3.7 Fields and their consequences (A-level only) — AQA 7408 spec coverage

3.7.1

Fields

A force field is a region in which a body experiences a non-contact force; gravitational fields act on mass and electric fields act on charge.
Field strength is a vector. Its direction is the direction of the force on a small positive test charge, or on a test mass for gravity.
Gravitational and electrostatic forces both follow inverse-square laws and can be represented by field lines, potentials and equipotential surfaces.
Masses always attract, whereas like charges repel and unlike charges attract. A common error is to infer a field direction from the sign of potential alone.

Tier 1 · Easy

1 mark

ORIGINAL

State one difference between a gravitational force field and an electric force field.

Tier 2 · Standard

3 marks

ORIGINAL

A small positive charge and a small mass are placed separately at the same point near a negatively charged metal sphere and a planet. Describe the direction of each force and explain how a field-line diagram shows field strength.

Tier 3 · Hard

5 marks

ORIGINAL

Two identical positive point charges are fixed a short distance apart. Discuss the electric field and electric potential at the midpoint, and describe the required relationship between field lines and equipotential surfaces near the charges.

3.7.2.1

Newton's law

Newton's law of gravitation gives the magnitude between point masses: $F=\dfrac{Gm_1m_2}{r^2}$ , where $r$ is their centre-to-centre separation.
Gravity is universal and attractive: the two bodies experience equal-magnitude, opposite-direction forces, consistent with Newton's third law.
For a spherically symmetric body, its mass may be treated as concentrated at its centre when calculating the external force.
Use $G=6.67\times10^{-11}\,\text{N m}^2\text{ kg}^{-2}$ . A common error is to use surface gap or diameter instead of centre-to-centre distance.

Tier 1 · Easy

2 marks

ORIGINAL

Calculate the gravitational force between masses $6.0\times10^{22}\,\text{kg}$ and $4.0\times10^{20}\,\text{kg}$ whose centres are $3.0\times10^7\,\text{m}$ apart.

Tier 2 · Standard

4 marks

ORIGINAL

An $850\,\text{kg}$ satellite is $7.20\times10^6\,\text{m}$ from the centre of a planet of mass $5.97\times10^{24}\,\text{kg}$ . Calculate the gravitational force on the satellite and its acceleration. Explain why the acceleration would be unchanged for a satellite of different mass at the same point.

Tier 3 · Hard

5 marks

ORIGINAL

Two approximately spherical bodies have masses $4.0\times10^{24}\,\text{kg}$ and $7.5\times10^{23}\,\text{kg}$ . Their mutual gravitational force is $8.0\times10^{20}\,\text{N}$ . Determine their centre-to-centre separation and the new force if that separation increases by $20\%$ .

3.7.2.2

Gravitational field strength

Gravitational field strength is force per unit mass: $g=F/m$ . Its unit is $\text{N kg}^{-1}$ , equivalent to $\text{m s}^{-2}$ .
Outside a spherical mass, the radial-field magnitude is $g=GM/r^2$ and the vector points towards the centre of mass.
Radial field lines point inward and become less dense with increasing $r$ , showing the inverse-square decrease in field strength.
Use distance from the centre, not altitude above the surface. A common error is to omit the inward direction when a vector answer is required.

Tier 1 · Easy

1 mark

ORIGINAL

Define gravitational field strength at a point.

Tier 2 · Standard

3 marks

ORIGINAL

Determine the gravitational field strength $2.40\times10^7\,\text{m}$ from the centre of a planet of mass $4.80\times10^{24}\,\text{kg}$ .

Tier 3 · Hard

5 marks

ORIGINAL

A planet has mass $6.40\times10^{24}\,\text{kg}$ and radius $7.00\times10^6\,\text{m}$ . Calculate the altitude above its surface where the gravitational field strength is $2.40\,\text{N kg}^{-1}$ .

3.7.2.3

Gravitational potential

Gravitational potential is work done per unit mass in bringing a small mass from infinity to a point; the zero is defined at infinity.
For a spherical mass, $V_g=-GM/r$ . The negative sign shows that work must be supplied to move a mass from a bound position to infinity.
For movement between two points, the external work done in a slow transfer is $\Delta W=m\Delta V_g=m(V_{\text{final}}-V_{\text{initial}})$ .
Equipotentials are perpendicular to field lines, and no work is done along one. Also $g=-\Delta V_g/\Delta r$ ; a common error is to treat zero field as necessarily zero potential.

Tier 1 · Easy

1 mark

ORIGINAL

State why gravitational potential near an isolated planet is negative when potential is defined as zero at infinity.

Tier 2 · Standard

4 marks

ORIGINAL

A $320\,\text{kg}$ probe moves slowly from $9.0\times10^6\,\text{m}$ to $1.8\times10^7\,\text{m}$ from the centre of a planet of mass $7.2\times10^{24}\,\text{kg}$ . Calculate the change in gravitational potential and the work done by the external force.

Tier 3 · Hard

6 marks

ORIGINAL

A $450\,\text{kg}$ craft is moved slowly from radius $8.0\times10^6\,\text{m}$ to radius $2.4\times10^7\,\text{m}$ from a body of mass $5.5\times10^{24}\,\text{kg}$ . Determine the external work done. The craft is then released from rest at the outer point; calculate its speed when it returns to the inner point.

3.7.2.4

Orbits of planets and satellites

For a circular orbit, gravity supplies the centripetal force: $GMm/r^2=mv^2/r$ , so $v=\sqrt{GM/r}$ .
Using $v=2\pi r/T$ with Newton's law gives $T^2=4\pi^2r^3/(GM)$ and hence $T^2\propto r^3$ for one central mass.
The total energy of a circular orbit is $E=-GMm/(2r)$ ; moving to a larger circular orbit makes the total energy less negative.
A geostationary satellite has a $24$ -hour period, travels west-to-east above the equator and stays over one point. A common error is to use altitude instead of orbital radius in equations.
Escape speed follows from setting total energy at infinity to zero: $v_e=\sqrt{2GM/r}$ .

Tier 1 · Easy

2 marks

ORIGINAL

A satellite completes a circular orbit of radius $7.5\times10^6\,\text{m}$ in $6.0\times10^3\,\text{s}$ . Calculate its orbital speed.

Tier 2 · Standard

4 marks

ORIGINAL

Starting from Newton's law of gravitation, show that $T^2\propto r^3$ for circular orbits around one planet. Hence determine the factor by which the period changes when orbital radius is multiplied by $1.60$ .

Tier 3 · Hard

6 marks

ORIGINAL

A planet of mass $6.0\times10^{24}\,\text{kg}$ and radius $7.0\times10^6\,\text{m}$ rotates once every $30\,\text{h}$ . Determine the radius and altitude of a synchronous circular orbit. Calculate the total energy of a $500\,\text{kg}$ satellite in that orbit.

3.7.3.1

Coulomb's law

For point charges in a vacuum, $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q_1Q_2|}{r^2}$ , where $r$ is the separation and $1/(4\pi\varepsilon_0)=8.99\times10^9\,\text{N m}^2\text{ C}^{-2}$ .
Like charges repel and unlike charges attract. Coulomb's law gives a magnitude, so direction must be determined separately from the charge signs and geometry.
Air may normally be treated as a vacuum, and the external field of a charged sphere may be modelled as if its charge were at the centre.
For several charges, calculate each force vector before adding components. A common error is to add force magnitudes when the forces are not collinear.

Tier 1 · Easy

2 marks

ORIGINAL

Point charges $+3.0\,\text{nC}$ and $-8.0\,\text{nC}$ are $0.120\,\text{m}$ apart in air. Calculate the force between them.

Tier 2 · Standard

4 marks

ORIGINAL

Determine the ratio of the electrostatic force to the gravitational force between a proton and an electron. Use $m_p=1.67\times10^{-27}\,\text{kg}$ and $m_e=9.11\times10^{-31}\,\text{kg}$ , and explain why their separation is not needed.

Tier 3 · Hard

6 marks

ORIGINAL

A charge $+2.0\,\text{nC}$ is at the origin. A charge $+5.0\,\text{nC}$ is at $(0.300\,\text{m},0)$ and a charge $-4.0\,\text{nC}$ is at $(0,0.400\,\text{m})$ . Determine the magnitude and direction of the resultant electrostatic force on the charge at the origin.

3.7.3.2

Electric field strength

Electric field strength is force per unit positive charge: $E=F/Q$ , with unit $\text{N C}^{-1}$ or $\text{V m}^{-1}$ .
For a point charge, $E=Q/(4\pi\varepsilon_0r^2)$ in magnitude; field lines point away from positive charge and towards negative charge.
Between parallel plates the field is approximately uniform and $E=V/d$ . From work, $Fd=Q\Delta V$ gives the same result.
A charged particle entering a uniform field perpendicular to its initial velocity has constant acceleration parallel to the field and constant velocity perpendicular to it, producing a parabolic path.
A common error is to reverse the field direction for an electron: the field follows force on positive charge, while an electron's force is opposite to $E$ .

Tier 1 · Easy

1 mark

ORIGINAL

Define electric field strength at a point.

Tier 2 · Standard

4 marks

ORIGINAL

Parallel plates have a potential difference of $1.80\,\text{kV}$ and separation $45.0\,\text{mm}$ . Calculate the uniform electric field strength, the force magnitude on an electron, and its acceleration magnitude.

Tier 3 · Hard

6 marks

ORIGINAL

A proton enters at $3.00\times10^6\,\text{m s}^{-1}$ perpendicular to a uniform electric field of strength $2.50\times10^4\,\text{V m}^{-1}$ . The field region is $80.0\,\text{mm}$ long in the initial direction of travel. Determine the proton's deflection and the angle of its velocity to its original direction as it leaves. Use $m_p=1.67\times10^{-27}\,\text{kg}$ .

3.7.3.3

Electric potential

Electric potential is work done per unit positive charge in bringing a small test charge from infinity; the zero of absolute potential is at infinity.
For a point charge, $V=Q/(4\pi\varepsilon_0r)$ . Potential is a scalar and carries the sign of the source charge.
The external work done in a slow transfer is $\Delta W=q\Delta V=q(V_{\text{final}}-V_{\text{initial}})$ ; the work done by the field is its negative.
No work is done along an equipotential. The field points towards decreasing potential, with $E=\Delta V/\Delta r$ in magnitude and potential difference given by area under an $E$ - $r$ graph.
A common error is to add electric-field magnitudes as scalars or to assume that $V=0$ implies $E=0$ .

Tier 1 · Easy

2 marks

ORIGINAL

Calculate the electric potential $0.200\,\text{m}$ from an isolated point charge of $+4.00\,\text{nC}$ .

Tier 2 · Standard

4 marks

ORIGINAL

A charge of $-3.0\,\text{nC}$ is moved slowly from a point at $+120\,\text{V}$ to a point at $-80\,\text{V}$ . Calculate the work done by the external force and state whether the charge's electric potential energy increases or decreases.

Tier 3 · Hard

6 marks

ORIGINAL

Charges $+8.0\,\text{nC}$ and $-2.0\,\text{nC}$ are fixed $0.600\,\text{m}$ apart. Find the point between them where the electric potential is zero. Determine the electric field strength there and the external work needed to bring a $+3.0\,\text{nC}$ charge slowly from infinity to that point.

3.7.4.1

Capacitance

Capacitance is charge stored per unit potential difference: $C=Q/V$ . One farad is one coulomb per volt.
For a fixed capacitor, a graph of $Q$ against $V$ is a straight line through the origin with gradient $C$ .
At fixed capacitance, increasing potential difference increases the equal and opposite charge magnitudes on the plates proportionally.
Convert units before substitution: $1\,\mu\text{F}=10^{-6}\,\text{F}$ . A common error is to confuse charge on one plate with the net charge of the whole capacitor.

Tier 1 · Easy

2 marks

ORIGINAL

A capacitor stores charge of $3.6\,\text{mC}$ at a potential difference of $12\,\text{V}$ . Calculate its capacitance.

Tier 2 · Standard

3 marks

ORIGINAL

A charge-potential-difference graph for a capacitor passes through the point $(16.0\,\text{V},4.80\,\text{mC})$ . Determine its capacitance and the charge stored at $27.0\,\text{V}$ .

Tier 3 · Hard

5 marks

ORIGINAL

A $470\,\mu\text{F}$ capacitor is initially at $9.0\,\text{V}$ . A charge-transfer process then moves $1.2\times10^{16}$ electrons from its negatively charged plate to its positively charged plate while capacitance remains constant. Determine the new charge magnitude on each plate and the new potential difference.

3.7.4.2

Parallel plate capacitor

For parallel plates, $C=\varepsilon_0\varepsilon_rA/d$ : capacitance increases with plate overlap area and relative permittivity, and decreases with separation.
A dielectric becomes polarised in the field. Bound charges create an opposing field, reducing the resultant field and potential difference for a given free charge.
If the capacitor remains connected to a fixed-voltage supply, adding a dielectric increases $C$ and therefore draws additional charge because $Q=CV$ .
Use the overlap area and separation in SI units. A common error is to say the dielectric increases charge when the isolated capacitor's free charge is actually fixed.

Tier 1 · Easy

2 marks

ORIGINAL

Two parallel plates in air have overlap area $2.50\times10^{-2}\,\text{m}^2$ and separation $1.20\,\text{mm}$ . Calculate their capacitance.

Tier 2 · Standard

4 marks

ORIGINAL

A parallel-plate capacitor has area $1.80\times10^{-2}\,\text{m}^2$ , separation $0.800\,\text{mm}$ and is connected to a $120\,\text{V}$ supply. A dielectric of relative permittivity $3.40$ completely fills the gap. Determine the new capacitance and charge stored.

Tier 3 · Hard

6 marks

ORIGINAL

A designer needs a $2.00\,\text{nF}$ parallel-plate capacitor using a dielectric with relative permittivity $4.50$ and thickness $0.750\,\text{mm}$ . Calculate the required overlap area. Explain at the molecular level why the dielectric increases capacitance.

3.7.4.3

Energy stored by a capacitor

The energy stored is the work done separating charge: $E=\tfrac12QV=\tfrac12CV^2=\tfrac12Q^2/C$ .
On a graph of potential difference against charge, stored energy is the area under the line; for a linear capacitor this is a triangle.
Choose the energy form that matches the quantities given and retain the square on $V$ or $Q$ where required.
When voltage falls, energy released is the difference between initial and final stored energies, not simply $Q\Delta V$ . This is a common marking error.

Tier 1 · Easy

2 marks

ORIGINAL

A capacitor stores $2.0\,\text{mC}$ at a potential difference of $9.0\,\text{V}$ . Calculate its stored energy.

Tier 2 · Standard

3 marks

ORIGINAL

The potential difference across a $150\,\mu\text{F}$ capacitor falls from $20.0\,\text{V}$ to $8.0\,\text{V}$ . Determine the energy transferred from the capacitor.

Tier 3 · Hard

5 marks

ORIGINAL

A $330\,\mu\text{F}$ capacitor is charged from zero to $24.0\,\text{V}$ through a resistor by an ideal constant-voltage supply. Calculate the work done by the supply, the final stored energy, and the mean power dissipated in the resistor if charging takes $0.800\,\text{s}$ .

3.7.4.4

Capacitor charge and discharge

The time constant is $\tau=RC$ . During discharge, $Q=Q_0e^{-t/RC}$ , with corresponding exponential equations for $V$ and the magnitude of $I$ .
After one time constant a discharging quantity is $e^{-1}=0.368$ of its initial value; the half-life is $T_{1/2}=0.69RC$ .
During charging, $Q=Q_0(1-e^{-t/RC})$ ; charge and voltage rise towards a limit while current falls towards zero.
For discharge, plotting $\ln Q$ against $t$ gives a straight line of gradient $-1/RC$ . A common error is to treat one time constant as the time for complete discharge.
On graph questions, a tangent gives instantaneous current through $I=\mathrm{d}Q/\mathrm{d}t$ , while area under an $I$ - $t$ graph gives transferred charge.

Tier 1 · Easy

2 marks

ORIGINAL

A $220\,\mu\text{F}$ capacitor discharges through a $47\,\text{k}\Omega$ resistor. Calculate the time constant and state the fraction of initial charge remaining after this time.

Tier 2 · Standard

3 marks

ORIGINAL

A capacitor initially stores $6.00\,\text{mC}$ and discharges through an $82.0\,\text{k}\Omega$ resistor. Its capacitance is $100\,\mu\text{F}$ . Determine the charge remaining after $12.0\,\text{s}$ .

Tier 3 · Hard

6 marks

ORIGINAL

During discharge, charge falls from $8.0\,\text{mC}$ to $1.6\,\text{mC}$ in $14.0\,\text{s}$ . The capacitance is $120\,\mu\text{F}$ . Determine the time constant, the resistance, and the charge after one further time constant. State the gradient of a graph of $\ln Q$ against $t$ .

3.7.5.1

Magnetic flux density

For a straight wire perpendicular to a magnetic field, $F=BIl$ . Magnetic flux density is therefore $B=F/(Il)$ .
One tesla is the flux density that gives a force of $1\,\text{N}$ on a $1\,\text{m}$ wire carrying $1\,\text{A}$ perpendicular to the field.
Fleming's left-hand rule gives force, field and conventional-current directions; electron flow is opposite to conventional current.
In a top-pan-balance experiment, convert the change in mass reading to force using $F=\Delta mg$ . A common error is to substitute grams directly into the force equation.

Tier 1 · Easy

2 marks

ORIGINAL

A $0.080\,\text{m}$ wire carries $3.2\,\text{A}$ perpendicular to a magnetic field of flux density $0.45\,\text{T}$ . Calculate the force on the wire.

Tier 2 · Standard

3 marks

ORIGINAL

A horizontal wire of length $0.120\,\text{m}$ carries $4.00\,\text{A}$ perpendicular to a magnetic field. Switching on the current changes the balance reading by $6.50\,\text{g}$ . Determine the magnetic flux density.

Tier 3 · Hard

5 marks

ORIGINAL

In a balance experiment, a perpendicular wire of active length $0.0850\,\text{m}$ gives a straight-line graph of balance-reading change against current with gradient $1.75\,\text{g A}^{-1}$ . Determine the flux density and predict the reading change at $5.20\,\text{A}$ . Explain what a non-zero vertical intercept would suggest.

3.7.5.2

Moving charges in a magnetic field

A charge moving perpendicular to a magnetic field experiences force $F=BQv$ . The force is perpendicular to both velocity and field.
Use Fleming's left-hand rule for positive conventional current; reverse the predicted force for a negative particle.
Because magnetic force is perpendicular to velocity, it does no work and changes direction without changing speed or kinetic energy.
For circular motion, $BQv=mv^2/r$ , so $r=mv/(BQ)$ . A common error is to include weight or use the particle's diameter as $r$ .
In a cyclotron, the non-relativistic orbital frequency is $f=BQ/(2\pi m)$ and is independent of radius and speed.

Tier 1 · Easy

2 marks

ORIGINAL

A proton moves at $4.0\times10^6\,\text{m s}^{-1}$ perpendicular to a $0.25\,\text{T}$ magnetic field. Calculate the magnetic force magnitude.

Tier 2 · Standard

3 marks

ORIGINAL

A proton of speed $3.20\times10^6\,\text{m s}^{-1}$ enters a uniform $0.480\,\text{T}$ magnetic field perpendicular to the field. Determine the radius of its path. Use $m_p=1.67\times10^{-27}\,\text{kg}$ .

Tier 3 · Hard

6 marks

ORIGINAL

An alpha particle of mass $6.64\times10^{-27}\,\text{kg}$ and charge $+3.20\times10^{-19}\,\text{C}$ reaches radius $0.450\,\text{m}$ in a cyclotron with magnetic flux density $0.800\,\text{T}$ . Determine its speed, kinetic energy in $\text{MeV}$ , and cyclotron frequency.

3.7.5.3

Magnetic flux and flux linkage

Magnetic flux is $\Phi=BA$ when the field is normal to area $A$ , and is measured in webers.
For a flat coil at angle $\theta$ between the field and the coil's normal, flux linkage is $N\Phi=BAN\cos\theta$ .
Flux linkage counts the flux through every turn; changing $N$ , $B$ , $A$ or orientation changes it.
A negative value of $N\Phi$ indicates that flux passes through the coil in the direction opposite to the chosen positive normal. A common error is to use the angle to the plane instead of the normal.

Tier 1 · Easy

2 marks

ORIGINAL

Magnetic flux passes normally through a flat surface of area $1.5\times10^{-2}\,\text{m}^2$ . The flux density is $0.32\,\text{T}$ . Calculate the flux.

Tier 2 · Standard

3 marks

ORIGINAL

A $240$ -turn coil of area $3.50\times10^{-3}\,\text{m}^2$ is in a $0.180\,\text{T}$ field. The field makes an angle of $35.0^\circ$ with the normal to the coil. Determine the flux linkage.

Tier 3 · Hard

5 marks

ORIGINAL

A $500$ -turn search coil of area $8.0\times10^{-4}\,\text{m}^2$ is in a uniform $0.120\,\text{T}$ field. Its chosen normal is rotated from $20.0^\circ$ to $110^\circ$ relative to the field. Calculate the initial and final flux linkages and their signed change.

3.7.5.4

Electromagnetic induction

Faraday's law states that the induced emf magnitude equals the rate of change of flux linkage: $|\varepsilon|=|\Delta(N\Phi)/\Delta t|$ .
Lenz's law gives the direction: the induced current produces effects that oppose the change in flux that caused it.
A straight conductor cutting field lines has an induced emf; a stationary search coil in a steady field has none unless its flux linkage changes.
For a coil rotating uniformly, $\varepsilon=BAN\omega\sin\omega t$ and the peak emf is $BAN\omega$ .
A common error is to say induced current opposes the field; it opposes the change in flux, and a current exists only when the circuit is complete.

Tier 1 · Easy

2 marks

ORIGINAL

State Faraday's law and Lenz's law for electromagnetic induction.

Tier 2 · Standard

3 marks

ORIGINAL

The magnetic flux through each turn of a $250$ -turn coil decreases uniformly from $3.2\,\text{mWb}$ to $0.80\,\text{mWb}$ in $40\,\text{ms}$ . Determine the induced emf magnitude and state how Lenz's law fixes its polarity.

Tier 3 · Hard

5 marks

ORIGINAL

A $180$ -turn coil of area $6.50\times10^{-3}\,\text{m}^2$ rotates at $50.0\,\text{Hz}$ in a uniform $0.240\,\text{T}$ field. At $t=0$ its flux linkage is maximum and positive. Calculate the peak emf and the emf magnitude at $t=2.50\,\text{ms}$ . Explain the origin and direction of the emf using Faraday's and Lenz's laws.

3.7.5.5

Alternating currents

For a sinusoidal waveform, $I_{\mathrm{rms}}=I_0/\sqrt2$ and $V_{\mathrm{rms}}=V_0/\sqrt2$ , where subscript $0$ denotes peak value.
Peak-to-peak value is twice the peak value. Mains voltage is quoted as an rms value, not a peak value.
An rms current has the same mean heating effect in a resistor as a direct current of that value; for a resistive load, mean power is $V_{\mathrm{rms}}I_{\mathrm{rms}}$ .
On an oscilloscope, amplitude equals vertical divisions times volts per division; period equals horizontal divisions times time per division, and $f=1/T$ .
A common error is to apply the $\sqrt2$ relationships to non-sinusoidal waveforms or to confuse peak with peak-to-peak.

Tier 1 · Easy

1 mark

ORIGINAL

A sinusoidal voltage has peak value $12.0\,\text{V}$ . Calculate its rms value.

Tier 2 · Standard

4 marks

ORIGINAL

A sinusoidal trace spans $6.4$ horizontal divisions per cycle at $0.50\,\text{ms div}^{-1}$ . Its peak-to-peak height is $5.6$ vertical divisions at $2.0\,\text{V div}^{-1}$ . Determine the frequency, peak voltage and rms voltage.

Tier 3 · Hard

6 marks

ORIGINAL

A resistive heater rated at $1.80\,\text{kW}$ operates from a sinusoidal $230\,\text{V rms}$ supply. Determine the peak and peak-to-peak supply voltages, the rms and peak currents, and the energy transferred in $12.0\,\text{min}$ .

3.7.5.6

The operation of a transformer

For an ideal transformer, $N_s/N_p=V_s/V_p$ . A step-up transformer has more secondary turns and a larger secondary voltage.
Efficiency is $\eta=P_{\mathrm{out}}/P_{\mathrm{in}}=I_sV_s/(I_pV_p)$ ; an ideal transformer conserves power, so stepping voltage up steps current down.
Energy losses include resistive heating in windings, eddy currents and hysteresis in the core, and flux leakage.
Laminating and electrically insulating the core restricts eddy-current loops; a suitable soft magnetic core reduces hysteresis loss.
High-voltage transmission reduces current for a given power and therefore reduces cable loss $P_{\mathrm{loss}}=I^2R$ . A common error is to calculate loss using transmitted voltage across the cable resistance.

Tier 1 · Easy

2 marks

ORIGINAL

An ideal transformer has $300$ primary turns and $1200$ secondary turns. The primary voltage is $24\,\text{V}$ . Calculate the secondary voltage.

Tier 2 · Standard

4 marks

ORIGINAL

A transformer takes $1.80\,\text{A}$ from a $230\,\text{V}$ supply and delivers $24.0\,\text{V}$ at $84.0\%$ efficiency. Determine the secondary current and the power dissipated in the transformer.

Tier 3 · Hard

6 marks

ORIGINAL

A station transmits $2.50\,\text{MW}$ through cables of total resistance $3.20\,\Omega$ . Compare the cable power losses when transmission voltage is $25.0\,\text{kV}$ and $250\,\text{kV}$ . Explain the transformer's role and how core construction reduces eddy-current loss.