, R₂ = individual branch resistances (Ω)

SI unit: Ω

In a parallel circuit, each branch is connected directly across the battery, so every branch has the same voltage across it. Current from the battery divides at the junction, with more current flowing through lower-resistance branches (I = V/R — smaller R means larger I). The total current drawn from the battery equals the sum of all branch currents: I_total = I₁ + I₂. Adding more branches gives the current more paths to flow, which REDUCES the total resistance — the total parallel resistance is always less than the smallest individual resistor.

Worked Example

A 6 Ω and a 12 Ω resistor are connected in parallel across a 12 V battery. Calculate (a) the total parallel resistance, (b) the total current from the battery, (c) the current through each branch.

Given

  • R₁ = 6 Ω
  • R₂ = 12 Ω
  • V = 12 V

Find

(a) R_p (b) I_total (c) I₁ and I₂

Solution

  1. 1(a) 1/R_p = 1/R₁ + 1/R₂ = 1/6 + 1/12 = 2/12 + 1/12 = 3/12
  2. 2 R_p = 12/3 = 4 Ω
  3. 3(b) I_total = V/R_p = 12/4 = 3 A
  4. 4(c) I₁ = V/R₁ = 12/6 = 2 A
  5. 5 I₂ = V/R₂ = 12/12 = 1 A
  6. 6Check: I₁ + I₂ = 2 + 1 = 3 A = I_total ✓
Answer: (a) R_p = 4 Ω (b) I_total = 3 A (c) I₁ = 2 A, I₂ = 1 A

Worked Example

Two identical 4 Ω resistors are connected in parallel. What is the total parallel resistance?

Given

  • R₁ = 4 Ω
  • R₂ = 4 Ω

Find

R_p

Solution

  1. 11/R_p = 1/4 + 1/4 = 2/4 = 1/2
  2. 2R_p = 2 Ω
Answer: R_p = 2 Ω (half of each individual resistor — as expected for two equal resistors in parallel)
⚠

Watch Out

The total resistance of a parallel combination is ALWAYS less than the smallest individual resistor. For example, 6 Ω and 12 Ω in parallel gives R_p = 4 Ω, which is less than 6 Ω. If you get a total parallel resistance larger than the smallest branch, you have made a calculation error.

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Practice Question

A 10 Ω and a 15 Ω resistor are connected in parallel across a 30 V battery. Calculate (a) the total parallel resistance, (b) the total current from the battery, (c) the current through each branch.

(6 marks)

Week 2

8.3 Internal Resistance and EMF

Define emfDefine internal resistanceε = V_terminal + V_internalV_terminal = ε − Ir
ε

Definition

Electromotive force (EMF, ε)

The electromotive force of a battery is the total energy supplied per unit charge by the battery; it equals the work done by the battery per coulomb of charge. ε = W/Q. SI unit: volt (V).

r

Definition

Internal resistance (r)

Internal resistance is the resistance of the battery or cell itself, caused by the electrolyte and internal components. It opposes the flow of current within the battery.

Definition

Terminal potential difference (V_terminal)

The terminal potential difference is the potential difference measured across the terminals of a battery when current is flowing through the circuit. It equals the EMF minus the voltage dropped across the internal resistance.

Formula

EMF and terminal voltage

ε=Vterminal+Ir\varepsilon = V_{\text{terminal}} + Ir

ε = EMF (V), V_terminal = terminal potential difference (V), I = current (A), r = internal resistance (Ω)

SI unit: V

A real battery is not a perfect voltage source — it has its own internal resistance r. When current I flows, the battery itself 'uses up' some of its energy driving current through r. The voltage lost inside the battery is V_internal = Ir. The voltage that appears across the external terminals is therefore V_terminal = ε − Ir. When no current flows (open circuit), V_terminal = ε. As the current drawn increases, V_terminal decreases — the battery's terminal voltage 'sags' under load.

Worked Example

A battery has EMF ε = 12 V and internal resistance r = 0,5 Ω. It is connected to an external resistance R = 2,5 Ω. Calculate (a) the current in the circuit, (b) the terminal potential difference.

Given

  • ε = 12 V
  • r = 0,5 Ω
  • R = 2,5 Ω

Find

(a) I (b) V_terminal

Solution

  1. 1(a) Total resistance = R + r = 2,5 + 0,5 = 3 Ω
  2. 2 I = ε / (R + r) = 12 / 3 = 4 A
  3. 3(b) V_terminal = ε − Ir = 12 − (4 × 0,5) = 12 − 2 = 10 V
  4. 4Check: V_terminal = IR = 4 × 2,5 = 10 V ✓
Answer: (a) I = 4 A (b) V_terminal = 10 V
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Practice Question

A battery has EMF ε = 9 V and internal resistance r = 1 Ω. It is connected to an external resistor R = 8 Ω. Calculate (a) the current in the circuit and (b) the terminal potential difference.

(4 marks)

🌍

Real World

Why do car headlights dim when you start the engine? The starter motor draws a very large current through the battery's internal resistance. This large Ir drop reduces the terminal voltage dramatically — sometimes from 12 V down to 8 V or less — causing the lights (connected across the terminals) to dim. Once the engine starts and the starter motor switches off, the current drops and the terminal voltage recovers.

Next

Matter and Its Classification

Electric Circuits — Series and Parallel Grade 10 Physical Sciences CAPS Notes | MathSciBuddy