3.3 - Electromotive Force (e.m.f.) and Internal Resistance

Electromotive Force

Electromotive force (e.m.f.), $ε$ , is the energy supplied or work done by an electrical source for each coulomb of charge moved around a complete circuit:

ε = \frac{E}{Q}

$E$ is energy in joules and $Q$ is charge in coulombs. The SI unit is volt, $V$ , equivalent to $J C^{- 1}$ .
A 1.5 V dry cell supplies 1.5 J per coulomb.
Sources include generators, dynamos, dry cells, lead-acid accumulators and lithium-ion batteries.
A voltmeter has high resistance, so its current is negligible. Across an unloaded cell, its reading gives the cell’s e.m.f.

Quantity	Meaning and measurement
E.m.f., $ε$	Work done by the source per coulomb around the complete circuit. Measure across the cell with the switch open, when no current flows through the external circuit.
Potential difference, $V$	Work done per coulomb between two points. With the switch closed, the voltmeter gives the potential difference across the bulb.

Both quantities have the same unit. In the cell-and-bulb circuit, the closed-circuit reading is smaller than the open-circuit e.m.f.

Internal resistance, $r$ , is the resistance caused by the electrolyte inside a dry cell. Its SI unit is ohm, $Ω$ .
Current flowing against this resistance transfers energy to heat inside the cell, reducing its terminal potential difference.
Comparing the open-circuit and closed-circuit voltmeter readings gives the voltage drop:

I r = ε - V

Alkaline batteries use potassium hydroxide as the electrolyte and are twice as durable as zinc-carbon batteries in this comparison. Different electrolytes give different internal resistances. Cell arrangement also affects effective internal resistance.
For external resistance $R$ , $V = I R$ . Hence:

ε = V + I r = I R + I r = I (R + r)

I = \frac{ε}{R + r}, V = ε - I r

With current flowing, $V < ε$ ; increasing current increases the voltage drop and decreases $V$ .

Aim: Determine the e.m.f. and internal resistance of a dry cell.
Apparatus: 1.5 V dry cell, cell holder, switch, connecting wires, ammeter, voltmeter and rheostat.

Connect the cell, switch, ammeter and rheostat in series; connect the voltmeter across the cell.
Obtain paired readings of current $I$ and potential difference $V$ for a set of currents.
Avoid parallax errors in meter readings and reduce energy loss from the cell during data collection.
Plot $V$ against $I$ , determine the gradient and vertical intercept, and compare with the linear equation:

V = - r I + ε

Result: A straight line with a negative gradient; $V$ falls as $I$ rises.
Gradient: $m = - r$ , so $r = - m$ .
Vertical intercept: At $I = 0$ , $V = ε$ ; the intercept gives the e.m.f.

For four identical dry cells, each with e.m.f. $ε$ and internal resistance $r$ :

Arrangement	Effective e.m.f. and internal resistance
Series	E.m.f. adds to $4 ε$ ; $r_{e} = r + r + r + r = 4 r$ .
Parallel	E.m.f. remains $ε$ ; $1/ r_{e} = 1/ r + 1/ r + 1/ r + 1/ r$ , so $r_{e} = r /4$ .

Series connection increases effective e.m.f.; parallel connection reduces effective internal resistance.
Compare current using both the effective e.m.f. and total resistance: $I = ε_{effective} / (R + r_{e})$ .

Each cell has $ε = 1.5 V$ and $r = 0.5 Ω$ ; external resistance is $10 Ω$ .
Two cells in series:

ε_{effective} = 3 V, r_{e} = 1.0 Ω, I = \frac{3}{10 + 1.0} \approx 0.27 A

ε_{effective} = 1.5 V, r_{e} = 0.25 Ω, I = \frac{1.5}{10 + 0.25} = 0.1463 A

Series gives greater current in this example because its effective e.m.f. is higher.
Three cells in parallel:

r_{e} = \frac{0.5}{3} \approx 0.167 Ω, I = \frac{1.5}{10 + 0.167} \approx 0.1475 A

Adding a parallel cell increases current because effective internal resistance decreases.

Electric vehicles use 100% electrical power, with rechargeable batteries supplying energy to electric motors; battery types include Li-Ion and Ni-MH. Hybrid cars combine rechargeable batteries with fossil fuels such as petrol.
The vehicle comparison uses 300–800 V batteries for electric vehicles and 100–200 V for hybrids, with batteries providing 25–40% of the hybrid’s power.
Batteries can receive energy from domestic supplies, charging stations and solar-cell charging stations.
Solar cells convert sunlight into electrical energy. Connecting solar cells in series forms a solar panel; panel arrangement provides the required voltage and current.
Electric and hybrid vehicles can reduce fossil-fuel consumption, improve energy efficiency and reduce air pollution.