Electric circuits are the backbone of all electrical devices. In this chapter you will apply Ohm's law, calculate equivalent resistance in series and parallel combinations, use Kirchhoff's laws, account for the internal resistance of a real battery, and calculate electrical power and energy.
5.1 Ohm's Law and Resistance
Definition
Ohm's Law
The amount of electric current through a metal conductor, at a constant temperature, in a circuit is proportional to the voltage across the conductor and can be described by I = V/R, where I is the current through the conductor, V is the voltage across the conductor and R is the resistance of the conductor. In other words, at constant temperature, the resistance of the conductor is constant.
Formula
Ohm's Law
V = potential difference (V), I = current (A), R = resistance (Ω)
SI unit: Ω (ohm)
Watch Out
COMMON MISTAKE — Ohm's law applies only to an OHMIC conductor at CONSTANT temperature. A light-bulb filament (which heats up) and a diode are non-ohmic — their resistance changes with temperature or direction of current. Do not apply V = IR to non-ohmic components without checking whether the question specifies constant temperature.
Definition
Equivalent resistance series
Rs = R₁ + R₂ + R₃ + ... + Rₙ
SERIES CIRCUITS: When resistors are connected end-to-end in a single path, the same current flows through each resistor. The total (equivalent) resistance is the sum of all individual resistances. Adding more resistors in series always INCREASES the total resistance.
Definition
Equivalent resistance parallel
1/Rp = 1/R₁ + 1/R₂ + ... + 1/Rₙ
PARALLEL CIRCUITS: When resistors are connected side-by-side (both ends joined at the same two nodes), the same potential difference appears across each resistor. The total current from the battery equals the sum of the currents through each branch. Adding more resistors in parallel always DECREASES the total resistance — there are more paths for current to flow.
Exam Tip
EXAM TIP — For TWO resistors in parallel only, use the shortcut formula: Rp = (R₁ × R₂)/(R₁ + R₂). For three or more, use the reciprocal formula 1/Rp = 1/R₁ + 1/R₂ + ... and take the reciprocal at the end. Never forget to take the final reciprocal — this is the most common error.
Worked Example
A 6 Ω and a 12 Ω resistor are connected in series to a 18 V battery. Calculate (a) the equivalent resistance, (b) the current through the circuit, and (c) the voltage across the 12 Ω resistor.
Given
- R₁ = 6 Ω
- R₂ = 12 Ω
- V_battery = 18 V (series circuit — no internal resistance)
Find
(a) Rs (b) I (c) V₂
Solution
- 1(a) Rs = R₁ + R₂ = 6 + 12 = 18 Ω
- 2(b) I = V/Rs = 18/18 = 1 A
- 3(c) V₂ = I × R₂ = 1 × 12 = 12 V
Worked Example
A 6 Ω and a 12 Ω resistor are connected in parallel to a 12 V battery. Calculate (a) the equivalent resistance, (b) the total current from the battery, and (c) the current through the 6 Ω resistor.
Given
- R₁ = 6 Ω
- R₂ = 12 Ω
- V = 12 V (same across both — parallel)
Find
(a) Rp (b) I_total (c) I₁
Solution
- 1(a) 1/Rp = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 → Rp = 12/3 = 4 Ω
- 2(b) I_total = V/Rp = 12/4 = 3 A
- 3(c) I₁ = V/R₁ = 12/6 = 2 A
Practice Question
Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel. (a) Calculate the equivalent resistance. (b) If this parallel combination is connected to a 24 V source, calculate the total current. (c) Which resistor carries the largest current? Explain.
(7 marks)
5.2 Kirchhoff's Laws and Internal Resistance
KIRCHHOFF'S CURRENT LAW (KCL): At any junction (node) in a circuit, the sum of currents entering the junction equals the sum of currents leaving. In symbols: ΣI_in = ΣI_out. This is a consequence of conservation of charge.
KIRCHHOFF'S VOLTAGE LAW (KVL): Around any closed loop in a circuit, the algebraic sum of all potential differences (emfs and voltage drops) equals zero. In symbols: ΣV = 0 (going around a loop). This is a consequence of conservation of energy.
Exam Tip
EXAM TIP — When applying KVL: choose a loop direction (clockwise or anti-clockwise), assign a sign convention (voltage rise across a battery = +, voltage drop across a resistor in your chosen direction = −), and sum all terms to zero. The sign convention must be consistent within each loop.
INTERNAL RESISTANCE: A real battery is not a perfect voltage source. It has an INTERNAL RESISTANCE (r) — the resistance of the materials inside the battery itself. When current flows, some voltage is 'lost' across this internal resistance. The terminal voltage (V_terminal) is always less than the EMF (E) when current is flowing.
Formula
Terminal voltage with internal resistance
V = terminal voltage (V), E = EMF of the battery (V), I = current (A), r = internal resistance (Ω)
SI unit: V
Note
KEY TERMS: • EMF (E): The total energy supplied per unit charge by the battery — measured in V when no current flows (open circuit). • Terminal voltage (V): The potential difference across the battery terminals when current is flowing — always ≤ EMF. • Lost voltage (Ir): The voltage 'wasted' internally — also called the 'voltage drop across the internal resistance'.
Watch Out
COMMON MISTAKE — EMF is NOT a force — it is measured in volts (V), not newtons (N). EMF is the energy per unit charge supplied by the source. When the circuit is OPEN (no current), the voltmeter reads the EMF directly. When the circuit is CLOSED, the voltmeter reads the terminal voltage (which is less than the EMF).
Worked Example
A battery has an EMF of 12 V and internal resistance of 0,5 Ω. An external resistance of 5,5 Ω is connected across its terminals. Calculate (a) the current in the circuit, (b) the terminal voltage, and (c) the voltage drop across the internal resistance.
Given
- E = 12 V
- r = 0,5 Ω
- R_ext = 5,5 Ω
Find
(a) I (b) V_terminal (c) V_r
Solution
- 1(a) Total resistance = R_ext + r = 5,5 + 0,5 = 6 Ω
- 2 I = E / (R + r) = 12 / 6 = 2 A
- 3(b) V = E − Ir = 12 − (2)(0,5) = 12 − 1 = 11 V
- 4(c) V_r = Ir = (2)(0,5) = 1 V
Practice Question
A battery of EMF 9 V and internal resistance 1 Ω is connected to two resistors of 4 Ω and 12 Ω in parallel. (a) Calculate the equivalent external resistance. (b) Calculate the current delivered by the battery. (c) Calculate the terminal voltage of the battery. (d) Apply Kirchhoff's current law to find the current through the 4 Ω resistor.
(10 marks)
5.3 Electrical Power and Energy
Definition
Electrical Power
Electrical power is the rate at which electrical energy is converted in an electric circuit. P = I·V
Formula
Electrical power (three equivalent forms)
P = power (W), I = current (A), V = potential difference (V), R = resistance (Ω)
SI unit: W (watt)
Formula
Electrical energy
W = energy (J), P = power (W), t = time (s)
SI unit: J (joule)
Exam Tip
CHOOSING WHICH POWER FORMULA TO USE: Use P = IV when you know both current and voltage. Use P = I²R when you know current and resistance. Use P = V²/R when you know voltage and resistance. All three are equivalent by Ohm's law — choose the one that avoids introducing an unknown.
Worked Example
A 60 W light bulb is connected to a 240 V supply. Calculate (a) the current through the bulb and (b) the resistance of the bulb filament.
Given
- P = 60 W
- V = 240 V
Find
(a) I (b) R
Solution
- 1(a) P = IV → I = P/V = 60/240 = 0,25 A
- 2(b) V = IR → R = V/I = 240/0,25 = 960 Ω
- 3 OR using P = V²/R → R = V²/P = 240²/60 = 57600/60 = 960 Ω ✓
Worked Example
An electric kettle has a resistance of 20 Ω and operates on 240 V. (a) Calculate the power of the kettle. (b) Calculate the electrical energy used if the kettle runs for 3 minutes.
Given
- R = 20 Ω
- V = 240 V
- t = 3 min = 180 s
Find
(a) P (b) W
Solution
- 1(a) P = V²/R = 240²/20 = 57 600/20 = 2 880 W
- 2(b) W = Pt = 2 880 × 180 = 518 400 J
Series vs Parallel: Effect on Power
| Property | Series connection | Parallel connection |
|---|---|---|
| Current | Same current through all components | Current splits — each branch has different I |
| Voltage | Voltage divides — each component has different V | Same voltage across all branches |
| Brightest bulb (identical bulbs) | All equally bright (same current, same R) | All equally bright (same voltage, same R) |
| If one bulb is removed | Circuit breaks — all bulbs go out | Remaining bulbs stay on at full brightness |
Practice Question
A toaster draws a current of 8 A from a 240 V supply. (a) Calculate the power of the toaster. (b) Calculate the resistance of the toaster's heating element. (c) Calculate the electrical energy used in 5 minutes. (d) If the cost of electricity is R2,50 per kWh, calculate the cost of using the toaster for 5 minutes.
(8 marks)