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]]. | ||
+ | |||
+ | ==Key Stage 4== | ||
+ | ===Meaning=== | ||
+ | [[Potential Difference]] is the amount of [[energy]] is [[Energy Transfer|transferred]] per unit [[Electrical Charge|charge]] between two points in a [[Electrical Circuit|circuit]]. | ||
+ | |||
+ | ===About Potential Difference=== | ||
+ | : '''Potential Difference''' is [[measure]]d using a [[Voltmeter]]. | ||
+ | : The [[SI Unit|units]] of '''potential difference''' are [[Volt]]s ([[V]]). | ||
+ | : '''Potential difference''' is the difference in '''potential''' between two points in a [[Electrical Circuit|circuit]]. | ||
+ | : '''Potential difference''' can be [[measure]]d between two points in a [[Electrical Circuit|circuit]] and is [[measure]]d across a [[Electrical Component|component]]. | ||
+ | : 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. | ||
+ | |||
+ | ===Equation=== | ||
+ | ''NB: You should remember this equation with energy transferred as the subject of the formula.'' | ||
+ | |||
+ | '''Charge''' = (Energy Transferred)/(Potential Difference) | ||
+ | |||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | Where: | ||
+ | |||
+ | <math>V</math> = The [[Potential Difference|potential difference]] between two points. | ||
+ | |||
+ | <math>Q</math> = The amount of [[Electrical Charge|charge]] that move between two points. | ||
+ | |||
+ | <math>E</math> = The [[Energy Transfer]]red by the '''charge'''. | ||
+ | |||
+ | ===Example Calculations=== | ||
+ | ====Finding Charge from Potential Difference and Energy Transferred==== | ||
+ | {| class="wikitable" | ||
+ | | 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]]. | ||
+ | | 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]]. | ||
+ | |- | ||
+ | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' | ||
+ | |||
+ | V = 12V | ||
+ | |||
+ | E = 3.7J | ||
+ | | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' | ||
+ | |||
+ | V = 1.5V | ||
+ | |||
+ | E = 2.8kJ = 2.8x10<sup>3</sup>J | ||
+ | |- | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
+ | |||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | <math>12=\frac{3.7}{Q}</math> | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | <math>1.5=\frac{2.8 \times 10^3}{Q}</math> | ||
+ | |- | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' | ||
+ | |||
+ | <math>Q=\frac{3.7}{12}</math> | ||
+ | |||
+ | <math>Q=0.3083C</math> | ||
+ | |||
+ | <math>Q\approx0.31C</math> | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' | ||
+ | |||
+ | <math>Q=\frac{2.8 \times 10^3}{1.5}</math> | ||
+ | |||
+ | <math>Q=1866.7C</math> | ||
+ | |||
+ | <math>Q\approx1900C</math> | ||
+ | |} | ||
+ | |||
+ | ====Finding Potential Difference from Charge and Energy Transferred==== | ||
+ | {| class="wikitable" | ||
+ | | 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]]. | ||
+ | | 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]]. | ||
+ | |- | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' | ||
+ | Q = 84C | ||
+ | |||
+ | E = 20kJ = 20x10<sup>3</sup>J | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' | ||
+ | |||
+ | Q = 92mC = 92x10<sup>-3<sup>C | ||
+ | |||
+ | E = 170J | ||
+ | |- | ||
+ | | style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' | ||
+ | |||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | <math>V=\frac{20 \times 10^3}{84}</math> | ||
+ | |||
+ | <math>V=238.0952V</math> | ||
+ | |||
+ | <math>V\approx 240V</math> | ||
+ | |||
+ | | style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' | ||
+ | |||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | <math>V=\frac{170}{92 \times 10^{-3}}</math> | ||
+ | |||
+ | <math>V=1847.826V</math> | ||
+ | |||
+ | <math>V\approx = 1800V</math> | ||
+ | |} | ||
+ | |||
+ | ====Finding Energy Transferred from Charge and Potential Difference==== | ||
+ | {| class="wikitable" | ||
+ | | 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]]. | ||
+ | | 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]]. | ||
+ | |- | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' | ||
+ | |||
+ | V = 31,000kV = 3.1x10<sup>7<sup>V | ||
+ | |||
+ | Q = 15C | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''1. State the known quantities in correct [[unit]]s.''' | ||
+ | |||
+ | V = 9V | ||
+ | |||
+ | Q = 4.3kC = 4.3x10<sup>3</sup> | ||
+ | |- | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
+ | |||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | <math>3.1 \times 10^7=\frac{E}{15}</math> | ||
+ | |||
+ | | style="height:20px; width: 300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
+ | |||
+ | <math>V=\frac{E}{Q}</math> | ||
+ | |||
+ | <math>9 =\frac{E}{4.3 \times 10^3}</math> | ||
+ | |- | ||
+ | | style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' | ||
+ | |||
+ | <math>E = 15 \times 3.1 \times 10^7</math> | ||
+ | |||
+ | <math>E = 4.65\times10^8 J</math> | ||
+ | |||
+ | <math>E\approx4.7\times10^8 J</math> | ||
+ | |||
+ | | style="height:20px; width: 300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].''' | ||
+ | |||
+ | <math>E = 4.3 \times 10^3 \times 9</math> | ||
+ | |||
+ | <math>E = 38700J</math> | ||
+ | |||
+ | <math>E \approx 39000 \times 10^4J</math> | ||
+ | |} |
Revision as of 13:31, 25 February 2019
Contents
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.
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.
\(E = 4.3 \times 10^3 \times 9\) \(E = 38700J\) \(E \approx 39000 \times 10^4J\) |