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

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.