Electric Power (VCE SSCE Physics): Revision Notes

Worked Example: Power in a kettle

Let's examine a practical example using an electric kettle rated at 2 kW.

Part (a): Calculate the thermal energy delivered to water in 1 minute.

Given:

• $P = 2 \text{ kW} = 2 \times 10^3 \text{ W}$
• $t = 1 \text{ min} = 60 \text{ s}$

Rearranging $P = \frac{E}{t}$ to find energy:

$E = P \times t$

$E = 2 \times 10^3 \times 60$

$E = 1.2 \times 10^5 \text{ J} = 120 \text{ kJ}$

The kettle delivers 120 kJ of energy to the water in one minute.

Part (b): How long would the kettle need to run to supply 4 kJ of energy?

Given:

• $P = 2 \times 10^3 \text{ W}$
• $E = 4 \text{ kJ} = 4 \times 10^3 \text{ J}$

Rearranging to find time:

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

$t = \frac{4 \times 10^3}{2 \times 10^3}$

$t = 2 \text{ s}$

The kettle takes only 2 seconds to deliver 4 kJ of energy.

Part (c): A different kettle supplies 3 kJ of energy per 30 seconds. Calculate its power.

Given:

• $E = 3 \text{ kJ} = 3 \times 10^3 \text{ J}$
• $t = 30 \text{ s}$

Using the power formula:

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

$P = \frac{3 \times 10^3}{30}$

$P = 100 \text{ W}$

This kettle has a power of only 100 W, making it much slower to boil water than the 2 kW kettle from parts (a) and (b).

Worked Example: Power in an electric stove

An electric stove uses 1.8 MJ of energy to heat its hotplate every 10 minutes.

Part (a): Calculate the power of the stove in watts.

Given:

• $E = 1.8 \text{ MJ} = 1.8 \times 10^6 \text{ J}$
• $t = 10 \text{ min} = 10 \times 60 = 600 \text{ s}$

Using the power formula:

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

$P = \frac{1.8 \times 10^6}{600}$

$P = 3000 \text{ W}$

The stove has a power of 3000 W or 3 kW.

Part (b): Calculate how long the stove takes to use 1.7 kJ of energy.

Given:

• $P = 3000 \text{ W}$
• $E = 1.7 \text{ kJ} = 1700 \text{ J}$

Rearranging for time:

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

$t = \frac{1700}{3000}$

$t = 0.57 \text{ s}$

The stove uses 1.7 kJ in just 0.57 seconds.

Part (c): If the stove's power increases by 500 W, calculate the energy used in 40 seconds.

New power: $P = 3000 + 500 = 3500 \text{ W}$

Given $t = 40 \text{ s}$:

$E = Pt$

$E = 3500 \times 40$

$E = 0.14 \times 10^6 \text{ J} = 0.14 \text{ MJ}$

At the higher power, the stove uses 0.14 MJ in 40 seconds.

Worked Example: Power in a light bulb

Consider a light bulb connected to a 12 V battery, carrying a current of 0.5 A.

Part (a): Calculate the power drawn by the light bulb.

Given:

• $V = 12 \text{ V}$
• $I = 0.5 \text{ A}$

Using $P = VI$:

$P = 12 \times 0.5$

$P = 6 \text{ W}$

The light bulb draws 6 W of power.

Part (b): If the bulb is switched on for 2 minutes, how much energy does it transform?

Given:

• $P = 6 \text{ W}$
• $t = 2 \text{ min} = 120 \text{ s}$

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

$E = Pt$

$E = 6 \times 120$

$E = 720 \text{ J}$

The bulb transforms 720 J of electrical energy (into light and heat) over 2 minutes.

Worked Example: Comparing power consumption

A kettle and an induction cooktop both connect to 230 V mains power. The kettle draws 10 A while the cooktop draws 32 A. Calculate the power used by each appliance.

For both appliances, $V = 230 \text{ V}$.

For the kettle ($I = 10 \text{ A}$):

$P = VI$

$P = 230 \times 10$

$P = 2300 \text{ W} = 2.3 \text{ kW}$

For the cooktop ($I = 32 \text{ A}$):

$P = VI$

$P = 230 \times 32$

$P = 7360 \text{ W} = 7.36 \text{ kW}$

The cooktop uses more than three times the power of the kettle. If both were used to heat the same mass of water from 20°C, the induction cooktop would reach boiling point much faster due to its higher power delivery.

Worked Example: Power loss due to resistance

A caravan needs connecting to a 240 V AC power supply using a 10 m extension cable. Two cables are available with resistances of 1 Ω and 5 Ω. The caravan appliances draw 10 A. Find the power loss in each cable.

Since we're calculating power loss due to heating in the wire, and we know current and resistance, we should use $P = I^2R$.

For cable 1 ($R = 1 \text{ Ω}$, $I = 10 \text{ A}$):

$P = I^2R$

$P = 10^2 \times 1$

$P = 100 \text{ W}$

Cable 1 wastes 100 W as heat.

For cable 2 ($R = 5 \text{ Ω}$, $I = 10 \text{ A}$):

$P = I^2R$

$P = 10^2 \times 5$

$P = 500 \text{ W}$

Cable 2 wastes 500 W as heat.

The 5 Ω cable transforms 400 W more power than the 1 Ω cable due to its higher resistance. You should choose the cable with lower resistance to minimise energy waste and ensure maximum power reaches the caravan appliances.

Worked Example: Resistance from power consumption

Calculate and compare the resistance of a 5 W LED bulb and a 9 W LED bulb when each is connected to 230 V mains supply.

Since we know power and voltage, we should use $P = \frac{V^2}{R}$ and rearrange to find resistance.

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

For the 5 W LED bulb ($V = 230 \text{ V}$, $P = 5 \text{ W}$):

$R = \frac{V^2}{P}$

$R = \frac{230^2}{5}$

$R = 10580 \text{ Ω} = 10.6 \text{ kΩ}$

For the 9 W LED bulb ($V = 230 \text{ V}$, $P = 9 \text{ W}$):

$R = \frac{230^2}{9}$

$R = 5878 \text{ Ω} = 5.9 \text{ kΩ}$

The 5 W LED bulb has almost twice the resistance of the 9 W LED bulb. This makes sense because at the same voltage, higher resistance means lower current, which produces lower power according to $P = VI$.

Worked Example: Power loss in series and parallel

Consider two light globes with resistances of 1000 Ω and 3000 Ω connected to a 24 V battery.

Part (a): Calculate the power used when the globes are connected in parallel.

First, draw a circuit diagram and calculate the equivalent resistance:

For resistors in parallel:

$\frac{1}{R_{\text{equivalent}}} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R_{\text{equivalent}}} = \frac{1}{1000} + \frac{1}{3000}$

$\frac{1}{R_{\text{equivalent}}} = \frac{1}{750}$

$R_{\text{equivalent}} = 750 \text{ Ω}$

The potential difference across the parallel combination is 24 V. Using $P = \frac{V^2}{R}$:

$P = \frac{V^2}{R}$

$P = \frac{24^2}{750}$

$P = 0.768 \text{ W} \approx 0.77 \text{ W}$

The parallel circuit uses 0.77 W of power.

Part (b): Calculate the power used if the same globes are connected in series.

For resistors in series:

$R_{\text{equivalent}} = R_1 + R_2$

$R_{\text{equivalent}} = 1000 + 3000 = 4000 \text{ Ω}$

The potential difference across the series combination is still 24 V. Using the same power formula:

$P = \frac{V^2}{R}$

$P = \frac{24^2}{4000}$

$P = 0.144 \text{ W}$

The series circuit uses only 0.144 W of power.

Comparison: The parallel circuit uses significantly more power (0.77 W) than the series circuit (0.144 W). This occurs because the parallel combination has lower equivalent resistance, allowing more current to flow and thus transforming energy at a faster rate. The parallel globes will be brighter (transforming more energy per second) but will drain the battery more quickly.

Worked Example: Step-by-step approach

Consider this example circuit:

Step 1: Simplify the circuit to find equivalent resistance

The circuit can be reduced through several stages:

• Combine series resistors: $R_{\text{series}} = R_1 + R_2$
• Combine identical parallel resistors: $R_{\text{parallel}} = \frac{R}{2}$ (for two identical resistors)
• Continue until you reach a single equivalent resistance

Following the diagram, the complex circuit with multiple 10 Ω, 16 Ω, and 8 Ω resistors simplifies to a single equivalent resistance of 10 Ω.

Step 2: Calculate total current

Using Ohm's law with the battery voltage (60 V) and equivalent resistance (10 Ω):

$I = \frac{V}{R_{\text{equivalent}}}$

$I = \frac{60}{10}$

$I = 6 \text{ A}$

Step 3: Calculate power supplied by the battery

Using $P = VI$:

$P = 60 \times 6$

$P = 360 \text{ W}$

This power (360 W) is dissipated among all the resistors in the circuit.