# Electromotive Force & Potential Difference

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## Electromotive Force

Electromotive Force (e.m.f.) of a source is the energy converted from non-electrical to electrical form when one coulomb of positive charge passes through the source.

• SI unit: Volt (V)

$$\epsilon = \frac{W}{Q}$$

, where $\epsilon$ = e.m.f., W = work done by source, Q = amount of positive charges

### Difference Between Electromotive Force And Potential Difference

• Electromotive Force of a source is the ability of the source to generate electrical energy from other forms of energy and supply it to the circuit whereas potential difference across a part of a circuit or device, measures the ability of the device or part of a circuit at converting electrical energy to other forms of energy.
• There will be no potential difference without electromotive force, but electromotive force exists whether a current flows or not.

## Potential Difference

When a torch bulb is connected to a battery, the torch bulb gets lit. The battery converts chemical energy into electrical energy and is therefore a source of electrical energy. The torch bulb converts electrical energy into heat and light and is therefore a sink of electrical energy.

• Dissipation of electrical energy between two points (e.g. across torch bulb) in an electrical circuit causes potential difference (p.d.) between those two points.

Potential difference between two points is defined as the energy converted from electrical to other forms when a coulomb of positive charge passes between the two points.

• SI unit: Volt (V). It is the same as that of e.m.f.. (Both are measures of electrical potential energy, e.m.f. is gained electrical energy while potential difference is lost electrical energy.)

$$V = \frac{W}{Q}$$

, where V = potential difference, W = work done in driving the charge between the two points, Q = amount of positive charges

One volt is the potential difference between two points in an electrical circuit when one joule of electrical energy is converted to other form of energy as one coulomb of charge passes from one point to the other.

IMPORTANT: There can be e.m.f. without a closed circuit. BUT there cannot be a potential difference without a closed circuit.

Analogy to waterfalls:

In order to help you understand the concept of potential difference better, you can think of a waterfall. In the case of a waterfall, the water flows due to a height difference. In electric circuits, current flows between two points due to the existence of potential difference between the two points. No potential difference = no current.

## Understanding Voltage: Visualizing Electrical Energy Transfer in Circuits

To enhance your comprehension of the concept of voltage, consider visualizing the electric current in a circuit as composed of individual “drops” of electricity. Each drop possesses a charge of 1 coulomb and carries a consistent quantity of electrical energy. Imagine a character, Mr. Coulomb, symbolizing one of these drops, journeying through the circuit, dispensing its energy to various components, with the majority being expended in the lamp. It’s crucial to note that it is the electrical energy that is consumed, not the charge or the current itself.

In this illustrative scenario, Mr. Coulomb circulates the circuit, shedding energy at different points, notably at the lamp, and is “recharged” with a new bundle of energy each time he traverses the battery. This analogy might imply rapid movement; however, we’ve previously established that electrons actually drift through the circuit at a slow pace. When the circuit is completed, energy instantaneously reaches the lamp, facilitated not by electrons directly from the battery but by those within the connecting wires. While this model aids in understanding, it doesn’t perfectly mirror reality.

Voltage is defined through demonstrations that illustrate how a power source with greater electromotive force (e.m.f.) imparts a larger bundle of energy to each coulomb, thus accelerating the rate of energy transfer, particularly visible in devices like lamps. Mathematically, the e.m.f. $E$ (in volts) of a supply is calculated by dividing the energy $W$ (in joules) transferred by the charge $Q$ (in coulombs) that moves through the circuit, expressed as $E = \frac{W}{Q}$.

The potential difference (p.d.) between two points in a circuit is defined as 1 volt when 1 joule of energy is transferred as 1 coulomb of charge moves between those two points. Simply put, 1 volt equals 1 joule per coulomb $1 \text{ V} = 1 \text{ J C}^{-1}$. If 2 joules are transferred per coulomb, the potential difference is 2 volts. More broadly, the potential difference $V$ (in volts) between two points is calculated as $V = \frac{W}{Q}$, where $W$ is the energy transferred in joules and $Q$ is the charge in coulombs. When dealing with a constant current $I$ (in amperes) over time $t$ (in seconds), the charge $Q$ can be represented as $I \times t$, leading to the formula for energy transferred being $W = I \times t \times V$. This approach offers a structured way to understand voltage and its implications on energy transfer within electrical circuits.

## Voltmeter Usage & Precision

A voltmeter, designed to measure electrical potential differences, must be connected in parallel to the component whose potential difference (p.d.) is being assessed. Ensure the positive terminal of the voltmeter connects to the component’s side where the current enters. Standard moving-coil voltmeters are limited to direct current (DC) measurements displayed on an analog scale.

The analog voltmeter features two distinct measurement ranges. The first range, 0–5 V, allows for readings up to 5.0 volts with each minor division indicating a 0.1 V increment, enabling measurements to be rounded to the nearest 0.1 V. Due to the human eye’s ability to discern even half a division, estimates can be made with significant accuracy to the nearest 0.05 V. The second range, 0–10 V, spans up to 10.0 volts, with each division representing 0.2 V, offering less precision compared to the 0–5 V scale.

Modern analog voltmeters, including multimeters, are evolved from moving-coil galvanometers, whereas digital multimeters utilize integrated circuits for enhanced functionality. On voltage settings, digital multimeters boast an exceptionally high input resistance (10 MΩ), minimizing their impact on circuits and ensuring highly accurate readings.

When employing a voltmeter, analog or digital, selecting an appropriate range is crucial based on the expected voltage. For voltages in the millivolt range, starting with the 10 mV setting is advisable, adjusting to a higher range if necessary to stay within scale.

Accuracy in measurement is underscored by the instrument’s calibrated scale. Documenting experimental procedures necessitates detailing the scales used, ensuring clarity and precision in reporting findings.

## Worked Examples

### Example 1

A circuit is set up with a lamp connected to a battery, allowing current to flow.

a. Find the voltage (potential difference) across the lamp when a work of 6 Joules is performed as 2 Coulombs of charge move through the lamp

b. If the voltage across the lamp rises to 5 Volts, calculate the energy transferred to the lamp with a current of 2 Amperes for a duration of 5 seconds

a. using the formula $V = \frac{W}{Q}$.

Therefore, the voltage across the lamp is $\frac{6 \, \text{J}}{2 \, \text{C}} = 3 \, \text{V}$.

b. First determine the charge using

\begin{aligned} Q &= I \times t \\ &= 2 \, \text{A} \times 5 \, \text{s} \\ &= 10 \, \text{C} \end{aligned}

By manipulating the formula $V = \frac{W}{Q}$ to find work,

\begin{aligned} W &= Q \times V \\ &= 10 \, \text{C} \times 5 \, \text{V} \\ &= 50 \, \text{J} \end{aligned}.

### Example 2

Determine the voltage (potential difference) across a lamp within an electrical circuit if 8 Joules of work is performed as 4 Coulombs of charge move through it.

Using the formula for voltage, $V = \frac{W}{Q}$, where $W$ is work done and $Q$ is the charge,

\begin{aligned}V &= \frac{8 \, \text{J}}{4 \, \text{C}} \\ &= 2 \, \text{V} \end{aligned}

### Example 3

If the potential difference across a lamp is 6 Volts, calculate the energy in joules transferred as 2 Coulombs of charge traverse the lamp.

The energy transferred, denoted by (W), when a charge, (Q), passes through a potential difference, (V), is given by $W = Q \times V$.

For a potential difference of 6 Volts and a charge of 2 Coulombs, the energy transferred is $W = 2 \, \text{C} \times 6 \, \text{V} = 12 \, \text{J}$.

### Example 4

With a potential difference of 6 Volts across a lamp, compute the work done while a current of 3 Amperes circulates through the lamp for a duration of 10 seconds.

To find the work done when a current of 3 Amperes flows for 10 seconds through a lamp with a potential difference of 6 Volts across it, first find the charge using $Q = I \times t$, where $I$ is current and $t$ is time. Then, calculate the work using $W = Q \times V$.

$$Q = 3 \, \text{A} \times 10 \, \text{s} = 30 \, \text{C}$$

$$W = 30 \, \text{C} \times 6 \, \text{V} = 180 \, \text{J}$$

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