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Difference between revisions of "Potential Difference"

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: The [[SI Unit|units]] of '''potential difference''' are [[Volt]]s ([[V]]).
 
: The [[SI Unit|units]] of '''potential difference''' are [[Volt]]s ([[V]]).
 
: '''Potential Difference''' is sometimes described as the 'push' that moves a [[Electrical Current|current]] around a [[circuit]].
 
: '''Potential Difference''' is sometimes described as the 'push' that moves a [[Electrical Current|current]] around a [[circuit]].
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==Key Stage 4==
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===Meaning===
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[[Potential Difference]] is the amount of [[energy]] is [[Energy Transfer|transferred]] per unit [[Electrical Charge|charge]] between two points in a [[Electrical Circuit|circuit]].
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===About Potential Difference===
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: '''Potential Difference''' is [[measure]]d using a [[Voltmeter]].
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: The [[SI Unit|units]] of '''potential difference''' are [[Volt]]s ([[V]]).
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: '''Potential difference''' is the difference in '''potential''' between two points in a [[Electrical Circuit|circuit]].
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: '''Potential difference''' can be [[measure]]d between two points in a [[Electrical Circuit|circuit]] and is [[measure]]d across a [[Electrical Component|component]].
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: If two points in a [[Electrical Circuit|circuit]] are at the same '''potential''' there is no '''potential difference''' between them so no [[energy]] is [[Energy Transfer|transferred]] between those two points.
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===Equation===
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''NB: You should remember this equation with energy transferred as the subject of the formula.''
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'''Charge''' = (Energy Transferred)/(Potential Difference)
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<math>V=\frac{E}{Q}</math>
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Where:
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<math>V</math> = The [[Potential Difference|potential difference]] between two points.
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<math>Q</math> = The amount of [[Electrical Charge|charge]] that move between two points.
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<math>E</math> = The [[Energy Transfer]]red by the '''charge'''.
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===Example Calculations===
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====Finding Charge from Potential Difference and Energy Transferred====
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{| class="wikitable"
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| style="height:20px; width:300px; text-align:center;" |The [[Potential Difference|potential difference]] of 12V is placed across a [[Electrical Resistor|resistor]] increasing its [[Thermal Energy Store|thermal energy store]] by 3.7J as a result. Calculate the '''charge''' that has flowed through the [[Electrical Resistor|resistor]] in this time correct to two [[Significant Figures|significant figures]].
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| style="height:20px; width:300px; text-align:center;" |A [[circuit]] transfers 2.8kJ of [[energy]] [[electrically]] to a [[motor]]. The [[Potential Difference|potential difference]] across the [[motor]] is 1.5V. Calculate the'''charge''' that has flowed through the [[motor]] in this time correct to two [[Significant Figures|significant figures]].
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|-
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| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''
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V = 12V
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E = 3.7J
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| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''
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V = 1.5V
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E = 2.8kJ = 2.8x10<sup>3</sup>J
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|-
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| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
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<math>V=\frac{E}{Q}</math>
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<math>12=\frac{3.7}{Q}</math>
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| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
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<math>V=\frac{E}{Q}</math>
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<math>1.5=\frac{2.8 \times 10^3}{Q}</math>
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|-
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| style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
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<math>Q=\frac{3.7}{12}</math>
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<math>Q=0.3083C</math>
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<math>Q\approx0.31C</math>
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| style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
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<math>Q=\frac{2.8 \times 10^3}{1.5}</math>
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<math>Q=1866.7C</math>
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<math>Q\approx1900C</math>
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|}
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====Finding Potential Difference from Charge and Energy Transferred====
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{| class="wikitable"
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| style="height:20px; width: 300px; text-align:center;" |A '''charge''' of 84C transfers an [[energy]] of 20kJ. Calculate the [[Potential Difference|potential difference]] correct to two [[Significant Figures|significant figures]].
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| style="height:20px; width: 300px; text-align:center;" |170J of [[energy]] is transferred by a '''charge''' of 92mC. Calculate the [[Potential Difference|potential difference]] correct to two [[Significant Figures|significant figures]].
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|-
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| style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''
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Q = 84C
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E = 20kJ = 20x10<sup>3</sup>J
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| style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''
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Q = 92mC = 92x10<sup>-3<sup>C
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E = 170J
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|-
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| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
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<math>V=\frac{E}{Q}</math>
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<math>V=\frac{20 \times 10^3}{84}</math>
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<math>V=238.0952V</math>
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<math>V\approx 240V</math>
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| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
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<math>V=\frac{E}{Q}</math>
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<math>V=\frac{170}{92 \times 10^{-3}}</math>
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<math>V=1847.826V</math>
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<math>V\approx = 1800V</math>
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|}
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====Finding Energy Transferred from Charge and Potential Difference====
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{| class="wikitable"
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| style="height:20px; width: 300px; text-align:center;" |A bolt of lightning with a [[Potential Difference|potential difference]] 31,000kV transfers a '''charge''' of 15C. Calculate the [[Energy Transfer|energy transferred]] by this bolt of [[lightning]] correct to two [[Significant Figures|significant figures]].
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| style="height:20px; width: 300px; text-align:center;" |A 9V [[Electrical Battery|battery]] is able to mobilise a '''charge''' of 4.3kC during its operation. Calculate the total amount of [[energy]] stored in this [[Electrical Battery|battery]] correct to two [[Significant Figures|significant figures]].
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|-
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| style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''
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V = 31,000kV = 3.1x10<sup>7<sup>V
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Q = 15C
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| style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.'''
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V = 9V
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Q = 4.3kC = 4.3x10<sup>3</sup>
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|-
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| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
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<math>V=\frac{E}{Q}</math>
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<math>3.1 \times 10^7=\frac{E}{15}</math>
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| style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
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<math>V=\frac{E}{Q}</math>
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<math>9 =\frac{E}{4.3 \times 10^3}</math>
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|-
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| style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
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<math>E = 15 \times 3.1 \times 10^7</math>
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<math>E = 4.65\times10^8 J</math>
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<math>E\approx4.7\times10^8 J</math>
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| style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
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<math>E = 4.3 \times 10^3 \times 9</math>
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<math>E = 38700J</math>
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<math>E \approx 39000 \times 10^4J</math>
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|}

Revision as of 13:31, 25 February 2019

Key Stage 2

Meaning

Voltage is how much push electricity has.

About Voltage

The bigger the voltage the more push the electricity has to go around the circuit.
Adding another cell to the circuit in series will increase the voltage.
With a bigger voltage a lamp will be brighter and a buzzer will be louder.
SeriesCircuit1Cell1Bulb.png
SeriesCircuit2Cell1Bulb.png
A series circuit with one cell and one bulb. A bulb will be brighter if the voltage is higher.

Key Stage 3

Meaning

Potential Difference is how much energy is transferred by a current.

About Potential Difference

Potential Difference is measured using a Voltmeter.
The units of potential difference are Volts (V).
Potential Difference is sometimes described as the 'push' that moves a current around a circuit.

Key Stage 4

Meaning

Potential Difference is the amount of energy is transferred per unit charge between two points in a circuit.

About Potential Difference

Potential Difference is measured using a Voltmeter.
The units of potential difference are Volts (V).
Potential difference is the difference in potential between two points in a circuit.
Potential difference can be measured between two points in a circuit and is measured across a component.
If two points in a circuit are at the same potential there is no potential difference between them so no energy is transferred between those two points.

Equation

NB: You should remember this equation with energy transferred as the subject of the formula.

Charge = (Energy Transferred)/(Potential Difference)

\(V=\frac{E}{Q}\)

Where\[V\] = The potential difference between two points.

\(Q\) = The amount of charge that move between two points.

\(E\) = The Energy Transferred by the charge.

Example Calculations

Finding Charge from Potential Difference and Energy Transferred

The potential difference of 12V is placed across a resistor increasing its thermal energy store by 3.7J as a result. Calculate the charge that has flowed through the resistor in this time correct to two significant figures. A circuit transfers 2.8kJ of energy electrically to a motor. The potential difference across the motor is 1.5V. Calculate thecharge that has flowed through the motor in this time correct to two significant figures.
1. State the known quantities in correct units.

V = 12V

E = 3.7J

1. State the known quantities in correct units.

V = 1.5V

E = 2.8kJ = 2.8x103J

2. Substitute the numbers and evaluate.

\(V=\frac{E}{Q}\)

\(12=\frac{3.7}{Q}\)

2. Substitute the numbers and evaluate.

\(V=\frac{E}{Q}\)

\(1.5=\frac{2.8 \times 10^3}{Q}\)

3. Rearrange the equation and solve.

\(Q=\frac{3.7}{12}\)

\(Q=0.3083C\)

\(Q\approx0.31C\)

3. Rearrange the equation and solve.

\(Q=\frac{2.8 \times 10^3}{1.5}\)

\(Q=1866.7C\)

\(Q\approx1900C\)

Finding Potential Difference from Charge and Energy Transferred

A charge of 84C transfers an energy of 20kJ. Calculate the potential difference correct to two significant figures. 170J of energy is transferred by a charge of 92mC. Calculate the potential difference correct to two significant figures.
1. State the known quantities in correct units.

Q = 84C

E = 20kJ = 20x103J

1. State the known quantities in correct units.

Q = 92mC = 92x10-3C

E = 170J

2. Substitute the numbers into the equation and solve.

\(V=\frac{E}{Q}\)

\(V=\frac{20 \times 10^3}{84}\)

\(V=238.0952V\)

\(V\approx 240V\)

2. Substitute the numbers into the equation and solve.

\(V=\frac{E}{Q}\)

\(V=\frac{170}{92 \times 10^{-3}}\)

\(V=1847.826V\)

\(V\approx = 1800V\)

Finding Energy Transferred from Charge and Potential Difference

A bolt of lightning with a potential difference 31,000kV transfers a charge of 15C. Calculate the energy transferred by this bolt of lightning correct to two significant figures. A 9V battery is able to mobilise a charge of 4.3kC during its operation. Calculate the total amount of energy stored in this battery correct to two significant figures.
1. State the known quantities in correct units.

V = 31,000kV = 3.1x107V

Q = 15C

1. State the known quantities in correct units.

V = 9V

Q = 4.3kC = 4.3x103

2. Substitute the numbers and evaluate.

\(V=\frac{E}{Q}\)

\(3.1 \times 10^7=\frac{E}{15}\)

2. Substitute the numbers and evaluate.

\(V=\frac{E}{Q}\)

\(9 =\frac{E}{4.3 \times 10^3}\)

3. Rearrange the equation and solve.

\(E = 15 \times 3.1 \times 10^7\)

\(E = 4.65\times10^8 J\)

\(E\approx4.7\times10^8 J\)

3. Rearrange the equation and solve.

\(E = 4.3 \times 10^3 \times 9\)

\(E = 38700J\)

\(E \approx 39000 \times 10^4J\)

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