Power and Energy (Grade 11 NSC Matric Physical Sciences): Revision Notes

Worked Example: Basic Power Calculation

Question: A circuit component has a voltage of 5 V across it and a resistance of 2 Ω. What is the power dissipated?

Solution:

Step 1: Identify what we know and what we need to find

• $V = 5$ V
• $R = 2$ Ω
• $P = ?$

Step 2: Choose the appropriate power formula

Since we have voltage and resistance, we use: $P = \frac{V^2}{R}$

Step 3: Substitute and solve

$P = \frac{V^2}{R} = \frac{(5)^2}{2} = \frac{25}{2} = 12.5 \text{ W}$

The power dissipated is 12.5 W.

Worked Example: Using Current and Resistance

Question: In a circuit, the resistance is 15 Ω and the current flowing through the resistor is 4 A. What is the power dissipated by the resistor?

Solution:

Step 1: Identify the known values

• $R = 15$ Ω
• $I = 4$ A
• $P = ?$

Step 2: Choose the appropriate formula

Since we know current and resistance, we use: $P = I^2R$

Step 3: Calculate the power

$P = I^2R = (4)^2 \times 15 = 16 \times 15 = 240 \text{ W}$

The power dissipated by the resistor is 240 W.

Worked Example: Power in a Series Circuit

Question: Two resistors (R₁ and R₂) are connected in series with a 6 V cell. The current flowing through both resistors is 0.25 A and R₁ = 1 Ω. Find the resistance and power of R₂.

Solution:

Step 1: Draw the circuit and identify known values

• $V = 6$ V (total voltage)
• $I = 0.25$ A (same current through both resistors)
• $R_1 = 1$ Ω
• $R_2 = ?$

Step 2: Find the total resistance

Using Ohm's Law for the complete circuit:

$R_{total} = \frac{V}{I} = \frac{6}{0.25} = 24 \text{ Ω}$

Step 3: Find R₂

In a series circuit: $R_{total} = R_1 + R_2$

$R_2 = R_{total} - R_1 = 24 - 1 = 23 \text{ Ω}$

Step 4: Calculate the power for R₂

$P_2 = I^2R_2 = (0.25)^2 \times 23 = 0.0625 \times 23 = 1.44 \text{ W}$

The resistance R₂ is 23 Ω and the power dissipated in R₂ is 1.44 W.

Worked Example: Complex Series-Parallel Circuit

Question: Given a complex circuit where the current leaving the battery is 1.07 A, the total power dissipated is 6.42 W, and various resistance ratios are provided. Determine the voltage of the battery and the power dissipated in each section.

Solution:

Step 1: Find the battery voltage

Using $P = V \times I$ for the entire circuit:

$V = \frac{P}{I} = \frac{6.42}{1.07} = 6.00 \text{ V}$

Step 2: Analyse the equivalent circuit

The complex parallel networks can be treated as equivalent resistors in series.

Step 3: Calculate total resistance

$R_{total} = \frac{V}{I} = \frac{6.00}{1.07} = 5.61 \text{ Ω}$

Step 4: Use the given ratios to find individual component values

Following the ratios provided ($R_{P1} : R_{P2} = 1:2$), we can determine:

• $R_{P1} = 1.87$ Ω
• $R_{P2} = 3.74$ Ω

Step 5: Calculate power in each parallel network

$P_{P1} = I^2R_{P1} = (1.07)^2 \times 1.87 = 2.14 \text{ W}$

$P_{P2} = I^2R_{P2} = (1.07)^2 \times 3.74 = 4.28 \text{ W}$

The battery voltage is 6.00 V, with powers of 2.14 W and 4.28 W in the respective parallel networks.

Worked Example: Energy Calculation

Question: A 30 W light bulb is left on for 8 hours overnight. How much energy was consumed?

Solution:

Step 1: Convert time to seconds

$8 \text{ hours} = 8 \times 3600 \text{ s} = 28\,800 \text{ s}$

Step 2: Calculate the energy

$E = P \times t = 30 \times 28\,800 = 864\,000 \text{ J}$

The light bulb consumed 864,000 J of energy.

Worked Example: Power and Energy Combined

Question: A resistor has a resistance of 27 Ω and a current of 3.3 A flows through it. What is the power dissipated and how much energy is consumed in 35 seconds?

Solution:

Step 1: Calculate the power

$P = I^2R = (3.3)^2 \times 27 = 10.89 \times 27 = 294.03 \text{ W}$

Step 2: Calculate the energy

$E = P \times t = 294.03 \times 35 = 10\,291.05 \text{ J}$

The power dissipated is 294.03 W and the energy consumed is 10,291.05 J.

Worked Example: Cost Calculation

Question: How much does it cost to run a 900 W microwave for 2.5 minutes if electricity costs 61.6 cents per kWh?

Solution:

Step 1: Convert to consistent units

• Power: $900 \text{ W} = 0.9 \text{ kW}$
• Time: $2.5 \text{ minutes} = \frac{2.5}{60} = 4.17 \times 10^{-2} \text{ hours}$

Step 2: Calculate energy usage

$E = P \times t = 0.9 \times 4.17 \times 10^{-2} = 3.75 \times 10^{-2} \text{ kWh}$

Step 3: Calculate cost

$\text{Cost} = \text{Energy} \times \text{Price} = 3.75 \times 10^{-2} \times 61.6 = 2.31 \text{ cents}$

It costs 2.31 cents to run the microwave for 2.5 minutes.