EMF, Internal Resistance and Terminal PD

Real batteries and cells are not ideal voltage sources. They have internal resistance that affects the voltage available to the external circuit. Understanding internal resistance is essential for circuit analysis and is a key topic in AQA A-Level Physics.

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The Concept of Internal Resistance

Every real source of EMF (battery, cell, generator) has some resistance within itself. This is called internal resistance (r).

When current flows through the source, some energy is dissipated as heat inside the source due to this internal resistance. This means the voltage available to the external circuit (the terminal p.d.) is less than the EMF.

The Circuit Model

A real battery can be modelled as an ideal EMF source (ε) in series with a small internal resistance (r):

Circuit description:

• Inside the battery: ideal EMF (ε) in series with internal resistance (r)
• External circuit: load resistance (R)
• Current I flows around the complete circuit

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The Key Equation

Applying conservation of energy around the circuit:

$\varepsilon = I(R + r)$ε=I(R+r)

or equivalently:

$\varepsilon = IR + Ir$ε=IR+Ir

where:

• ε = EMF of the source (V)
• I = current in the circuit (A)
• R = external (load) resistance (Ω)
• r = internal resistance (Ω)
• IR = p.d. across the external resistance (terminal p.d., V)
• Ir = "lost volts" (p.d. across internal resistance, V)

Terminal p.d. and Lost Volts

Rearranging: V_terminal = ε − Ir

Term	Meaning
ε (EMF)	Total energy supplied per coulomb by the source
V_terminal = IR	Energy per coulomb delivered to the external circuit ("useful" voltage)
Ir (lost volts)	Energy per coulomb dissipated inside the source as heat

Key Point: The terminal p.d. is what you would measure with a voltmeter connected across the battery terminals when current is flowing. It is always less than the EMF when current flows.

When is V_terminal = ε?

The terminal p.d. equals the EMF only when no current flows (I = 0):

• When the circuit is open (no external resistance connected)
• When measured with a very high-resistance voltmeter (which draws negligible current)

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Worked Examples

Worked Example 1 — Basic Internal Resistance

A battery has an EMF of 9.0 V and an internal resistance of 0.50 Ω. It is connected to a 5.5 Ω external resistor. Calculate (a) the current, (b) the terminal p.d., and (c) the lost volts.

Solution:

(a) ε = I(R + r)

I = ε / (R + r) = 9.0 / (5.5 + 0.50) = 9.0 / 6.0 = 1.5 A

(b) V_terminal = IR = 1.5 × 5.5 = 8.25 V

Or: V_terminal = ε − Ir = 9.0 − (1.5 × 0.50) = 9.0 − 0.75 = 8.25 V ✓

(c) Lost volts = Ir = 1.5 × 0.50 = 0.75 V

Check: V_terminal + lost volts = 8.25 + 0.75 = 9.0 V = ε ✓

Worked Example 2 — Finding EMF and Internal Resistance

A student connects a variable resistor to a cell and measures the following:

R (Ω)	I (A)	V_terminal (V)
10.0	0.138	1.38
5.0	0.250	1.25
2.0	0.500	1.00

Use two sets of readings to determine the EMF and internal resistance.

Solution:

Using V = ε − Ir:

From reading 1: 1.38 = ε − 0.138r ... (1) From reading 3: 1.00 = ε − 0.500r ... (2)

Subtracting (2) from (1): 0.38 = 0.362r

r = 0.38 / 0.362 = 1.05 Ω ≈ 1.0 Ω

Substituting back into (1): ε = 1.38 + (0.138 × 1.05) = 1.38 + 0.145 = 1.52 V ≈ 1.5 V

Check with reading 2: V = 1.52 − (0.250 × 1.05) = 1.52 − 0.263 = 1.26 V ≈ 1.25 V ✓

Worked Example 3 — Power Delivered to External Resistance