Real batteries are not perfect — they have internal resistance that causes the terminal voltage to drop when current flows. In this chapter you will analyse series-parallel circuits with internal resistance, interpret V-I graphs, and calculate power and energy in complex circuits using Kirchhoff's laws.
9.1 Internal Resistance and EMF
Every real battery has some resistance inside it called internal resistance (r). When the battery drives current through a circuit, this internal resistance causes a voltage drop inside the battery itself. The electromotive force (emf, symbol E) is the total energy per coulomb supplied by the battery, while the terminal voltage (V) is the voltage available to the external circuit.
Definition
Electromotive Force (E)
The emf of a source is the work done (energy transferred) per unit charge by the source in driving charge around the complete circuit. SI unit: volt (V).
Definition
Internal Resistance (r)
The resistance of the internal components of a battery or source that opposes current flow within the source itself. SI unit: ohm (Ω).
Definition
Load
The external resistance in the circuit is referred to as the load. V_load = I·R
Formula
EMF equation
E = emf (V), V = terminal voltage (V), I = current (A), r = internal resistance (Ω)
SI unit: V
Formula
Terminal voltage
V = terminal voltage (V), E = emf (V), I = current (A), r = internal resistance (Ω)
SI unit: V
On a V vs I graph, the terminal voltage V is plotted on the y-axis and current I on the x-axis. The y-intercept (when I = 0) gives the emf E because no current means no internal voltage drop. The graph has a negative gradient equal to −r (the internal resistance). This makes it easy to read off both E and r directly from the graph.
Exam Tip
V-I graph exam tip: y-intercept = E (emf). Gradient = −r. Always quote units: E in volts, r in ohms. If the graph is not a straight line, the internal resistance is not constant.
Worked Example
A battery has an emf of 12 V and internal resistance of 0.5 Ω. It drives a current of 4 A through an external resistor. Calculate (a) the terminal voltage and (b) the voltage drop across the internal resistance.
Given
- E = 12 V
- r = 0.5 Ω
- I = 4 A
Find
Terminal voltage V and voltage drop Vr
Solution
- 1Voltage drop across internal resistance: Vr = Ir = 4 × 0.5 = 2 V
- 2Terminal voltage: V = E − Ir = 12 − 2 = 10 V
- 3Check: E = V + Ir → 10 + 2 = 12 V ✓
Practice Question
A cell has emf 6 V and internal resistance 0.8 Ω. When connected to a load, the terminal voltage is 5.2 V. Calculate the current drawn from the cell.
(3 marks)
Watch Out
Common mistake: students use V = E instead of V = E − Ir. The terminal voltage is ALWAYS less than the emf when current flows. Only when I = 0 (open circuit) is V = E.
9.2 Series and Parallel Circuits with Internal Resistance
When a battery with internal resistance drives current through a circuit containing both series and parallel elements, you must first find the total external resistance before applying E = V + Ir. For series elements, add resistances directly. For parallel elements, use 1/R_p = 1/R₁ + 1/R₂. The total external resistance R_ext combined with r gives you the total circuit resistance.
Kirchhoff's Laws (summary)
- KCL (Current Law): The sum of currents entering a junction equals the sum of currents leaving it. ΣI_in = ΣI_out
- KVL (Voltage Law): The algebraic sum of all voltage drops around any closed loop equals zero. ΣV = 0
Exam Tip
Exam strategy for series-parallel circuits: (1) Find the equivalent resistance of parallel branches first. (2) Add series resistances including r. (3) Use E = I·R_total to find total current. (4) Work backwards to find branch voltages and currents.
Worked Example
A battery (E = 24 V, r = 1 Ω) is connected to R₁ = 3 Ω in series with a parallel combination of R₂ = 6 Ω and R₃ = 12 Ω. Calculate (a) the total current from the battery and (b) the terminal voltage.
Given
- E = 24 V
- r = 1 Ω
- R₁ = 3 Ω
- R₂ = 6 Ω
- R₃ = 12 Ω
Find
Total current I and terminal voltage V
Solution
- 1Parallel combination: 1/R_p = 1/6 + 1/12 = 2/12 + 1/12 = 3/12, so R_p = 4 Ω
- 2Total external resistance: R_ext = R₁ + R_p = 3 + 4 = 7 Ω
- 3Total circuit resistance: R_total = R_ext + r = 7 + 1 = 8 Ω
- 4Total current: I = E / R_total = 24 / 8 = 3 A
- 5Terminal voltage: V = E − Ir = 24 − (3)(1) = 21 V
Practice Question
A battery (E = 18 V, r = 0.5 Ω) drives current through two resistors in series: R₁ = 5 Ω and R₂ = 3 Ω. Calculate (a) the current in the circuit and (b) the terminal voltage.
(5 marks)
Series vs Parallel Circuits
| Property | Series Circuit | Parallel Circuit |
|---|---|---|
| Current | Same in all components | Splits between branches |
| Voltage | Splits across components | Same across all branches |
| Total resistance | R_T = R₁ + R₂ + … | 1/R_T = 1/R₁ + 1/R₂ + … |
| If one component fails | Entire circuit breaks | Other branches still work |
| Kirchhoff's law | KVL: ΣV = E | KCL: ΣI = I_total |
Watch Out
Common mistake: forgetting to include r when calculating R_total. Internal resistance is in series with all external resistors. Always write R_total = R_ext + r.
9.3 Power and Energy in Complex Circuits
Power is the rate at which energy is transferred. In electric circuits, power dissipated in a resistor can be calculated using any of three equivalent formulas depending on which quantities are known. Energy is simply power multiplied by time.
Formula
Electrical power
P = power (W), I = current (A), V = voltage across component (V), R = resistance (Ω)
SI unit: W
Formula
Electrical energy
W = energy (J), P = power (W), t = time (s)
SI unit: J
Exam Tip
Power dissipated by the internal resistance is 'wasted' — it heats the battery. The useful power delivered to the external circuit is P_ext = IV = I(E − Ir). Total power from battery = EI.
Worked Example
A battery (E = 12 V, r = 0.4 Ω) drives a current of 5 A through an external resistor. Calculate (a) the power dissipated in the external resistor, (b) the power wasted in the internal resistance, and (c) the total power output of the battery.
Given
- E = 12 V
- r = 0.4 Ω
- I = 5 A
Find
P_external, P_internal, P_total
Solution
- 1Terminal voltage: V = E − Ir = 12 − (5)(0.4) = 12 − 2 = 10 V
- 2External resistance: R = V/I = 10/5 = 2 Ω
- 3Power in external resistor: P_ext = I²R = 25 × 2 = 50 W (or P = IV = 5 × 10 = 50 W)
- 4Power wasted internally: P_int = I²r = 25 × 0.4 = 10 W
- 5Total power: P_total = EI = 12 × 5 = 60 W
- 6Check: P_ext + P_int = 50 + 10 = 60 W ✓
Practice Question
An electric heater with resistance 20 Ω is connected to a battery (E = 9 V, r = 1 Ω). Calculate the energy dissipated in the heater in 5 minutes.
(6 marks)
Practice Question
A 60 W light bulb operates at 220 V. Calculate its resistance and the current it draws.
(4 marks)
Watch Out
When using P = V²/R, V must be the voltage ACROSS that specific resistor, not the emf. Use P = I²R when you know the current, and P = IV when you know both.
Real World
Real-world connection: The internal resistance of a car battery increases as it ages. On cold mornings, the starter motor draws a very large current (up to 200 A), causing a large voltage drop across r. This is why old batteries struggle to start cars in winter.