Electrical Current
Contents
- 1 Key Stage 2
- 2 Key Stage 3
- 3 Key Stage 4
- 3.1 Meaning
- 3.2 About Electrical Current
- 3.3 Equation
- 3.4 Example Calculations
- 3.4.1 Finding Current from Charge and Time
- 3.4.2 Finding Charge from Current and Time
- 3.4.3 Finding Time from Current and Charge
- 3.4.4 Finding Current from Potential Difference and Resistance
- 3.4.5 Finding Resistance from Potential Difference and Current
- 3.4.6 Finding Potential Difference from Current and Resistance
- 3.5 References
Key Stage 2
Meaning
Electrical Current is the amount of electricity flowing through a wire.
About Electrical Current
- The bigger the electrical current the brighter a bulb and the louder a buzzer.
- If an electrical current goes through an animal it is called an electrical shock.
Key Stage 3
Meaning
An electrical current is a flow of charge.
About Electrical Current
- Current is measured using an Ammeter.
- The units of current are amperes, which are also called amps (A).
- A current in a wire is a flow of electrons which are negatively charged particles.
- Conventional Current flows from positive to negative. This is because electricity was discovered before scientists knew about electrons.
- In a salt solution current is the flow of both positive and negative ions.
Key Stage 4
Meaning
Electrical current is the rate of flow of charge.
About Electrical Current
- Current is measured using an Ammeter.
- The SI Units of current are amperes, which are also called amps (A).
- A current in a wire is a flow of electrons which are negatively charged particles.
- Conventional Current flows from positive to negative. This is because electricity was discovered before scientists knew about electrons.
- In a salt solution current is the flow of both positive and negative ions.
- In a series circuit the current is the same everywhere in the circuit.
- In a parallel circuit the current splits at a junction.
Equation
Equation Linking Current, Charge and Time
NB: You should remember this equation with charge as the subject of the formula.
Current = (Charge)/(time)
\(I=\frac{Q}{t}\)
Where
\(I\) = The electrical current
\(Q\) = The amount of charge flowing past a point.
\(t\) = The time taken for the charge to flow.
- This can give the definition "Current (I) is the (=) amount of charge flowing past a point (Q) per (÷) unit time (t)."
Equation Linking Current, Potential Difference and Resistance
NB: You should remember this equation.
Current = (Potential Difference)/(Resistance)
\(I=\frac{V}{R}\)
Where
\(I\) = The electrical current
\(V\) = The potential difference across a component.
\(R\) = The resistance of an component.
Example Calculations
Finding Current from Charge and Time
A charge of 15 Coulombs passes through a point in a circuit ever 0.52 seconds. Calculate the current flowing past this point correct to two significant figures. | A capacitor stores a charge of 10C. It discharges in 12ms. Calculate the current flowing out of the capacitor correct to two significant figures. |
1. State the known quantities in correct units.
Q = 15C t = 0.52s |
1. State the known quantities in correct units.
Q = 10C t = 12ms = 12x10^{-3}s |
2. Substitute the numbers into the equation and solve.
\(I=\frac{Q}{t}\) \(I=\frac{15}{0.52}\) \(I=28.846153A\) \(I\approx29A\) |
2. Substitute the numbers into the equation and solve.
\(I=\frac{Q}{t}\) \(I=\frac{10}{12 \times 10^{-3}}\) \(I=833.3A\) \(I\approx830A\) |
Finding Charge from Current and Time
A battery supplies 4.7Amps to a bulb over a period of 14 seconds. Calculate the charge leaving the battery in this time correct to two significant figures. | A hairdryer uses a current of 7.2A for 5 minutes to dry a person’s hair. Calculate the charge flowing through the hairdryer in this time correct to two significant figures. |
1. State the known quantities in correct units.
I = 4.7A t = 14s |
1. State the known quantities in correct units.
I = 7.2A t = 5min = 300s |
2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(4.7=\frac{Q}{14}\) |
2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(7.2=\frac{Q}{300}\) |
3. Rearrange the equation and solve.
\(Q=4.7 \times 14\) \(Q = 65.8C\) \(Q \approx 66C\) |
3. Rearrange the equation and solve.
\(Q=7.2 \times 300\) \(Q = 2160C\) \(Q \approx 2200C\) |
Finding Time from Current and Charge
A battery charger uses a current of 150mA to deliver a charge of 245 Coloumbs to a battery. Calculate the time taken to charge this battery correct to two significant figures. | A cloud in a thunderstorm loses 15C in one lightening strike. At a current of 31,000kA. Calculate how long this lightning strike lasts correct to two significant figures. |
1. State the known quantities in correct units.
I = 150mA = 150x10^{-3}A Q = 245C |
1. State the known quantities in correct units.
I = 31,000kA = 3.1x10^{7}A Q = 15C |
2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(150 \times 10^{-3} = \frac{245}{t}\) |
2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(3.1 \times 10^7 = \frac{15}{t}\) |
3. Rearrange the equation and solve.
\(t=\frac{245}{150 \times 10^{-3}}\) \(t=1633.3s\) \(t\approx1633.3s\) |
3. Rearrange the equation and solve.
\(t=\frac{15}{3.1 \times 10^7}\) \(t = 4.8387 \times 10^{-7}s\) \(t\approx4.8 \times 10^{-7}s\) |
Finding Current from Potential Difference and Resistance
A potential difference of 9.9V is placed across an 19 Ohm resistor. Calculate the current flowing through the resistor correct to two significant figures. | A toaster has a resistance of 27 Ohms is plugged into the mains which has a potential difference of 230V. Calculate the current flowing through the toaster correct to two significant figures. |
1. State the known quantities in correct units.
V = 9.9V R = 19Ω |
1. State the known quantities in correct units.
V = 230V R = 27Ω |
2. Substitute the numbers into the equation and solve.
\(I=\frac{V}{R}\) \(I=\frac{9.9}{19}\) \(I=0.52105A\) \(I\approx0.52A\) |
2. Substitute the numbers into the equation and solve.
\(I=\frac{V}{R}\) \(I=\frac{230}{27}\) \(I=8.519A\) \(I\approx8.5A\) |
Finding Resistance from Potential Difference and Current
A student measures a potential difference of 5.4V and a current of 0.13mA across a component. Calculate the resistance of the component. | Calculating the resistance of a buzzer connected in series to a 9V battery with an ammeter reading of 23mA. |
1. State the known quantities in correct units.
V = 5.4V I = 0.13mA = 0.13x10^{-3} |
1. State the known quantities in correct units.
V = 9V I = 23mA = 23x10^{-3} |
2. Substitute the numbers and evaluate.
\(I=\frac{V}{R}\) \(0.13 \times 10^{-3}=\frac{5.4}{R}\) |
2. Substitute the numbers and evaluate.
\(23 \times 10^{-3}=\frac{9}{R}\) |
3. Rearrange the equation and solve.
\(R=\frac{5.4}{0.13 \times 10^{-3}}\) \(R=41538.46\Omega\) \(R\approx 42000\Omega\) |
3. Rearrange the equation and solve.
\(R=\frac{9}{23 \times 10^{-3}}\) \(R=391.3043\Omega\) \(R\approx 390\Omega\) |
Finding Potential Difference from Current and Resistance
A current of 55mA flows through a component with a resistance of 93 Ohms. Calculate the potential difference across this component correct to two significant figures. | A 22kΩ resistor has a current flowing through it of 6mA. Calculate the potential difference across the resistor correct to two significant figures. |
1. State the known quantities in correct units.
I = 55mA = 55x10^{-3}A R = 93Ω |
1. State the known quantities in correct units.
I = 6mA = 6x10^{-3}A R = 22kΩ = 22x10^{3}Ω |
2. Substitute the numbers and evaluate.
\(I=\frac{V}{R}\) \(55 \times 10^{-3}=\frac{V}{93}\) |
2. Substitute the numbers and evaluate.
\(I=\frac{V}{R}\) \(6 \times 10^{-3}=\frac{V}{22 \times 10^3}\) |
3. Rearrange the equation and solve.
\(V= 55 \times 10^{-3} \times 93\) \(V= 5.115V\) \(V\approx5.1V\) |
3. Rearrange the equation and solve.
\(V= 6 \times 10^{-3} \times 22 \times 10^3\) \(V= 132V\) \(V\approx130V\) |
References
AQA
- Current (electrical), pages 180, 181, 190, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical), pages 24, 33, GCSE Physics; The Revision Guide, CGP, AQA
- Current (electrical); alternating, page 188, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); alternating, pages 31, 96, 97, GCSE Physics; The Revision Guide, CGP, AQA
- Current (electrical); direct, page 188, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); direct, pages 31, 97, GCSE Physics; The Revision Guide, CGP, AQA
- Current (electrical); energy transferred, page 190, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); I-V characteristics, page 183, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); I-V characteristics, page 26, GCSE Physics; The Revision Guide, CGP, AQA
- Current (electrical); magnetic field around a current-carrying wire, page 230, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); magnetic field around a current-carrying wire, pages 93, 94, GCSE Physics; The Revision Guide, CGP, AQA
- Current (electrical); mains supply, page 188, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); national grid, page 191, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); parallel circuits, pages 186-187, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current (electrical); series circuits, pages 185-187, GCSE Combined Science; The Revision Guide, CGP, AQA
- Current, pages 46-7, 52-69, 72-5, 247-9, GCSE Physics; Student Book, Collins, AQA
- Current, pages 52-53, 58-61, 64-65, 68-71, 216-229, GCSE Physics; Third Edition, Oxford University Press, AQA
- Current, pages 61-67, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current, pages 63-69, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; alternating (ac), pages 86, 203, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current; alternating (ac), pages 89, 245, 306, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; direct (dc), page 86, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current; direct (dc), pages 89, 306, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; in parallel, pages 73, 74, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current; in parallel, pages 75, 76, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; in series, page 69, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current; in series, page 71, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; induced, pages 303, 304, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; magnetic field around a current carrying wire, pages 291, 292, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current; magnetic field around a current-carrying wire, pages 218, 219, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current; measuring, pages 235, 236, GCSE Combined Science Trilogy; Physics, CGP, AQA
- Current; measuring, pages 331, 332, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
- Current-potential difference (I-V characteristic) graphs, pages 43-5, GCSE Physics, Hodder, AQA
- Current-potential difference (I-V) graphs, pages 298-9, GCSE Combined Science Trilogy 1, Hodder, AQA
- Current-potential difference (I-V) graphs; investigating the characteristics of a circuit component, page 300, GCSE Combined Science Trilogy 1, Hodder, AQA
- Current-potential difference graphs, page 55, GCSE Physics; Third Edition, Oxford University Press, AQA