Electric Circuits — Internal Resistance

EMF and Internal Resistance

A real battery is not a perfect source of voltage. It has an internal resistance (r) — the resistance of the chemical cells inside the battery itself.

EMF (ε): the maximum voltage a battery can provide when no current flows (open circuit). Also called electromotive force.
Terminal potential difference (V_terminal): the actual voltage across the battery terminals when current flows.
Lost voltage (V_lost): voltage "lost" due to internal resistance = Ir

ε = V_terminal + Ir
V_terminal = ε − Ir
V_lost = Ir

Symbol	Meaning	Unit
ε	EMF (electromotive force)	V
V_terminal	Terminal potential difference	V
I	Current in circuit	A
r	Internal resistance of battery	Ω
R	External resistance (load)	Ω

As current I increases (smaller external load R), the terminal PD drops. When I = 0 (open circuit), V_terminal = ε.

Series Circuit with Internal Resistance

For a battery (ε, r) connected to an external resistor R in series:

ε = I(R + r)
I = ε / (R + r)
V_terminal = IR = ε − Ir

Worked Example 1: A battery has EMF = 12 V and internal resistance r = 0.5 Ω. It is connected to an external resistor R = 5.5 Ω. Find I, V_terminal, and V_lost.

I = ε/(R+r) = 12/(5.5+0.5) = 12/6 = 2 A
V_terminal = IR = 2×5.5 = 11 V
V_lost = Ir = 2×0.5 = 1 V
Check: 11 + 1 = 12 V = ε ✓

Kirchhoff's Current Law (KCL)

KCL: The sum of all currents entering a junction equals the sum of all currents leaving that junction. (Conservation of charge.)

ΣI_in = ΣI_out

Example: If 5 A enters a junction and splits into two branches, and one branch carries 2 A, the other carries 3 A.

Kirchhoff's Voltage Law (KVL)

KVL: The sum of all potential differences (EMFs and voltage drops) around any closed loop in a circuit is zero. (Conservation of energy.)

Σε = ΣIR (around any closed loop)

Sign conventions for KVL:
• Travelling through a battery from − to +: add +ε (gain in potential)
• Travelling through a battery from + to −: add −ε (loss in potential)
• Travelling through a resistor in the direction of current: add −IR (loss in potential)
• Travelling through a resistor against current: add +IR (gain in potential)

Parallel Circuit with Internal Resistance

When resistors R₁ and R₂ are in parallel, their combined resistance is:

1/R_ext = 1/R₁ + 1/R₂ (or R_ext = R₁R₂/(R₁+R₂))
I_total = ε / (R_ext + r)
V_terminal = I_total × R_ext

Adding more parallel resistors reduces R_ext → increases I_total → V_terminal drops further as Ir increases.

Power and Energy in Circuits

P = VI = I²R = V²/R
E = Pt = VIt

Power delivered by the EMF source: P_total = εI
Power dissipated externally: P_ext = V_terminal × I = I²R
Power dissipated internally: P_int = I²r

Worked Example 2 — Finding Internal Resistance from Data

Problem: A battery drives a current of 3 A through an 8 Ω resistor, and the terminal PD is 24 V. Find ε and r.

V_terminal = IR → 24 = 3×8 ✓ (consistent)
ε = V_terminal + Ir → we need r, but let's say the battery also shows 25.5 V on open circuit → ε = 25.5 V
r = (ε − V_terminal)/I = (25.5 − 24)/3 = 1.5/3 = 0.5 Ω

IEB Extension — Multi-branch Kirchhoff Analysis

IEB circuits often have two or more loops and multiple unknown currents. Apply KVL to each independent loop:

Label a current direction for each branch (guess if unsure — a negative answer means the actual direction is opposite)
Apply KCL at each junction to express currents in terms of fewer unknowns
Apply KVL to each independent loop to generate simultaneous equations
Solve the system of equations

Loop 1: ε₁ − I₁R₁ − (I₁−I₂)R₃ = 0
Loop 2: ε₂ − I₂R₂ + (I₁−I₂)R₃ = 0

Efficiency of energy transfer from battery: η = P_ext/P_total = (I²R)/(εI) = IR/ε = V_terminal/ε × 100%

Internal Resistance Circuit Simulator

Battery and Load

EMF ε (V) 12 V

Internal r (Ω) 1.0 Ω

External R (Ω) 5 Ω

Calculated Values

Current I

—A

V terminal

—V

V lost (Ir)

—V

P external

—W

P internal

—W

Efficiency

—%

Question 1 of 8

A battery has an EMF of 9 V and internal resistance of 1 Ω. When connected to an external resistor, the current is 3 A. The terminal potential difference is:

Question 2 of 8

What happens to the terminal potential difference of a battery as the external resistance decreases (more current is drawn)?

Question 3 of 8

Kirchhoff's Current Law states that at any junction in a circuit:

Question 4 of 8

A battery (ε = 6 V, r = 0.5 Ω) is connected to two 5 Ω resistors in parallel. The current drawn from the battery is:

Question 5 of 8

A battery with EMF 12 V and internal resistance 2 Ω drives a current of 2 A. The power dissipated in the internal resistance is:

Question 6 of 8

The terminal PD of a battery equals its EMF when:

Question 7 of 8

A student plots terminal potential difference (V) against current (I) for a battery, obtaining a straight-line graph with y-intercept 9.0 V and gradient −1.5 Ω. What current would produce a terminal PD of 3.0 V?

Question 8 of 8

A learner connects the same battery to different values of external resistance R and measures the current I each time: when R = 8 Ω, I = 1.0 A; when R = 3 Ω, I = 2.0 A. Using ε = I(R + r) for both readings, calculate the internal resistance r and EMF ε of the battery.

IEB Extended Questions

IEB Question 1

A circuit contains two batteries in series: ε₁ = 6 V (r₁ = 0.5 Ω) and ε₂ = 4 V (r₂ = 0.5 Ω) opposing each other, and an external resistor R = 4 Ω. Using KVL, the current in the circuit is:

Show all formulae, substitutions, units, and a labelled circuit diagram where appropriate. Apply Kirchhoff's laws systematically, stating each law before using it.

Question 1 — Internal Resistance from Graph (6 marks)

A student measures the terminal PD of a battery for different currents and plots a V vs I graph. The graph is a straight line with y-intercept at 8 V and x-intercept at 4 A. (a) What does the y-intercept represent? (b) What does the x-intercept represent physically? (c) Determine the EMF of the battery. (d) Calculate the internal resistance from the gradient of the graph. (e) Write the equation of the line in the form V = ε − Ir. (f) Calculate V_terminal when I = 1.5 A.

Question 2 — Series Circuit (7 marks)

A battery (ε = 15 V, r = 0.8 Ω) is connected to two resistors in series: R₁ = 4 Ω and R₂ = 6.2 Ω. (a) Calculate the total resistance of the circuit. (b) Calculate the current drawn from the battery. (c) Calculate the terminal potential difference. (d) Calculate the potential difference across R₁ only. (e) Calculate the potential difference across R₂ only. (f) Verify your answers using KVL around the complete loop. (g) Calculate the power delivered to R₂.

Question 3 — Parallel Circuit with Internal Resistance (7 marks)

A battery (ε = 12 V, r = 1 Ω) is connected to two resistors in parallel: R₁ = 6 Ω and R₂ = 12 Ω. (a) Calculate the equivalent external resistance. (b) Calculate the total current from the battery. (c) Calculate the terminal potential difference. (d) Calculate the current through R₁. (e) Calculate the current through R₂. (f) Verify using KCL. (g) Calculate the efficiency of the circuit (useful power / total power).

Question 4 — Determining EMF and Internal Resistance Experimentally (5 marks)

A learner connects a battery to a variable resistor and measures the following: when R = 10 Ω, I = 0.9 A and V_terminal = 9 V. When R = 4 Ω, I = 1.8 A and V_terminal = 7.2 V. (a) Using the equation ε = V + Ir, set up two simultaneous equations. (b) Solve for ε and r. (c) Calculate the short-circuit current (R = 0). (d) Explain why it is dangerous to short-circuit a battery with low internal resistance. (e) At what value of external resistance will exactly half the EMF appear across the terminals?