Understanding Electricity and Circuits for Grade 11 Physical Science

Electricity is one of the highest-mark topics in Grade 11 Physical Science, and it builds directly into Grade 12. Students who understand circuits well score well in Paper 1. Students who don't — even those who are strong in other areas — consistently lose marks here.

The good news: electricity is completely logical. Once you understand what's happening physically, the maths follows naturally. This guide builds from fundamentals up to exam-level questions, with worked examples at each stage.

The Three Fundamentals: Current, Voltage, Resistance

Before touching a single circuit, you need a firm grasp of these three quantities.

Electric current ($I$) is the rate of flow of charge through a conductor. Think of it as how many electrons are passing a point per second. Measured in amperes (A).

Potential difference / voltage ($V$) is the energy per unit charge — essentially the "push" driving the current around the circuit. Measured in volts (V). A battery is an energy source that maintains a potential difference between its terminals.

Resistance ($R$) is how much a component opposes the flow of current. A thick copper wire has almost no resistance. A light bulb filament has significant resistance — which is exactly how it produces heat and light. Measured in ohms ($\Omega$).

These three quantities are connected by Ohm's Law — the single most important equation in this section.

Ohm's Law: $V = IR$

$V = IR$

(equivalently: $R = \dfrac{V}{I}$, or $I = \dfrac{V}{R}$ — rearrange as needed)

Ohm's Law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant.

That last part matters. Ohm's Law only applies to ohmic conductors at constant temperature. A light bulb is non-ohmic because its resistance changes as it heats up — but Grade 11 exam questions usually specify "assume constant resistance."

Worked Example: Basic Ohm's Law

A resistor has a resistance of $15\ \Omega$. A current of 2 A flows through it. What is the potential difference across it?

$V = IR = 2 \times 15 = \mathbf{30\ \text{V}}$

Simple — but in circuit questions, the challenge is knowing which voltage and which current to use. That's where series and parallel circuits come in.

Series Circuits

In a series circuit, components are connected one after another in a single loop. There is only one path for current to flow.

Rules for series circuits:

| Quantity | Rule | |---|---| | Current ($I$) | Same through all components: $I_{\text{total}} = I_1 = I_2 = I_3$ | | Voltage ($V$) | Divides between components: $V_{\text{total}} = V_1 + V_2 + V_3$ | | Resistance ($R$) | Adds up: $R_{\text{total}} = R_1 + R_2 + R_3$ |

Think of it like a single pipe — the same water flows through every section, but pressure drops at each narrowing.

Worked Example: Series Circuit

Two resistors, $R_1 = 4\ \Omega$ and $R_2 = 6\ \Omega$, 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

a)

$R_{\text{total}} = R_1 + R_2 = 4 + 6 = \mathbf{10\ \Omega}$

b)

$I = \frac{V}{R} = \frac{20}{10} = \mathbf{2\ \text{A}}$

(This same 2 A flows through both resistors.)

c)

$V_1 = IR_1 = 2 \times 4 = \mathbf{8\ \text{V}}$

$V_2 = IR_2 = 2 \times 6 = \mathbf{12\ \text{V}}$

Check: $8 + 12 = 20\ \text{V}$ ✓ (Voltages must add up to the battery voltage.)

Parallel Circuits

In a parallel circuit, components are connected across the same two nodes, giving current multiple paths to flow.

Rules for parallel circuits:

| Quantity | Rule | |---|---| | Current ($I$) | Divides between branches: $I_{\text{total}} = I_1 + I_2 + I_3$ | | Voltage ($V$) | Same across all branches: $V_{\text{total}} = V_1 = V_2 = V_3$ | | Resistance ($R$) | $\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$ |

Think of parallel branches like lanes merging on a highway — more lanes means less total congestion (lower total resistance). This is why adding more appliances to a parallel circuit draws more current from the supply.

Worked Example: Parallel Circuit

$R_1 = 6\ \Omega$ and $R_2 = 12\ \Omega$ are connected in parallel. The supply voltage is 12 V. Calculate: a) The total resistance b) The current through each resistor c) The total current drawn

a)

$\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$

$R_{\text{total}} = \mathbf{4\ \Omega}$

(Total resistance is always less than the smallest individual resistor — if your answer is larger than $6\ \Omega$, something went wrong.)

b)

$I_1 = \frac{V}{R_1} = \frac{12}{6} = \mathbf{2\ \text{A}}$

$I_2 = \frac{V}{R_2} = \frac{12}{12} = \mathbf{1\ \text{A}}$

c)

$I_{\text{total}} = I_1 + I_2 = 2 + 1 = \mathbf{3\ \text{A}}$

Check: $I_{\text{total}} = \dfrac{V}{R_{\text{total}}} = \dfrac{12}{4} = 3\ \text{A}$ ✓

Electrical Power

Power ($P$) measures how quickly electrical energy is converted — to light, heat, motion, sound. The unit is the watt (W).

There are three equivalent formulas — which one you use depends on what information is given:

$P = VI$

$P = I^2 R$

$P = \frac{V^2}{R}$

• If you're given $V$ and $I$ → use $P = VI$
• If you're given $I$ and $R$ → use $P = I^2 R$
• If you're given $V$ and $R$ → use $P = \dfrac{V^2}{R}$

Worked Example: Power

An electric kettle draws 8 A from a 230 V supply. What is its power rating? How much energy does it use in 5 minutes?

$P = VI = 230 \times 8 = \mathbf{1840\ \text{W}\ (1.84\ \text{kW})}$

$E = P \times t = 1840 \times (5 \times 60) = 1840 \times 300 = \mathbf{552\,000\ \text{J}\ (552\ \text{kJ})}$

The unit of electrical energy is the joule (J). The kilowatt-hour (kWh) is used on electricity bills: $1\ \text{kWh} = 3\,600\,000\ \text{J}$.

EMF, Internal Resistance, and Terminal PD (Grade 11/12)

This is where many students first hit a wall. Let's demystify it.

A real battery isn't perfect. It has electromotive force (EMF, symbol $\varepsilon$) — the total energy per unit charge it can provide — but it also has internal resistance ($r$), a small resistance inside the battery itself.

When current flows, some voltage is "lost" inside the battery. What's left for the external circuit is the terminal potential difference ($V_{\text{terminal}}$).

$\varepsilon = V_{\text{terminal}} + Ir$

Or equivalently:

$V_{\text{terminal}} = \varepsilon - Ir$

The term $Ir$ is called the "lost volts" or voltage drop across the internal resistance.

What This Means in Practice

When a battery is connected to nothing (open circuit), no current flows, so $Ir = 0$ and $V_{\text{terminal}} = \varepsilon$. The voltmeter reads the full EMF.

When a battery is connected to a load (external resistance $R$), current flows, internal resistance creates a voltage drop, and $V_{\text{terminal}} < \varepsilon$.

Worked Example: Internal Resistance

A battery has $\varepsilon = 12\ \text{V}$ and internal resistance $r = 0.5\ \Omega$. It is connected to an external resistor $R = 11.5\ \Omega$. Calculate: a) The current b) The terminal potential difference c) The "lost volts"

a)

$I = \frac{\varepsilon}{R + r} = \frac{12}{11.5 + 0.5} = \frac{12}{12} = \mathbf{1\ \text{A}}$

b)

$V_{\text{terminal}} = \varepsilon - Ir = 12 - (1 \times 0.5) = \mathbf{11.5\ \text{V}}$

c)

$\text{Lost volts} = Ir = 1 \times 0.5 = \mathbf{0.5\ \text{V}}$

Check: $V_{\text{terminal}} + \text{lost volts} = 11.5 + 0.5 = 12\ \text{V} = \varepsilon$ ✓

Common Exam Mistakes

Mixing up series and parallel rules. Label each component in the circuit as series or parallel with respect to the others. Don't apply series current rules to a parallel branch.

Using total current in a parallel branch. In a parallel circuit, each branch carries only its share of the current. Only the total current (before the split) equals $I_{\text{total}}$.

Forgetting to include internal resistance in the total circuit resistance. The full circuit resistance is $R_{\text{external}} + r$. Many students use only $R_{\text{external}}$.

Incorrect reciprocal calculation for parallel resistance. Calculate $\dfrac{1}{R_{\text{total}}}$, then take the reciprocal. This is where calculators save time — use them carefully.

Exam Strategy for Circuits

1. Draw the circuit — even if one is given, redraw it neatly. Mark what you know.
2. Identify the structure — which components are in series? Which in parallel?
3. Find total resistance — starting from the most "nested" group outward.
4. Find total current using $I = \dfrac{V}{R}$ (or $I = \dfrac{\varepsilon}{R+r}$ if internal resistance applies).
5. Work backwards — use the total current to find branch currents and individual voltages.

This systematic approach prevents the most common circuit errors.

Practice Problems

1. $R_1 = 8\ \Omega$ and $R_2 = 4\ \Omega$ in series, connected to 18 V. Find total $R$, total $I$, and voltage across each.
2. $R_1 = 10\ \Omega$ and $R_2 = 15\ \Omega$ in parallel, connected to 30 V. Find total $R$, total $I$, and current through each.
3. A 1500 W heater runs on 230 V. Find the current it draws and its resistance.
4. A battery ($\varepsilon = 9\ \text{V}$, $r = 1\ \Omega$) is connected to $R = 8\ \Omega$. Find the terminal PD and lost volts.

Want to Work Through This Together?

Electricity and circuits are the kind of topic that clicks quickly once you see someone work through questions in real time. If you're struggling with circuits — or if the combination of series, parallel, and internal resistance is getting confusing — book a tutoring session. I cover all of Grade 11 and Grade 12 Physical Science, including exam paper walk-throughs for both NSC and IEB.