Contents
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.
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 0.5 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
Q = 15C t = 0.5s |
1. State the known quantities
Q = 10C t = 12ms = 12x10-3s |
| 2. Substitute the numbers into the equation and solve.
\(I=\frac{Q}{t}\) \(I=\frac{15}{0.5}\) \(I=30A\) |
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 4Amps 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 7A 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
I = 4A t = 14s |
1. State the known quantities
I = 7A t = 5min = 300s |
| 2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(4=\frac{Q}{14}\) |
2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(7=\frac{Q}{300}\) |
| 3. Rearrange the equation and solve.
\(Q=4 \times 14\) \(Q = 56C\) |
3. Rearrange the equation and solve.
\(Q=7 \times 300\) \(Q = 2100C\) |
Finding Time from Current and Charge
| A battery charger uses a current of 150mA to deliver a charge of 240 Coloumbs to a battery. Calculate the time taken to charge this battery correct to two significant figures. | A cloud in a thunderstorm loses 15kC in one lightening strike. At a current of 30,000kA. Calculate how long this lightning strike lasts correct to two significant figures. |
| 1. State the known quantities
I = 150mA = 150x10-3A Q = 240C |
1. State the known quantities
I = 30,000kA = 3x107A Q = 15kC = 15x103 |
| 2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(150 \times 10^{-3} = \frac{240}{t}\) |
2. Substitute the numbers and evaluate.
\(I=\frac{Q}{t}\) \(3 \times 10^7 = \frac{15 \times 10^3}{t}\) |
| 3. Rearrange the equation and solve.
\(t=\frac{240}{150 \times 10^{-3}}\) \(t=1600s\) |
3. Rearrange the equation and solve.
\(t=\frac{15 \times 10^3}{3 \times 10^7}\) \(t = 5.0 \times 10^{-4}s\) |