2.13-2.16 - Voltage, current, resistance and charge

Current As Charge Flow

Electric current is not a substance that gets used up. It is a rate: it tells you how much electric charge passes a point each second.

The symbol for current is $I$. The unit is the ampere, $\mathrm{A}$. A current of $1\,\mathrm{A}$ means $1\,\mathrm{C}$ of charge passes a point every second, so an ampere is the same as a coulomb per second.

Charge has the symbol $Q$ and is measured in coulombs, $\mathrm{C}$. Time has the symbol $t$ and is measured in seconds, $\mathrm{s}$. If more charge passes the point in the same time, the current is larger. If the same charge takes longer to pass, the current is smaller.

When an Edexcel question asks what current means, the important phrase is "rate of flow of charge". A common near miss is to say "number of electrons per second". In a metal, electrons do move, but the definition of current is about charge per second, not number of particles.

Charge, Current And Time

Because current is charge per second, charge can be calculated from current and time.

In this equation:

The equation can be rearranged depending on the unknown:

For example, if a current of $0.40\,\mathrm{A}$ flows for $30\,\mathrm{s}$, then:

If a question gives time in minutes, convert to seconds before substituting. For example, $4.0\,\mathrm{min} = 240\,\mathrm{s}$, so a current of $2.5\,\mathrm{A}$ transfers:

Keep the final unit with the answer. Include the appropriate unit unless the question already supplies it on the answer line.

Voltage, Current And Resistance

Voltage is measured in volts, $V$. In circuit calculations, it is the potential difference across the component you are considering. Resistance is measured in ohms, $\Omega$, and describes how much a component opposes the current through it.

For this cluster, the required relationship is:

In this equation:

The same equation can be rearranged:

For example, if the voltage across a resistor is $6.0\,\mathrm{V}$ and the current through it is $0.50\,\mathrm{A}$, then:

Convert current to amperes before substituting. For example, 250 mA is 0.250 A, so a 6.0 V resistor carrying that current has resistance $R=6.0/0.250=24\,\Omega$, not 0.024 ohm.

Use the relationship as a sense check: for the same voltage, a larger resistance would make the calculated current smaller. For example, the same $6.0\,\mathrm{V}$ across a $24\,\Omega$ resistor gives $I = 6.0 \div 24 = 0.25\,\mathrm{A}$: double the resistance, half the current.

Electrons In Metals

In a solid metallic conductor, such as a copper wire, the moving charge carriers are negatively charged electrons. The positive metal ions are arranged in a fixed lattice and do not travel along the wire. Some electrons in a metal are not held by any one atom; these free electrons can move between the fixed ions, so they are the charge carriers.

[DIAGRAM: electron_flow_in_metal: Voltage, current, resistance and charge - diagram 01; asset_slug: p17_voltage_current_resistance_and_charge__diagram_01; recommended_method: image_gen; description: Monochrome conductor cutaway showing fixed positive metal ions and negatively charged electrons flowing from a negative end towards a positive end, with a cross-section labelled as charge passing a point.]

The diagram links two ideas that are often confused:

The current is still measured in amperes, even though the individual moving particles are electrons. You do not need to count electrons in ordinary IGCSE current calculations; you use the total charge in coulombs.

Some later circuit diagrams may use conventional current direction, which is defined as the direction positive charge would move. Electron flow in a metal is opposite to conventional current: in the diagram, the electrons drift from the negative end towards the positive end, so the conventional current is from the positive end to the negative end. For this lesson's specification point, the key statement is simply: current in solid metallic conductors is a flow of negatively charged electrons.

Choosing The Equation

Most mistakes in this topic come from choosing the wrong relationship or forgetting to convert the time unit. Convert prefixed units too: a current in milliamperes must be changed to amperes, as above, and a resistance of $2.0\,\mathrm{k\Omega}$ is $2000\,\Omega$. Start by identifying the three quantities in the question.

Use the units as a check. Coulombs belong with charge, amperes with current, seconds with time, volts with voltage, and ohms with resistance. If your answer to a resistance calculation is in coulombs, you have chosen the wrong equation.

Here is a linked example.

A charge of $36\,\mathrm{C}$ passes through a resistor in $12\,\mathrm{s}$. The resistor has a resistance of $5.0 \Omega$. Calculate the current and then the voltage across the resistor.

First find the current:

Then use the current in the voltage equation:

Notice the order: $Q = I \times t$ gives the current, then $V = I \times R$ uses that current. The equations are connected by $I$, but they do different jobs.

Quick recap

• Current is the rate of charge flow: $I=Q/t$, so $Q=It$. One ampere means one coulomb per second.
• Voltage, current and resistance are related by $V=IR$; rearrange to find the unknown quantity.
• Convert time to seconds and current to amperes before substituting; charge is measured in C and resistance in $\Omega$.
• In solid metals, electrons carry charge through the fixed positive-ion lattice. Electron flow is opposite to conventional current.