Unit outcomes

By the end of this unit you will be able to:

• Define and calculate energy transfer in an electrical circuit.
• Define and calculate electrical power.
• Identify the application of power on tools, etc.
• Determine the relationship between current, pd, resistance and power.

What you should know

Before you start this unit, make sure you can:

• Identify components in electrical circuits. Refer to level 3 subject outcome 4.3 unit 2 if you need help with this.
• Identify the relationship between current, voltage and resistance in electrical circuits. Refer to level 3 subject outcome 4.3 unit 1 if you need help with this.

Introduction

In this unit you will be learning about energy transfer in electrical circuits as well as applying the principle of power to electrical circuits.

Energy transfer in electrical circuits

A source of energy is required to drive current round a complete circuit. This is provided by batteries or an electrical generator. The energy is used to do work on the electrons in the circuit. The emf of the battery or the induced emf in a generator is the total energy supplied per coulomb of charge. It is also referred to as potential difference as the electrical potential energy of the charge changes as it moves through the battery or coils of the generator.

This can be written as a formula: [latex]\scriptsize V=\displaystyle \frac{W}{q}[/latex]

Where:

[latex]\scriptsize V[/latex] is the voltage (emf) in volts
[latex]\scriptsize W[/latex] is the work (energy) in joules (J)
[latex]\scriptsize q[/latex] is the charge in coulombs (C)

For example:

• A [latex]\scriptsize 4.5\text{ V}[/latex] cell supplies [latex]\scriptsize 4.5\text{ J}[/latex] of electrical energy to every coulomb of charge that passes through the cell.
• A [latex]\scriptsize 9\text{ V}[/latex] cell supplies [latex]\scriptsize 9\text{ J}[/latex] of electrical potential energy to every coulomb of charge that passes through the cell.

Example 2.1

Calculate the energy transferred by [latex]\scriptsize 10\text{ C}[/latex] of charge as it passes through a [latex]\scriptsize 12\text{ V}[/latex] battery.

Solution

Step 1: Write down the given information and check the units

[latex]\scriptsize \begin{align*}q&=10\text{ C}\\\text{V}&= \text{12 V}\\\text{W}&= \text{?}\end{align*}[/latex]

Step 2: Write down the formula, substitute the values and solve

[latex]\scriptsize \begin{align*}W&=Vq\\&=\text{ 12 x 10}\\&=\text{ 120 J}\end{align*}[/latex]

Exercise 2.1

• Calculate the emf of a battery if it supplies [latex]\scriptsize 250\text{ J}[/latex] of energy when [latex]\scriptsize 10\text{ C}[/latex] of charge pass through it.
• Calculate how much energy is transferred when [latex]\scriptsize 0.5\text{ C}[/latex] of charge passes through a cell of [latex]\scriptsize 6\text{ V}[/latex].

The full solutions are at the end of the unit.

Power in electrical circuits

As charges pass through resistors in an electrical circuit the electrical energy they carry is transformed into other types of energy in the resistors. Power is a measure of how rapidly this energy transformation or work is done, and is defined as the rate at which the work is done, or work done per unit time. Work is measured in joules [latex]\scriptsize \displaystyle (\text{J)}[/latex] and time in seconds ([latex]\scriptsize \text {s}[/latex]) so power will be [latex]\scriptsize \text{J}\text{.s}[/latex] which we call a watt ([latex]\scriptsize \text {W}[/latex]). The following formula can be used to calculate power:

[latex]\scriptsize P=\displaystyle \frac{W}{{\Delta t}}[/latex]

Where:

[latex]\scriptsize W[/latex] is the work/energy in joules ([latex]\scriptsize \text {J}[/latex])
[latex]\scriptsize \Delta t[/latex] is the time in seconds ([latex]\scriptsize \text {s}[/latex])
[latex]\scriptsize P[/latex] is the power in watts ([latex]\scriptsize \text {W}[/latex])

Application of power rating

Electrical appliances have a power rating which is an indication of how fast they can convert electrical energy into other types of energy. A heater that has a power rating of [latex]\scriptsize 850\text{ W}[/latex] means that [latex]\scriptsize 850\text{ J}[/latex] of electrical energy is converted into heat energy every second.

When purchasing an electrical appliance, it is important to choose the correct power rating for the purpose. For example, if you need to heat a large room, it would be better to buy a heater with a power rating of [latex]\scriptsize 2\text{ 000 W}[/latex] rather than one with a power rating of [latex]\scriptsize 1\text{ 000 W}[/latex] as the rate at which electrical energy is transformed into heat energy is double in the heater with the higher power rating. However, you need to bear in mind that the electricity consumption and therefore the cost of running the higher rated heater will be double that of the one with the lower power rating.

As another example, a battery-operated drill may have a power rating of [latex]\scriptsize 400\text{ W}[/latex] whereas one using mains supply may have a power rating of [latex]\scriptsize 1\text{ 000 W}[/latex]. If you are drilling holes in soft wood, the battery-operated drill may supply sufficient mechanical energy per second, however, if you are drilling into concrete, the energy provided per second would not be sufficient and you would be better advised to use the higher rated drill.

Using multiple appliances with high power ratings at the same time may overload circuits and cause the electricity supply to trip.

Example 2.2

Calculate the power of a hairdryer that can convert [latex]\scriptsize \text{192 kJ}[/latex] of energy in two minutes.

Solution

Step 1: Write down the given information and check the units

[latex]\scriptsize \begin{align*}W&=192\text{ kJ = 192 x 1}{{\text{0}}^{3}}\text{ J}\\\Delta \text{t}&= \text{ 2 mins}\text{. = 120 s}\\\text{P}&= \text{?}\end{align*}[/latex]

Step 2: Write down the formula, substitute the values and solve

[latex]\scriptsize \begin{align*}P&=\displaystyle \frac{W}{{\Delta t}}\\&=\displaystyle \frac{{\text{192 x 1}{{\text{0}}^{3}}}}{{120}}\\&=\text{ 1 600 W}\end{align*}[/latex]

Exercise 2.2

1. How long will it take a kettle to produce [latex]\scriptsize \text{3 x 1}{{\text{0}}^{6}}\text{ J}[/latex] of energy if it has a power rating of [latex]\scriptsize 1\text{ 500 W}[/latex]?
2. How much mechanical energy is produced by a fan with a power rating of [latex]\scriptsize 400\text{ W}[/latex] in one hour?
3. Two light bulbs, one rated at [latex]\scriptsize 200\text{ W}[/latex] and the other at [latex]\scriptsize 60\text{ W}[/latex] are available to light a dark parking area. Explain which light bulb should be chosen.

The full solutions are at the end of the unit.

The relationship between current, voltage (PD), resistance and power

The energy used by a component in a circuit with a particular resistance ([latex]\scriptsize R[/latex]) depends on the potential difference ([latex]\scriptsize V[/latex]) across the component, the current ([latex]\scriptsize I[/latex]) through it and the length of time ([latex]\scriptsize \Delta t[/latex]) for which the current flows. The relationship between these variables is given in the following formulae:

[latex]\scriptsize W=VI\Delta[/latex]    [latex]\scriptsize W={{I}^{2}}R\Delta t[/latex]    [latex]\scriptsize W\displaystyle \frac{{{{V}^{2}}}}{R}\Delta t[/latex]

The relationship of these variables with the power of the component can be determined by combining these formulae with the defining formula for power:

[latex]\scriptsize P=\displaystyle \frac{W}{{\Delta t}}\text{ or }W\text{ }=\text{ }P\Delta t[/latex]

This gives three formulae that can be used to calculate the power of an electrical component:

[latex]\scriptsize P=VI[/latex]     [latex]\scriptsize P={{I}^{2}}R[/latex]     [latex]\scriptsize P=\displaystyle \frac{{{{V}^{2}}}}{R}[/latex]

All these formulae are very useful when determining the relationship between the variables in an electrical appliance.

These relationships are:

• From [latex]\scriptsize P=VI[/latex], the power is directly proportional to the voltage and the current.
• From [latex]\scriptsize P={{I}^{2}}R[/latex], the power is directly proportional to the square of the current and the resistance.
• From [latex]\scriptsize P=\displaystyle \frac{{{{V}^{2}}}}{R}[/latex], power is directly proportional to voltage and inversely proportional to resistance.

Example 2.3

The resistance of a resistor is [latex]\scriptsize 27\text{ }\Omega[/latex] and the current is [latex]\scriptsize 3.3\text{ A}[/latex]. How much energy is dissipated in the resistor in [latex]\scriptsize 35\text{ s}[/latex]?

Solution

Step 1: List the given information and check the units

[latex]\scriptsize \begin{align*}I&=3.3\text{ A}\\R&=27\text{ }\Omega \\\Delta t&=35\text{ s}\\W&=\text{?}\end{align*}[/latex]

Step 2: Choose the correct formula

[latex]\scriptsize W={{I}^{2}}R\Delta t\text{ }[/latex]

Step 3: Substitute the values and calculate the answer

[latex]\scriptsize \begin{align*}W&={{(3.3)}^{2}}(27)(35)\\&=\text{ 10 291}\text{.05 J }\end{align*}[/latex]

Example 2.4

A kettle connected to a [latex]\scriptsize 220\text{ V}[/latex] power supply and a resistance of [latex]\scriptsize 50\text{ }\Omega[/latex] takes three minutes to boil water. How much energy is used?

Solution

Step 1: List the given information and check the units

[latex]\scriptsize \begin{align*}V&=220\text{ V}\\R&=100\text{ }\Omega \\\Delta t&=\text{ 3 minutes = 180 s}\\W&=?\end{align*}[/latex]

Step 2: Choose the correct formula

[latex]\scriptsize W=\displaystyle \frac{{{{V}^{2}}}}{R}\Delta t[/latex]

Step 3: Substitute the values and calculate the answer

[latex]\scriptsize \begin{align*}W&=\displaystyle \frac{{{{{220}}^{2}}}}{{50}}(180)\\&=\text{ 174 240 J}\end{align*}[/latex]

Example 2.5

Given that a circuit component has a voltage of [latex]\scriptsize 5\text{ V}[/latex] and a resistance of [latex]\scriptsize 2\text{ }\Omega[/latex], calculate the power of the component.

Solution

Step 1: List the given information and check the units

[latex]\scriptsize \begin{align*}V&=5\text{ V}\\R&=2\text{ }\Omega \\P&=?\end{align*}[/latex]

Step 2: Choose the correct formula

[latex]\scriptsize P=\displaystyle \frac{{{{V}^{2}}}}{R}[/latex]

Step 3: Substitute the values and calculate the answer

[latex]\scriptsize \begin{align*}P&=\displaystyle \frac{{{{5}^{2}}}}{2}\\&=\text{ 12}\text{.5 W}\end{align*}[/latex]

Example 2.6

A resistor has a resistance of [latex]\scriptsize 15\text{ }\Omega[/latex] and the current through it is [latex]\scriptsize 400\text{ mA}[/latex]. What is the power of the resistor?

Solution

Step 1: List the given information and check the units

[latex]\scriptsize \begin{align*}I&=400\text{ mA = 0}\text{.4 A}\\R&=15\text{ }\Omega \\P&=?\end{align*}[/latex]

Step 2: Choose the correct formula

[latex]\scriptsize P={{I}^{2}}R[/latex]

Step 3: Substitute the values and calculate the answer

[latex]\scriptsize \begin{align*}P&=({{0.4}^{2}})(15)\\&=\text{ 2}\text{.4 W}\end{align*}[/latex]

Summary

• Batteries and generators supply energy to electrical circuits.
• The emf of a battery or generator is defined as the total energy supplied per coulomb of charge. It can be calculated using the formula: [latex]\scriptsize V=\displaystyle \frac{W}{q}[/latex].
• Power is the rate at which energy is transferred or transformed. It can be calculated using the formula: [latex]\scriptsize P=\displaystyle \frac{W}{t}[/latex].
• Power is measured in the units watts ([latex]\scriptsize \text{W}[/latex]) which are equivalent to [latex]\scriptsize \text{J}\text{.}{{\text{s}}^{{-1}}}[/latex].
• The energy transformed by a component in an electrical circuit is determined by the voltage across it, the current flowing through it and its resistance. The following formulae can be used to calculate this energy transformation:
  [latex]\scriptsize W=VI\Delta t[/latex]     [latex]\scriptsize W={{I}^{2}}R\Delta t[/latex]     [latex]\scriptsize W=\displaystyle \frac{{{{V}^{2}}}}{R}\Delta t[/latex]
• The following formulae show the relationship between the power, the voltage, the current and the resistance of an electrical device:
  [latex]\scriptsize P=VI[/latex]     [latex]\scriptsize P={{I}^{2}}R[/latex]     [latex]\scriptsize P=\displaystyle \frac{{{{V}^{2}}}}{R}[/latex]

Unit 2: Assessment

Suggested time to complete: 30 minutes

1. Energy of [latex]\scriptsize 4\text{ x 1}{{\text{0}}^{3}}\text{ J}[/latex] is used to move [latex]\scriptsize 2\text{ x 1}{{\text{0}}^{5}}\text{ C}[/latex] of charge from one point in a circuit to another point. What is the potential difference between the two points?
2. A battery has a potential difference of [latex]\scriptsize 6\text{ V}[/latex]. When it is connected into a circuit, [latex]\scriptsize 3\text{ C}[/latex] of charge flows through a conductor. Calculate the energy transferred to the conductor.
3. A current of [latex]\scriptsize 0.5\text{ A}[/latex] flows through a conductor for [latex]\scriptsize 5\text{ s}[/latex]. During that time, [latex]\scriptsize 5\text{ J}[/latex] of energy are transferred to the conductor. Calculate the potential difference across the conductor.
4. A resistor of [latex]\scriptsize 5\text{ }\Omega[/latex] has a current of [latex]\scriptsize 0.2\text{ A}[/latex] passing through it for five minutes. Calculate the energy transferred to the resistor, by the current, during that time.
5. A microwave has a power rating of [latex]\scriptsize 800\text{ W}[/latex].
   1. What does this mean?
   2. If the microwave is connected to a [latex]\scriptsize 220\text{ V}[/latex] power source, calculate its resistance.
6. A light bulb is rated [latex]\scriptsize 60\text{ W: 220 V}[/latex]
   1. What does this mean?
   2. What current passes through the bulb when it is operating under correct conditions?
   3. How much energy does the bulb transfer in [latex]\scriptsize 30\text{ s}[/latex]?
7. What is the power of a [latex]\scriptsize 1\text{ x 1}{{\text{0}}^{2}}\text{ MV}[/latex] lightning bolt having a current of [latex]\scriptsize 2\text{ x 1}{{\text{0}}^{4}}\text{ A}[/latex]?
8. How much work is done in the starter motor of a large truck that draws [latex]\scriptsize 250\text{ A}[/latex] of current from a [latex]\scriptsize 24\text{ V}[/latex] battery in [latex]\scriptsize 0.5\text{ s}[/latex]?
9. A charge of [latex]\scriptsize 4\text{ C}[/latex] passes through a pocket calculator’s solar cells in four hours. What is the power output, given the calculator’s voltage output is [latex]\scriptsize \text{3 V}[/latex]?
10. Find the power dissipated in each of these extension cords through which [latex]\scriptsize 5\text{ A}[/latex] passes:
    1. an extension cord having a [latex]\scriptsize 0.06\text{ }\Omega[/latex] resistance
    2. a cheaper cord with a resistance of [latex]\scriptsize 0.03\text{ }\Omega[/latex].
11. A circuit transfers [latex]\scriptsize 3000\text{ J}[/latex] energy to a heater. If the current through the heater is [latex]\scriptsize 15\text{ A}[/latex] and the potential difference across it is [latex]\scriptsize 200\text{ V }[/latex] calculate the time required for the energy to be transferred.
12. The power dissipated in a resistor is given by [latex]\scriptsize P=\displaystyle \frac{{{{V}^{2}}}}{R}[/latex], which means power decreases if resistance increases. Yet power is also given by [latex]\scriptsize P={{I}^{2}}R[/latex], which means power increases if resistance increases. Explain why there is no contradiction here.

The full solutions are at the end of the unit.

 Unit 2: Solutions

Exercise 2.1

1. .
   [latex]\scriptsize \begin{align*}v&=\displaystyle \frac{W}{q}\\&=\displaystyle \frac{{250}}{{10}}\\&=\text{ 25 V}\end{align*}[/latex]
2. .
   [latex]\scriptsize \begin{align*}W&=Vq\\&=\text{ 6 x 0}\text{.5}\\&=\text{ 3 J}\end{align*}[/latex]

Back to Exercise 2.1

Exercise 2.2

1. .
   [latex]\scriptsize \begin{align*}\Delta t&=\displaystyle \frac{W}{P}\\&=\displaystyle \frac{{3\text{ x 1}{{\text{0}}^{6}}}}{{1\text{ 500}}}\\&=\text{ 2 000 s}\end{align*}[/latex]
2. .
   [latex]\scriptsize \begin{align*}W&=P\text{ x }\Delta t\\&=\text{ 400 x 3 600}\\&=\text{ 1 440 000J}\end{align*}[/latex] (hours must be converted to seconds)
3. The [latex]\scriptsize 200\text{ W}[/latex] light bulb would be better as it converts [latex]\scriptsize 200\text{ J}[/latex] of electrical energy into light energy per second so the light would be brighter. The [latex]\scriptsize 60\text{ W}[/latex] bulb would only convert [latex]\scriptsize 60\text{ J}[/latex] of energy per second.

Back to Exercise 2.2

Unit 2: Assessment

1. .
   [latex]\scriptsize \begin{align*}V&=\displaystyle \frac{W}{q}\\&=\displaystyle \frac{{4\text{ x 1}{{\text{0}}^{3}}}}{{2\text{ x 1}{{\text{0}}^{5}}}}\\&=\text{ 0}\text{.02 V}\end{align*}[/latex]
2. .
   [latex]\scriptsize \begin{align*}W&=Vq\\&=\text{ 6 x 3 }\\&=\text{ 18 J}\end{align*}[/latex]
3. .
   [latex]\scriptsize \begin{align*}W&=VI\Delta t\\V&=\displaystyle \frac{W}{{I\Delta t}}\\&=\displaystyle \frac{5}{{0.5\text{ x 5}}}\\&=\text{ 2 V}\end{align*}[/latex]
4. .
   [latex]\scriptsize \begin{align*}W&={{I}^{2}}R\Delta t\\&=\text{ (0}\text{.2}{{\text{)}}^{2}}(5)(3000)\\&=\text{ 60 J}\end{align*}[/latex] (convert minutes to seconds)
5. .
   1. [latex]\scriptsize 800\text{ J}[/latex] of energy is transformed per second
   2. .
      [latex]\scriptsize \begin{align*}P&=\displaystyle \frac{{{{V}^{2}}}}{R}\\R&=\displaystyle \frac{{{{V}^{2}}}}{P}\\&=\displaystyle \frac{{{{{220}}^{2}}}}{{800}}\\&=\text{ 60}\text{.5 }\Omega \end{align*}[/latex]
6. .
   1. When connected to a [latex]\scriptsize 220\text{ V}[/latex] power supply, [latex]\scriptsize 60\text{ J}[/latex] of energy will be transformed per second.
   2. .
      [latex]\scriptsize \begin{align*}P&=VI\\I&=\displaystyle \frac{P}{V}\\&=\displaystyle \frac{{60}}{{220}}\\&=\text{ 0}\text{.27 A}\end{align*}[/latex]
   3. .
      [latex]\scriptsize \begin{align*}W&=P\Delta t\\&=\text{ 60 x 30}\\&=\text{ 1 800 J}\end{align*}[/latex]
7. .
   [latex]\scriptsize \begin{align*}P&=VI\\&=\text{ (1 x 1}{{\text{0}}^{5}})(2\text{ x 1}{{\text{0}}^{4}})\\&=\text{ 2 x 1}{{\text{0}}^{9}}\text{ W }\end{align*}[/latex] (convert [latex]\scriptsize \text{ MV to V}[/latex]
8. .
   [latex]\scriptsize \begin{align*}W&=VI\Delta t\\&=\text{ (24)(250)(0}\text{.5)}\\&=\text{ 3 000 J}\end{align*}[/latex]
9. .
   [latex]\scriptsize \begin{align*}V&=\displaystyle \frac{W}{q}\\W&=Vq\\&=\text{ 3 x 4}\\&=\text{ 12 J}\\\text{P}&=\displaystyle \frac{W}{{\Delta t}}\\&=\displaystyle \frac{{12}}{{(4\text{ x 60 x 60)}}}\\&=\text{ 8}\text{.33 x 1}{{\text{0}}^{{-4}}}\text{ W}\end{align*}[/latex]
10. .
    1. .
       [latex]\scriptsize \begin{align*}P&={{I}^{2}}R\\&=\text{ (}{{\text{5}}^{2}})(0.06)\\&=\text{ 1}\text{.5 W}\end{align*}[/latex]
    2. .
       [latex]\scriptsize \begin{align*}P&={{I}^{2}}R\\&=\text{ (}{{\text{5}}^{2}})(0.03)\\&=\text{ 0}\text{.75 W}\end{align*}[/latex]
11. .
    [latex]\scriptsize \begin{align*}W&=VI\Delta t\\\Delta t&=\displaystyle \frac{W}{{VI}}\\&=\displaystyle \frac{{3\text{ 000}}}{{(200)(15)}}\\&=\text{ 1 s}\end{align*}[/latex]
12. Power increases if the resistance decreases only when voltage is kept constant. Power increases if resistance increases, only when the current is kept constant.

Back to Unit 2: Assessment