GCSE Physics (9-1)

6.1 Current, potential difference and resistance — Learn

Prerequisites: Chapter 1, especially energy stores and energy transfers; Section 5.1, especially the prefixes milli, kilo and mega; multiplying and dividing decimals and numbers in standard form.

By the end of this section you can: describe an electric current as a flow of charge; use the equation Q=ItQ = It; use the equation V=IRV = IR; and describe how the resistance of a lamp, a diode, a thermistor and an LDR changes.

1. Electric charge and current

Electric charge is a property of matter. There are two kinds of charge: positive and negative. An electron carries a negative charge. Charge is measured in coulombs (C).

For charge to flow, a circuit needs two things: a source of potential difference, and a closed loop of conductor. A battery or a mains supply is a source of potential difference. Break the loop, and the charge stops flowing.

An electric current is the rate of flow of electric charge. In one second, a current of one ampere carries one coulomb of charge past a point. Current is measured in amperes (A). The ampere is often called the amp.

Q=ItQ = It

where QQ is the charge flow in coulombs, II is the current in amperes and tt is the time in seconds.

Current is measured with an ammeter. The ammeter is connected in series with the component, so that the current through the ammeter is the current through the component.

In a single closed loop, the current has the same value at every point. Charge is not used up as it flows around the circuit. The energy it carries is transferred to the components, but the charge itself keeps moving.

2. Potential difference, current and resistance

Potential difference (pd) is the energy transferred to or from each unit of charge as it moves between two points in a circuit. It is measured in volts (V). One volt means one joule of energy is transferred by each coulomb of charge. Many books and exam papers also use the word voltage for potential difference.

The potential difference across a component is measured with a voltmeter. The voltmeter is connected in parallel with the component, so that it measures the difference between the two ends.

The current through a component depends on two things: the potential difference across it, and the resistance of the component. Resistance is measured in ohms (Ω\Omega). For a given potential difference, a larger resistance gives a smaller current. The three quantities are linked by

V=IRV = IR

where VV is the potential difference in volts, II is the current in amperes and RR is the resistance in ohms.

3. Circuit symbols and diagrams

Circuit diagrams use standard symbols, so that every physicist reads the same circuit. These are the components this chapter uses.

Component What it does
Cell and battery Supplies a potential difference
Switch Opens or closes the circuit
Lamp Transfers energy to the thermal and light stores
Resistor Opposes the flow of charge
Variable resistor A resistor whose value can be changed
Ammeter Measures current; connected in series
Voltmeter Measures potential difference; connected in parallel
Diode Allows current in one direction only
Thermistor Resistance falls as temperature rises
LDR Resistance falls as light intensity rises

A circuit diagram shows the components joined by straight lines that represent the connecting wires. The long line of the cell symbol is the positive terminal, and the short thick line is the negative terminal.

4. Worked examples

Worked example 1. A current of 3.0 A flows through a lamp for 20 s. Calculate the charge that flows.

Q=It=3.0×20=60 C\begin{aligned} Q &= It \\ &= 3.0 \times 20 \\ &= 60\,\text{C} \end{aligned}

Worked example 2. A charge of 120 C passes a point in a circuit in 40 s. Calculate the current.

I=Qt=12040=3.0 A\begin{aligned} I &= \frac{Q}{t} \\ &= \frac{120}{40} \\ &= 3.0\,\text{A} \end{aligned}

Worked example 3. A current of 0.50 A flows through a resistor of resistance 12 Ω\Omega. Calculate the potential difference across the resistor.

V=IR=0.50×12=6.0 V\begin{aligned} V &= IR \\ &= 0.50 \times 12 \\ &= 6.0\,\text{V} \end{aligned}

Worked example 4. The potential difference across a resistor is 9.0 V, and the current through it is 0.30 A. Calculate the resistance.

R=VI=9.00.30=30 Ω\begin{aligned} R &= \frac{V}{I} \\ &= \frac{9.0}{0.30} \\ &= 30\,\Omega \end{aligned}

5. Resistors whose resistance changes

For some resistors, the resistance stays the same as the current changes. For others, it changes.

You can find the resistance of a linear component from its VV–II graph. The gradient of a graph of VV against II is the resistance RR, because V=IRV = IR rearranges to V=R×IV = R \times I. Take care: if II is on the vertical axis instead, the gradient is 1R\dfrac{1}{R}.

6. Changing the current

The figure shows a simple circuit with a cell, a resistor and an ammeter. You can change the potential difference of the cell and the resistance of the resistor.

Predict first. Do not move the sliders yet. Write down your answers.

  1. If the resistance doubles and the potential difference stays the same, what happens to the current?
  2. If the potential difference doubles and the resistance stays the same, what happens to the current?

Test. Move the sliders one at a time. Watch the current as you change the resistance, then as you change the potential difference.

Explain. The current is directly proportional to the potential difference when the resistance is constant. Why does doubling the potential difference double the current? Write your answer in two or three sentences.

Watch out. The current is not used up as it goes around a circuit. In a single closed loop the current is the same at every point. What the components change is the energy carried by each coulomb, not the number of coulombs flowing each second. Also keep current and potential difference apart: current is measured in amperes with an ammeter in series, and potential difference is measured in volts with a voltmeter in parallel. Re-check which one a question asks for before you choose an equation.

7. Progress check

Return to the chapter problem. The cabin heater takes a current of 5.0 A from the 230 V supply.

  1. Calculate the resistance of the heater.
  2. The heater is switched on for 30 minutes. Calculate the charge that flows through it.

You should now be able to

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