Potential difference and power

1. Overview

Potential difference (p.d.) and power are the fundamental quantities used to describe the transfer of energy in electrical circuits. In any circuit, energy is provided by a source (like a battery) and transferred to components (like resistors or lamps). The potential difference across a component quantifies the amount of electrical energy converted into other forms—such as thermal energy, light, or sound—per unit of charge passing through it. Electrical power measures the rate at which this energy transfer occurs. Mastering these concepts requires a precise understanding of the relationship between work done, charge, current, and resistance.

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Key Definitions

• Potential Difference ($V$): The energy transferred per unit charge from electrical energy to other forms as the charge passes through a component.
• The Volt ($\text{V}$): The unit of potential difference, defined as one joule per coulomb ($1\text{ V} = 1\text{ J C}^{-1}$).
• Electrical Power ($P$): The rate at which energy is transferred or work is done within a circuit.
• The Watt ($\text{W}$): The unit of power, defined as the transfer of energy at a rate of one joule per second ($1\text{ W} = 1\text{ J s}^{-1}$).

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Content

3.1 The Nature of Potential Difference ($V$)

When a source of electromotive force (e.m.f.) is connected to a circuit, it creates an electric field within the conductors. This field exerts a force on the free electrons (charge carriers), causing them to drift through the circuit.

As these electrons move through a component with resistance (such as a filament lamp), they collide with the positive ions of the metal lattice. During these collisions, the electrons transfer some of their kinetic energy to the ions, increasing the lattice's vibrational energy. Macroscopically, this is observed as a rise in temperature (thermal energy) or the emission of light.

The potential difference $V$ across the component is the measure of this energy "loss" from the perspective of the charge carriers. It is defined by the equation:

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

Where:

• $V$ is the potential difference in volts ($\text{V}$)
• $W$ is the work done or energy transferred in joules ($\text{J}$)
• $Q$ is the charge in coulombs ($\text{C}$)

3.2 Electrical Power ($P$)

Power is defined generally as the rate of doing work ($P = \frac{W}{t}$). In an electrical context, we can derive a specific formula for power by considering the definitions of potential difference and current.

Derivation of $P = VI$:

1. Start with the definition of power: $P = \frac{W}{t}$
2. From the definition of potential difference, the work done is: $W = VQ$
3. Substitute this into the power equation: $P = \frac{VQ}{t}$
4. Recall the definition of electric current: $I = \frac{Q}{t}$
5. Substitute $I$ into the equation: $P = VI$

This equation shows that the power dissipated by a component depends on both the potential difference across it and the current flowing through it.

3.3 Power, Resistance, and Ohm's Law

For components that obey Ohm’s Law ($V = IR$), we can express power in terms of resistance. This is particularly useful when one of the variables ($V$ or $I$) is not directly measured.

Derivation of $P = I^2R$:

1. Start with $P = VI$
2. Substitute the Ohmic relationship $V = IR$
3. $P = (IR) \times I$
4. $P = I^2R$ Application: This formula is often used to calculate "Joule heating." It shows that for a fixed resistance, the power dissipated is proportional to the square of the current. This is why high-voltage transmission lines use low current to minimize power losses in the wires.

Derivation of $P = \frac{V^2}{R}$:

1. Start with $P = VI$
2. Substitute the Ohmic relationship $I = \frac{V}{R}$
3. $P = V \times \left(\frac{V}{R}\right)$
4. $P = \frac{V^2}{R}$ Application: This formula is useful when components are connected in parallel to a constant voltage source (like household mains). It shows that for a fixed voltage, a component with lower resistance will dissipate more power.

3.4 Energy Transfer and Time

Since $P = \frac{W}{t}$, the total energy transferred ($W$) over a period of time ($t$) can be calculated by: $$W = Pt$$ By substituting the various power formulas, we get:

• $W = VIt$
• $W = I^2Rt$
• $W = \frac{V^2}{R}t$

In exam questions, ensure that time is always converted to seconds before performing these calculations.

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Key Equations

Equation	Meaning	Status
$V = \frac{W}{Q}$	Potential difference = Work done / Charge	Must Memorise
$P = VI$	Power = Potential difference $\times$ Current	Must Memorise
$P = I^2R$	Power = Current$^2$ $\times$ Resistance	Must Memorise
$P = \frac{V^2}{R}$	Power = Potential difference$^2$ / Resistance	Must Memorise
$W = VIt$	Energy transferred = P.D. $\times$ Current $\times$ Time	Must Memorise

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Common Mistakes to Avoid

• ❌ Wrong: Confusing Potential Difference (p.d.) with Electromotive Force (e.m.f.). ✓ Right: While both are measured in Volts, p.d. is energy transferred from electrical to other forms (used by a load), whereas e.m.f. is energy transferred to electrical form (provided by a source).
• ❌ Wrong: Forgetting to square the values in $I^2R$ or $V^2/R$. ✓ Right: Always write the formula first and use brackets during substitution, e.g., $P = (2.0 \times 10^{-3})^2 \times 50$.
• ❌ Wrong: Using time in minutes or hours when calculating energy. ✓ Right: The Watt and Joule are defined per second. Always convert time to seconds ($1\text{ hour} = 3600\text{ s}$).
• ❌ Wrong: Assuming $P$ is always proportional to $V$. ✓ Right: $P \propto V$ only if $I$ is constant. In most circuits, $R$ is the constant. If $V$ doubles, $I$ also doubles (since $I = V/R$), so $P$ actually increases by a factor of four ($P = V^2/R$).
• ❌ Wrong: Using the total circuit voltage for a single component's power. ✓ Right: When using $P = VI$ or $P = V^2/R$, the $V$ must be the potential difference across that specific component, not the total e.m.f. of the battery (unless it is the only component).

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Exam Tips

1. Unit Conversions: Cambridge examiners frequently use non-SI prefixes. Convert $\text{mA}$ to $10^{-3}\text{ A}$, $\text{k}\Omega$ to $10^3\text{ }\Omega$, and $\text{MJ}$ to $10^6\text{ J}$ immediately.
2. The "Show That" Requirement: If a question asks you to "show that $P = I^2R$", you must show the starting point ($P=VI$), the substitution ($V=IR$), and the final algebraic step. Skipping the middle step may lose marks.
3. Efficiency Links: You may be asked to calculate the efficiency of a device. Remember: $$\text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\%$$ The "Total Power Input" is often calculated using $P = VI$.
4. Internal Resistance: In later topics, you will learn about internal resistance. For now, assume wires and batteries are "ideal" unless the question states otherwise.
5. Significant Figures: Always provide your final answer to the same number of significant figures as the least precise data point given in the question (usually 2 or 3 s.f.).

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7. Worked Examples

Worked Example 1 — Energy and Charge

A battery-powered torch is switched on for $20\text{ minutes}$. During this time, the battery maintains a constant potential difference of $4.5\text{ V}$ and a current of $0.60\text{ A}$ flows through the bulb. Calculate the total energy transferred by the bulb.

Step 1: List the known quantities in SI units

• $V = 4.5\text{ V}$
• $I = 0.60\text{ A}$
• $t = 20\text{ minutes} = 20 \times 60 = 1200\text{ s}$

Step 2: Select the appropriate equation

• $W = VIt$

Step 3: Substitute and calculate

• $W = 4.5 \times 0.60 \times 1200$
• $W = 2.7 \times 1200$
• $W = 3240\text{ J}$

Step 4: Final answer with units and s.f.

• $W = 3200\text{ J}$ (to 2 s.f., matching the input data)

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Worked Example 2 — Power Dissipation in a Resistor

A resistor is labeled "$10\text{ }\Omega, 2.5\text{ W}$". This means the resistor has a resistance of $10\text{ }\Omega$ and can safely dissipate a maximum power of $2.5\text{ W}$ without overheating. Calculate the maximum potential difference that can be applied across this resistor.

Step 1: List the known quantities

• $R = 10\text{ }\Omega$
• $P_{max} = 2.5\text{ W}$

Step 2: Select the appropriate equation

• $P = \frac{V^2}{R}$

Step 3: Rearrange for $V$

• $V^2 = P \times R$
• $V = \sqrt{P \times R}$

Step 4: Substitute and calculate

• $V = \sqrt{2.5 \times 10}$
• $V = \sqrt{25}$
• $V = 5.0\text{ V}$

Step 5: Final answer

• The maximum potential difference is $5.0\text{ V}$.

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Worked Example 3 — Comparing Power in Series

Two resistors, $R_1 = 100\text{ }\Omega$ and $R_2 = 200\text{ }\Omega$, are connected in series to a power supply. The current in the circuit is $0.050\text{ A}$. Calculate the ratio of the power dissipated in $R_1$ to the power dissipated in $R_2$.

Step 1: Identify the constant variable

• In a series circuit, the current ($I$) is the same through all components.

Step 2: Select the power formula involving $I$ and $R$

• $P = I^2R$

Step 3: Set up the ratio

• $\frac{P_1}{P_2} = \frac{I^2R_1}{I^2R_2}$
• Since $I$ is the same, it cancels out: $\frac{P_1}{P_2} = \frac{R_1}{R_2}$

Step 4: Calculate the ratio

• $\frac{P_1}{P_2} = \frac{100}{200} = 0.5$

Step 5: Final answer

• The ratio of power $P_1 : P_2$ is $1 : 2$. (Note: In series, the larger resistor dissipates more power).