EMF and Internal Resistance — AQA A-level Physics

SPEC 3.5.1.6 • ELECTRICITY

Electromotive Force and Internal Resistance

Understanding EMF, terminal pd, and energy losses in real power sources

What is EMF?

Electromotive force (EMF) is not actually a force - it's a potential difference measured in volts.

Definition

Electromotive Force (EMF) is the energy transferred per unit charge by a power source.

ε = \frac{E}{Q}

SymbolQuantityUnit

ε

Electromotive force (EMF)Volts (V)

E

Energy transferredJoules (J)

Q

ChargeCoulombs (C)

Key Point

EMF represents the total energy the power source can provide to each coulomb of charge.

EMF vs Terminal pd

When a cell is connected to a circuit, the voltage measured across its terminals is less than the EMF.

EMF

The maximum potential difference a cell can provide

Measured when no current flows (open circuit)

Terminal pd

The actual potential difference across the cell terminals

Measured when current flows (closed circuit)

Important

Terminal pd is always less than EMF when current flows, because some energy is dissipated inside the cell itself.

The Relationship

Terminal pd = ε - Lost volts

Internal Resistance

All power sources have internal resistance - resistance within the cell itself due to the chemicals and electrodes.

SymbolMeaningUnit

r

Internal resistance of cellOhms (Ω)

R

External (load) resistanceOhms (Ω)

The EMF Equation

The EMF of a cell equals the sum of all potential differences around the circuit.

EMF Equation

ε = I (R + r)

or equivalently:

ε = I R + I r

IR

Potential difference across external circuit (terminal pd)

V = I R

Ir

Potential difference "lost" inside the cell

v = I r

Rearranging

The equation can also be written as: $V = ε - I r$

This shows that terminal pd = EMF minus lost volts

01Worked Example

A cell has EMF 1.5 V and internal resistance 0.50 Ω. It is connected to a 2.5 Ω resistor. Calculate:

(a) the current in the circuit

(b) the terminal pd

(c) the lost volts

a)

Using

ε = I (R + r)

:

I = \frac{ε}{R + r} = \frac{1.5}{2.5 + 0.50} = \frac{1.5}{3.0} = 0.50 A

b)

Terminal pd:

V = I R = 0.50 \times 2.5 = 1.25 V

c)

Lost volts:

v = I r = 0.50 \times 0.50 = 0.25 V

Check:

V + v = 1.25 + 0.25 = 1.50 V = ε

✓

Lost Volts

The "lost volts" represent energy dissipated as heat inside the cell due to its internal resistance.

Lost Volts

v = I r

where

v

= lost volts,

I

= current,

r

= internal resistance

Energy Conservation

The EMF provides energy. This energy is split between:

Useful energy delivered to external circuit (

V = I R

)

Wasted energy dissipated in internal resistance (

v = I r

)

ε = V + v

Important

Higher current = more lost volts

This is why batteries get warm during heavy use - the internal resistance dissipates energy as heat.

Worked Example02

A battery has EMF 12 V. When connected to an external resistance, a current of 2.0 A flows and the terminal pd is 11 V. Calculate the internal resistance of the battery.

1

Calculate lost volts:

v = ε - V = 12 - 11 = 1.0 V

2

Using

v = I r

:

r = \frac{v}{I} = \frac{1.0}{2.0} = 0.50 Ω

AnswerInternal resistance = 0.50 Ω

Worked Example03

A cell with EMF 1.5 V and internal resistance 0.80 Ω is connected to a lamp of resistance 4.2 Ω. Calculate the power dissipated in the lamp.

1

Calculate current:

I = \frac{ε}{R + r} = \frac{1.5}{4.2 + 0.80} = \frac{1.5}{5.0} = 0.30 A

2

Power in lamp:

P = I^{2} R = (0.30)^{2} \times 4.2 = 0.090 \times 4.2 = 0.378 W

AnswerPower = 0.38 W (2 s.f.)

Circuit Analysis with Internal Resistance

When analysing circuits with internal resistance, treat the internal resistance as a resistor in series with the EMF source.

Total Resistance

The total resistance in the circuit is:

R_{t o t a l} = R + r

where

R

= external resistance,

r

= internal resistance

Problem-Solving Strategy

Find total resistance:

R_{t o t a l} = R + r

Find current:

I = \frac{ε}{R + r}

Find terminal pd:

V = I R

or

V = ε - I r

Find lost volts:

v = I r

Check:

ε = V + v

Common Mistake

Don't forget to include the internal resistance when calculating total resistance!

Worked Example04

Two identical cells, each with EMF 1.5 V and internal resistance 0.40 Ω, are connected in series with a 5.2 Ω resistor. Calculate the current and terminal pd across both cells combined.

1

Total EMF (cells in series):

ε_{t o t a l} = 1.5 + 1.5 = 3.0 V

2

Total internal resistance:

r_{t o t a l} = 0.40 + 0.40 = 0.80 Ω

3

Current:

I = \frac{ε _{t o t a l}}{R + r _{t o t a l}} = \frac{3.0}{5.2 + 0.80} = \frac{3.0}{6.0} = 0.50 A

4

Terminal pd:

V = ε_{t o t a l} - I r_{t o t a l} = 3.0 - (0.50 \times 0.80) = 3.0 - 0.40 = 2.6 V

AnswerCurrent = 0.50 A, Terminal pd = 2.6 V

Graphical Analysis

A graph of terminal pd (V) against current (I) gives a straight line that can be used to find EMF and internal resistance.

Linear Form

V = ε - I r

Comparing with

y = m x + c

:

Y-Intercept

EMF (ε)

When I = 0, V = ε

Gradient

−r (negative)

Gradient =

- \frac{Δ V}{Δ I} = - r

Worked Example05

A V-I graph for a cell shows that when I = 0, V = 1.50 V, and when I = 0.40 A, V = 1.30 V. Determine the EMF and internal resistance.

1

EMF = y-intercept = 1.50 V (value of V when I = 0)

2

Gradient =

\frac{Δ V}{Δ I} = \frac{1.30 - 1.50}{0.40 - 0} = \frac{- 0.20}{0.40} = - 0.50

3

Since gradient = −r: internal resistance r = 0.50 Ω

Answerε = 1.50 V, r = 0.50 Ω

06Worked Example

A student measures the following data for a cell:

I / A0.100.200.300.400.50

V / V1.421.341.261.181.10

Determine the EMF and internal resistance of the cell.

1

Calculate gradient:

\frac{Δ V}{Δ I} = \frac{1.10 - 1.42}{0.50 - 0.10} = \frac{- 0.32}{0.40} = - 0.80

2

Internal resistance: r = 0.80 Ω

3

Using

V = ε - I r

:

1.42 = ε - (0.10 \times 0.80)

4

ε = 1.42 + 0.08 = 1.50 V

Answerε = 1.50 V, r = 0.80 Ω

Practice

Question 1: A cell has EMF 1.55 V and internal resistance 0.25 Ω. When connected to a resistor, the current is 0.50 A. Calculate (a) the terminal pd, (b) the resistance of the resistor.

a)

Lost volts:

v = I r = 0.50 \times 0.25 = 0.125 V

•

Terminal pd:

V = ε - v = 1.55 - 0.125 = 1.425 V \approx 1.43 V

b)

R = \frac{V}{I} = \frac{1.425}{0.50} = 2.85 Ω

Answer(a) 1.43 V (b) 2.85 Ω

Practice

Question 2: A battery with EMF 9.0 V and internal resistance 1.5 Ω is connected to a 6.0 Ω resistor. Calculate the power dissipated (a) in the external resistor, (b) in the internal resistance.

1

Current:

I = \frac{ε}{R + r} = \frac{9.0}{6.0 + 1.5} = \frac{9.0}{7.5} = 1.2 A

a)

Power in external resistor:

P_{R} = I^{2} R = (1.2)^{2} \times 6.0 = 8.64 W

b)

Power in internal resistance:

P_{r} = I^{2} r = (1.2)^{2} \times 1.5 = 2.16 W

Answer(a) 8.6 W (b) 2.2 W

Summary

Key Equations

EquationDescription

ε = \frac{E}{Q}

EMF = energy per unit charge

ε = I (R + r)

EMF = current × total resistance

V = ε - I r

Terminal pd = EMF − lost volts

v = I r

Lost volts = current × internal resistance

Key Exam Skills

Calculate current, terminal pd, and lost volts in circuits with internal resistance

Determine EMF and internal resistance from V-I graphs

Analyse circuits with cells in series

Remember

EMF = maximum pd (no current) | Terminal pd = actual pd (with current) | Lost volts = energy wasted inside cell