recall and use P = VI, P = I 2R and P = V 2 / R

Potential Difference and Power

Learning Objective

Recall and use the power relationships:

• $P = VI$
• $P = I^{2}R$
• $P = \dfrac{V^{2}}{R}$

1. Potential Difference (Voltage)

The potential difference between two points is the work done per unit charge in moving a charge from one point to the other.

Mathematically,

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

where $W$ is the work done (in joules) and $Q$ is the charge (in coulombs). The SI unit is the volt (V).

2. Electrical Power

Power is the rate at which electrical energy is transferred or converted.

The basic definition is

$$P = \frac{E}{t}$$

where $E$ is energy (J) and $t$ is time (s). The SI unit is the watt (W).

2.1 Deriving the Power Formulas

1. Start from the definition of work done by a charge moving through a potential difference: $$W = Q\Delta V$$
2. Since $I = \dfrac{Q}{t}$, substitute $Q = It$ into the work expression: $$W = (It)\Delta V$$
3. Divide by $t$ to obtain power: $$P = \frac{W}{t} = I\Delta V$$
4. Using Ohm’s law, $\Delta V = IR$, replace $\Delta V$: $$P = I(IR) = I^{2}R$$
5. Alternatively, replace $I$ with $\dfrac{\Delta V}{R}$: $$P = \Delta V\left(\frac{\Delta V}{R}\right) = \frac{\Delta V^{2}}{R}$$

2.2 Summary of Power Relationships

Form	Expression	When to Use
$P = VI$	Power in terms of voltage and current	Both $V$ and $I$ are known
$P = I^{2}R$	Power in terms of current and resistance	Current and resistance are known (e.g., heating of a resistor)
$P = \dfrac{V^{2}}{R}$	Power in terms of voltage and resistance	Voltage across and resistance are known

3. Practical Applications

Understanding these relationships helps in:

• Calculating the power rating required for resistors and fuses.
• Estimating energy consumption of appliances: $E = Pt$.
• Analyzing heating effects in conductors (Joule heating).

4. Example Problems

Example 1

A resistor of $10\ \Omega$ carries a current of $2\ \text{A}$. Find the power dissipated.

Solution using $P = I^{2}R$:

$$P = (2\ \text{A})^{2}\times 10\ \Omega = 4\times10 = 40\ \text{W}$$

Example 2

A lamp is connected across a $240\ \text{V}$ supply and draws $0.5\ \text{A}$. Determine the power consumed and the resistance of the lamp.

Power:

$$P = VI = 240\ \text{V}\times0.5\ \text{A}=120\ \text{W}$$

Resistance using $R = \dfrac{V}{I}$:

$$R = \frac{240\ \text{V}}{0.5\ \text{A}} = 480\ \Omega$$

Example 3

A heater is rated at $1500\ \text{W}$ when connected to a $120\ \text{V}$ supply. Find the current through the heater and its resistance.

Current from $P = VI$:

$$I = \frac{P}{V} = \frac{1500\ \text{W}}{120\ \text{V}} = 12.5\ \text{A}$$

Resistance from $P = \dfrac{V^{2}}{R}$:

$$R = \frac{V^{2}}{P} = \frac{(120\ \text{V})^{2}}{1500\ \text{W}} = \frac{14400}{1500}=9.6\ \Omega$$

5. Practice Questions

1. A $5\ \Omega$ resistor has $3\ \text{A}$ flowing through it. Calculate the voltage across it and the power dissipated.
2. A device operates at $12\ \text{V}$ and consumes $24\ \text{W}$. Determine the current drawn and the equivalent resistance.
3. Two resistors, $R_{1}=8\ \Omega$ and $R_{2}=12\ \Omega$, are connected in series across a $24\ \text{V}$ battery. Find the total power supplied by the battery.
4. A heating element is designed to produce $2000\ \text{W}$ when connected to a $240\ \text{V}$ supply. What resistance must the element have?
5. If a current of $0.2\ \text{A}$ flows through a $50\ \Omega$ resistor, what is the rate of energy conversion (power) in kilojoules per hour?

6. Key Points to Remember

• Power can be expressed in three interchangeable forms: $P = VI$, $P = I^{2}R$, $P = \dfrac{V^{2}}{R}$.
• Choose the form that uses the quantities you already know.
• Always keep track of units: volts (V), amperes (A), ohms (Ω), watts (W).
• Energy consumed over time is $E = Pt$ (joules), or $E\text{(kWh)} = \dfrac{P\text{(kW)}\times t\text{(h)}}{1}$ for practical electricity billing.