B.5 Current and circuits

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.

Two ways to supply emf

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

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.

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.

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 $230\,\mathrm{V}$.

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.

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.

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.

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.

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 $V=12-4.0=8.0\,\mathrm V$.

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.

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.

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.

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.