B.5 Current and circuits

A cell as an energy source

A cell transfers energy from a non-electrical source, such as chemical or solar energy, to charge carriers. The energy source establishes an electromotive force (emf) that can drive charge around a circuit.

Meaning of emf

The emf is the energy supplied by the source per unit charge when charge passes through the source. Its unit is the volt, $$1\,\mathrm V=1\,\mathrm{J\,C^{-1}}$$.

Follow the energy

The cell is not a reservoir of charge that gets used up. Charge circulates; the cell supplies energy that is transferred in circuit components such as lamps, motors and resistors.

Common trap

Emf is not the same as current. Emf is energy per charge supplied by the source; current is charge flow per unit time.

Two ways to supply emf

Chemical and solar cells both supply energy per unit charge, but they obtain that energy differently.

Boundary

Compare the energy source and operating conditions, not just the external circuit. A separate battery may store energy produced by a solar cell.

Resistance

Resistance is the ratio of potential difference across a component to current through it:

$$R=\frac{V}{I}$$

Its SI unit is the ohm, Ω.

Conductors, insulators and the origin of resistance

A conductor has mobile charge carriers that can drift when an electric field is applied. In a metal these carriers are electrons. In an insulator, charge carriers are not sufficiently mobile for a sustained current under ordinary conditions. Resistance arises because moving carriers interact with the material's lattice and transfer energy to it.

Interpret the ratio

For a given current, a larger potential difference means larger resistance. Resistance describes how strongly a component opposes charge flow under the stated operating conditions.

Worked example from the mapped local textbook

A component carries $0.78\,\mathrm{A}$ when the potential difference across it is $4.4\,\mathrm{V}$.

$$R=\frac{V}{I}=\frac{4.4}{0.78}=5.6\,\Omega$$

This is its resistance at that operating point; it should not be assumed constant unless the component is ohmic under fixed conditions.

Unit check

From $$R=V/I$$, $$1\,\Omega=1\,\mathrm{V\,A^{-1}}$$. Use the voltage across the component, not the emf of the whole source unless they are equal in the circuit.

Common trap

Resistance is not the same as current. A component can have high resistance and a small current for a given voltage.

Ohm’s law

At constant temperature, an ohmic conductor has $$V\propto I$$, so $$R=V/I$$ is constant. Its I–V graph is a straight line through the origin when plotted with V and I consistently.

Non-ohmic behaviour

A non-ohmic component has a changing resistance, so current is not directly proportional to potential difference. Filament lamps, diodes and thermistors can be non-ohmic.

Why temperature matters

Heating can change a conductor’s resistance. Apply Ohm’s law only under the stated constant-temperature condition; otherwise the slope or ratio changes as the component operates.

Common trap

A curved I–V graph is not automatically wrong. It is evidence that the component is non-ohmic under those operating conditions.

Electrical power

Power is the rate of electrical energy transfer. For a resistor,

$$P=IV=I^2R=\frac{V^2}{R}$$

Choose the convenient form

Use $$P=IV$$ when current and voltage are given, $$P=I^2R$$ when current and resistance are given, and $$P=V^2/R$$ when voltage and resistance are given.

Interpret the unit

A watt is a joule per second: $$1\,\mathrm W=1\,\mathrm{J\,s^{-1}}$$. In a resistor, the transferred electrical energy becomes mainly internal energy and may produce heating.

Worked example from the mapped local textbook

A heater is rated $230\,\mathrm{V}$ and $1100\,\mathrm{W}$. Since voltage and power are known, use $P=V^2/R$:

$$R=\frac{V^2}{P}=\frac{230^2}{1100}=48\,\Omega$$

The rating means the heater transfers about $1100\,\mathrm{J}$ each second when operated at $230\,\mathrm{V}$.

Common trap

For alternating-current questions, distinguish peak values from mean or rms values. Use the convention and data supplied by the question.

Energy over time

For a steady direct current,

$$E=Pt=IVt$$

where E is electrical energy transferred in time t.

Build the relation

Potential difference is energy per charge, $$V=E/q$$, and current is charge per time, $$I=q/t$$. Combining them gives $$E=VIt$$.

Units and billing

Use seconds for t to obtain joules. Electricity billing may use kWh: $$1\,\mathrm{kWh}=3.6\times10^6\,\mathrm J$$.

Worked example from the mapped local textbook

A resistor carries $3\,\mathrm A$ with a potential difference of $6\,\mathrm V$. In one second, $3\,\mathrm C$ passes and each coulomb transfers $6\,\mathrm J$.

$$E=VIt=(6)(3)(1)=18\,\mathrm J$$

Therefore $P=E/t=18\,\mathrm W$: the resistor transfers $18\,\mathrm J$ each second.

Common trap

Do not use power in place of energy. Power is the rate of transfer; multiply by time for total energy.

Energy conversion in a cell

A chemical cell converts chemical potential energy into electrical energy. Internal chemical processes separate charge and maintain an emf between the terminals.

In a complete circuit

When the circuit is closed, charge flows and energy supplied by the cell is transferred to components. The cell’s chemical energy decreases as electrical energy is delivered.

Source versus load

The cell is the source of energy; a resistor, lamp or motor is a load where electrical energy is transferred to other forms. The charges circulate through both.

Common trap

A cell supplies energy, not a continuous supply of new electrons. The same charge carriers circulate through the circuit.

Read the circuit arrangement

A circuit diagram shows which components are connected in series and which share the same two nodes in parallel. Use the standard symbols supplied in the Physics data booklet; do not infer a connection merely because drawn lines cross unless a junction is shown.

Ideal meters

Place an ideal ammeter in series with the branch whose current is measured; its resistance is zero. Place an ideal voltmeter in parallel across the component whose potential difference is measured; its resistance is infinite. If a meter is stated to be non-ideal, use its stated constant resistance.

Conventional current

Conventional current follows the direction positive charge would move: through the external circuit from the source's positive terminal toward its negative terminal. In a metal, electrons drift in the opposite direction.

Common trap

Do not put an ideal ammeter directly across a source or an ideal voltmeter in series: those connections change or interrupt the intended circuit.

Current

Electric current is charge passing a point per unit time:

$$I=\frac{\Delta q}{\Delta t}$$

The SI unit is the ampere, $$1\,\mathrm A=1\,\mathrm{C\,s^{-1}}$$.

Find charge or carrier count

$$\Delta q=I\Delta t$$

If each carrier has charge magnitude e, the number of carriers passing is $$N=\Delta q/e$$.

Microscopic picture

Metal electrons move randomly with a small drift superimposed when current flows. Current measures net charge flow, not the total random motion of every electron.

Worked example from the mapped local textbook

A lamp carries $50\,\mathrm{mA}=0.050\,\mathrm{A}$ for $1.0\,\mathrm{min}=60\,\mathrm{s}$.

$$q=It=(0.050)(60)=3.0\,\mathrm{C}$$

$$N=\frac{q}{e}=\frac{3.0}{1.60\times10^{-19}}=1.9\times10^{19}\ \text{electrons}$$

Use the electron charge magnitude for the carrier count; direction is handled separately by current convention.

Common trap

Use time in seconds and charge in coulombs. Do not use I/t for charge; current is already charge divided by time.

Defining distinction and boundary

A changing current remains DC if it never reverses. The syllabus requires this distinction only; waveform calculations, rms values, rectification and detailed AC-circuit analysis are outside this objective.

Parallel junctions: conservation of charge

Charge does not accumulate at a steady circuit junction. The total current entering therefore equals the total current leaving:

$$I_{total}=I_1+I_2+\cdots$$

This explains why branch currents add in parallel.

Series path: conservation of energy

Each coulomb receives energy from the source and transfers it through series components. The potential differences across those components therefore add to the supply potential difference:

$$V_{supply}=V_1+V_2+\cdots$$

Use the conservation statements

At a two-branch junction, a missing branch current is $I_2=I_{total}-I_1$. In local practice question 4, the series supply is $12\,\mathrm V$ and the lamp drop is $4.0\,\mathrm V$, so the other series component has $V=12-4.0=8.0\,\mathrm V$.

Boundary

These are the simple series/parallel consequences required here. Do not extend this card to arbitrary multi-loop equation solving.

Resistance of a uniform conductor

$$R=\rho\frac{L}{A}$$

where ρ is resistivity, L is length and A is cross-sectional area. Resistivity is a material property at the stated conditions.

Read the scaling

At fixed material, doubling length doubles R. Doubling diameter makes area four times larger and reduces R to one quarter. A longer, thinner wire has greater resistance.

Units

Resistivity has SI unit Ω m. Use $$A=\pi r^2$$ for a circular wire and convert radius/diameter to metres before calculating.

Worked example from the mapped local textbook

Nichrome has $\rho=1.1\times10^{-6}\,\Omega\,\mathrm m$. A wire has $L=1.96\,\mathrm m$ and radius $r=0.21\,\mathrm{mm}=0.21\times10^{-3}\,\mathrm m$.

$$A=\pi r^2=1.39\times10^{-7}\,\mathrm{m^2}$$

$$R=\frac{\rho L}{A}=\frac{(1.1\times10^{-6})(1.96)}{1.39\times10^{-7}}=16\,\Omega$$

Converting the radius before squaring prevents a factor-of-$10^6$ error.

Common trap

Do not treat resistivity as the resistance of every sample of a material. Geometry changes resistance even when ρ is unchanged.

Series rules

In series, the same current passes through each component:

$$I=I_1=I_2$$

Potential differences and resistances add: $$V=V_1+V_2$$ and $$R_s=R_1+R_2$$.

Parallel rules

In parallel, each branch has the same potential difference:

$$V=V_1=V_2$$

Currents add at the junction and reciprocal resistances add:

$$I=I_1+I_2,\quad \frac1{R_p}=\frac1{R_1}+\frac1{R_2}$$

Solve systematically

Identify junctions and branches, replace simple groups with equivalent resistance, then use Ohm’s law and conservation rules to recover branch currents and voltage drops.

Worked comparison from the mapped local textbook

For $5.0\,\mathrm{k\Omega}$ and $8.0\,\mathrm{k\Omega}$ resistors:

$$R_s=5.0+8.0=13\,\mathrm{k\Omega}$$

$$R_p=\left(\frac1{5000}+\frac1{8000}\right)^{-1}=3.1\,\mathrm{k\Omega}$$

The parallel equivalent is smaller than either branch resistance, which is a useful check.

Common trap

Do not use the series current rule in a parallel branch or add parallel resistances directly.

Real-cell model

A real cell has emf ε and internal resistance r. With external resistance R and current I,

$$\varepsilon=I(R+r)$$

The internal resistance accounts for energy transferred inside the cell.

Terminal potential difference

The terminal voltage across the external load is

$$V=IR=\varepsilon-Ir$$

As current increases, the internal voltage drop Ir increases and terminal voltage falls.

Use a graph

A graph of terminal V against I has intercept ε and gradient −r. A graph of ε against I with total resistance has slope R+r.

Worked example from the mapped local textbook

A cell has $\varepsilon=1.5\,\mathrm V$, internal resistance $r=0.82\,\Omega$ and load $R=5.6\,\Omega$.

$$I=\frac{\varepsilon}{R+r}=\frac{1.5}{5.6+0.82}=0.23\,\mathrm A$$

$$V_{terminal}=IR=(0.23)(5.6)=1.3\,\mathrm V$$

The loaded terminal voltage is below the emf because energy is also transferred in the internal resistance.

Common trap

The emf is not always the same as the terminal voltage. They are equal only when current is zero or internal resistance is negligible.

Variable resistance

A variable resistor lets the resistance in a circuit be changed. Increasing the resistance of a series variable resistor reduces the current for a fixed supply voltage. A rheostat normally uses two terminals to control current; a potentiometer uses three terminals as a potential divider.

Predict the circuit response

For a fixed supply, use $I=V/R_{total}$. If the variable resistance increases, total resistance increases and current decreases. In a series circuit, the potential difference across the variable resistor increases while the potential difference across a fixed series component decreases.

Sensor examples

An LDR has resistance that depends on incident light intensity. An NTC thermistor has lower resistance at higher temperature. These components allow a circuit to respond to its surroundings, but the resistance–stimulus relationship must be obtained from data or a stated model.

Common trap

Do not assume that “more resistance” means a larger current. First decide whether the supply voltage is fixed and whether the component is in series or parallel. For an internal-resistance investigation, changing a variable resistor is useful because it creates multiple $V$-$I$ data points.