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Electrical Current

Revision as of 13:27, 7 June 2019 by NRJC (talk | contribs) (Equation Linking Current, Potential Difference and Resistance)

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 = 12x10-3s

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 = 150x10-3A

Q = 245C

1. State the known quantities in correct units.

I = 31,000kA = 3.1x107A

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 = 55x10-3A

R = 93Ω

1. State the known quantities in correct units.

I = 6mA = 6x10-3A

R = 22kΩ = 22x103Ω

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.

\(V= 6 \times 10^{-3} \times 22 \times 10^3\)

\(V= 132V\)

\(V\approx130V\)