Skip to content

AQA A-level Physics revision notes

Fields and their consequences (A-level only)

Section 3.7
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
18 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.7

Checked against AQA 7408 section 3.7. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

In the exam: Data and formulae booklet provided · calculator allowed in every paper

Open the printable pack
3.7.1

Fields

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A force field is a region in which a body experiences a non-contact force. Gravitational fields arise from masses, electric fields from static charges, and magnetic fields from interactions involving moving charges.
  • Field strength is a vector whose direction is found by inspecting the force on a test mass or positive test charge.
  • Gravitational and electrostatic fields both obey inverse-square laws and can be represented using field lines, potential and equipotential surfaces.
  • Their key difference is that masses always attract, while charges may attract or repel.
  • Examiners expect comparisons to distinguish vector field strength from scalar potential and to determine direction from the interaction, not merely from the sign of a potential value.
Field directions around a mass and a positive charge, showing gravitational attraction and electric repulsion.
Worked example

Two identical positive charges are fixed symmetrically about a midpoint. Compare the resultant electric field and electric potential at the midpoint.

  1. 1.The two equal electric-field vectors point in opposite directions.
  2. 2.Vector addition gives zero resultant field.
  3. 3.The two positive potential contributions are scalars and add.

Answer: The electric field is zero, but the electric potential is positive and non-zero.

Common mistakes

  • Don't assume zero field at a point must also mean zero potential.
  • Don't state that gravitational interaction can repel like an electric interaction.
  • Don't use the force direction on an electron to define electric field direction.

Exam tip

In a comparison, state the shared inverse-square behaviour before contrasting attraction-only gravity with electric attraction or repulsion.

Tier 1 · Easy

ORIGINAL

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

[1 mark]

Total for this question: 1

Tier 2 · Standard

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.

[3 marks]

Total for this question: 3

Tier 3 · Hard

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.

[5 marks]

Total for this question: 5

Your progress and exam materials

This section: Evidence from your answers: 0/18 secureYour confidence: 0 self-rated secureTracker status: 0/18 secure, 0 shaky, 18 unseen

Overall: Evidence from your answers: 0/147 secureYour confidence: 0 self-rated secureTracker status: 0/147 secure, 0 shaky, 147 unseen

Progress is saved on this device for guests and accounts right now; cross-device account sync is not live yet.

3.7.2.1

Newton's law

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Newton’s law of gravitation gives the magnitude of the universal attractive force between point masses: F=Gm1m2/r2F=Gm_1m_2/r^2, where GG is the gravitational constant and rr is their centre-to-centre separation.
  • Each mass experiences an equal force directed towards the other mass.
  • Outside a spherically symmetric body, its mass may be treated as concentrated at its centre.
  • The inverse-square dependence means multiplying separation by a factor kk divides force by k2k^2.
  • Examiners expect masses in kilograms, separation from centre to centre, a direction as well as a magnitude, and sensible estimation of the extremely small force between ordinary laboratory objects compared with astronomical bodies.
Two point masses separated centre to centre, with equal attractive forces directed towards each other.
Worked example

Masses 5.0×1024kg5.0\times10^{24}\,\text{kg} and 8.0×1022kg8.0\times10^{22}\,\text{kg} are 4.0×108m4.0\times10^8\,\text{m} apart. Calculate their gravitational force.

  1. 1.Use centre-to-centre separation in F=Gm1m2/r2F=Gm_1m_2/r^2.
  2. 2.Substitute F=(6.67×1011)(5.0×1024)(8.0×1022)/(4.0×108)2F=(6.67\times10^{-11})(5.0\times10^{24})(8.0\times10^{22})/(4.0\times10^8)^2.
  3. 3.Evaluate and give the attractive direction.

Answer: 1.7×1020N1.7\times10^{20}\,\text{N}, attractive.

Common mistakes

  • Don't use the gap between two surfaces rather than centre-to-centre separation.
  • Don't apply an inverse 1/r1/r relationship instead of the inverse square 1/r21/r^2.
  • Don't call the force on the second mass a cancelling force on the first mass.

Exam tip

Write the separation definition beside rr before substituting values from a radius or altitude diagram.

Tier 1 · Easy

ORIGINAL

Calculate the gravitational force between masses 6.0×1022kg6.0\times10^{22}\,\text{kg} and 4.0×1020kg4.0\times10^{20}\,\text{kg} whose centres are 3.0×107m3.0\times10^7\,\text{m} apart.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An 850kg850\,\text{kg} satellite is 7.20×106m7.20\times10^6\,\text{m} from the centre of a planet of mass 5.97×1024kg5.97\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.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Two approximately spherical bodies have masses 4.0×1024kg4.0\times10^{24}\,\text{kg} and 7.5×1023kg7.5\times10^{23}\,\text{kg}. Their mutual gravitational force is 8.0×1020N8.0\times10^{20}\,\text{N}. Determine their centre-to-centre separation and the new force if that separation increases by 20%20\%.

[5 marks]

Total for this question: 5

3.7.2.2

Gravitational field strength

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Gravitational field strength is the gravitational force per unit mass on a small test mass, g=F/mg=F/m, with units N kg1\text{N kg}^{-1}, equivalent to m s2\text{m s}^{-2}.
  • Field lines point in the direction of force on the test mass.
  • Around a spherical mass they are radial and directed inward; their increasing separation with radius represents decreasing strength.
  • The magnitude outside the mass is g=GM/r2g=GM/r^2, where rr is measured from the centre.
  • Examiners expect altitude to be converted to radial distance, the inward vector direction to be stated, and field-line density to be interpreted qualitatively rather than treated as a literal count of physical lines.
A radial gravitational field with arrows directed towards the central mass.
Worked example

Determine the gravitational field strength 8.0×106m8.0\times10^6\,\text{m} from the centre of a planet of mass 6.0×1024kg6.0\times10^{24}\,\text{kg}.

  1. 1.Use g=GM/r2g=GM/r^2 with distance measured from the centre.
  2. 2.g=(6.67×1011)(6.0×1024)/(8.0×106)2g=(6.67\times10^{-11})(6.0\times10^{24})/(8.0\times10^6)^2.
  3. 3.Evaluate and state the radial direction.

Answer: 6.3N kg16.3\,\text{N kg}^{-1} towards the planet.

Common mistakes

  • Don't use altitude above the surface directly as rr.
  • Don't quote only the magnitude although gravitational field strength is a vector.
  • Don't draw radial gravitational arrows pointing away from the mass.

Exam tip

If a radius and altitude are both given, write r=R+hr=R+h before using g=GM/r2g=GM/r^2.

Tier 1 · Easy

ORIGINAL

Define gravitational field strength at a point.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Determine the gravitational field strength 2.40×107m2.40\times10^7\,\text{m} from the centre of a planet of mass 4.80×1024kg4.80\times10^{24}\,\text{kg}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A planet has mass 6.40×1024kg6.40\times10^{24}\,\text{kg} and radius 7.00×106m7.00\times10^6\,\text{m}. Calculate the altitude above its surface where the gravitational field strength is 2.40N kg12.40\,\text{N kg}^{-1}.

[5 marks]

Total for this question: 5

3.7.2.3

Gravitational potential

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Gravitational potential is work done per unit mass in bringing a small mass from infinity to a point, with zero potential defined at infinity. For a spherical mass, V=GM/rV=-GM/r; the negative sign shows that a mass at finite radius is gravitationally bound.
  • Potential difference determines external work in a slow transfer, ΔW=mΔV\Delta W=m\Delta V.
  • No work is done moving along an equipotential surface, which is perpendicular to field lines.
  • The radial graphs satisfy $g=-\Delta V/\Delta r$, so field strength is the magnitude of the potential gradient, while potential difference is obtained from the signed area under a ggrr graph.
  • Examiners expect signs, the zero reference and the distinction between scalar potential and vector field strength.
Gravitational potential outside a spherical mass, remaining negative and approaching zero as radius increases.
Worked example

A 200kg200\,\text{kg} probe moves slowly from 8.0×106m8.0\times10^6\,\text{m} to 1.6×107m1.6\times10^7\,\text{m} from a mass of 6.0×1024kg6.0\times10^{24}\,\text{kg}. Find the external work done.

  1. 1.Calculate Vi=GM/ri=5.00×107J kg1V_i=-GM/r_i=-5.00\times10^7\,\text{J kg}^{-1}.
  2. 2.Calculate Vf=GM/rf=2.50×107J kg1V_f=-GM/r_f=-2.50\times10^7\,\text{J kg}^{-1}.
  3. 3.ΔW=m(VfVi)=200(2.50×107)=5.0×109J\Delta W=m(V_f-V_i)=200(2.50\times10^7)=5.0\times10^9\,\text{J}.

Answer: The external work done is +5.0×109J+5.0\times10^9\,\text{J}.

Common mistakes

  • Don't drop the negative sign from V=GM/rV=-GM/r.
  • Don't use mVimV_i rather than m(VfVi)m(V_f-V_i) for work between two finite points.
  • Don't claim that work is done while moving along one equipotential surface.

Exam tip

Write both potentials with signs before forming ΔV=VfVi\Delta V=V_f-V_i.

Tier 1 · Easy

ORIGINAL

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

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A 320kg320\,\text{kg} probe moves slowly from 9.0×106m9.0\times10^6\,\text{m} to 1.8×107m1.8\times10^7\,\text{m} from the centre of a planet of mass 7.2×1024kg7.2\times10^{24}\,\text{kg}. Calculate the change in gravitational potential and the work done by the external force.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A 450kg450\,\text{kg} craft is moved slowly from radius 8.0×106m8.0\times10^6\,\text{m} to radius 2.4×107m2.4\times10^7\,\text{m} from a body of mass 5.5×1024kg5.5\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.

[6 marks]

Total for this question: 6

3.7.2.4

Orbits of planets and satellites

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a circular orbit, gravity supplies the centripetal force, giving v=GM/rv=\sqrt{GM/r}. Combining this with v=2πr/Tv=2\pi r/T gives T2=4π2r3/(GM)T^2=4\pi^2r^3/(GM) and therefore T2r3T^2\propto r^3 for one central mass.
  • A circular satellite’s total energy is GMm/(2r)-GMm/(2r); escape speed follows by setting total energy at infinity to zero. A synchronous satellite matches the body’s rotation period.
  • A geostationary satellite is a special synchronous satellite in a circular equatorial orbit, moving in the rotation direction, with orbital radius measured from the planet’s centre.
  • Low orbits give shorter periods and different coverage.
  • Examiners expect energy, period, speed, orbit plane and radius to be connected to the stated satellite use.
Circular satellite orbits at different radii around a planet.
Worked example

A satellite orbits a planet of mass 6.0×1024kg6.0\times10^{24}\,\text{kg} at radius 4.0×107m4.0\times10^7\,\text{m}. Determine its speed and period.

  1. 1.v=GM/r=(6.67×1011)(6.0×1024)/(4.0×107)v=\sqrt{GM/r}=\sqrt{(6.67\times10^{-11})(6.0\times10^{24})/(4.0\times10^7)}.
  2. 2.Evaluate v=3.16×103m s1v=3.16\times10^3\,\text{m s}^{-1}.
  3. 3.T=2πr/v=7.95×104s=22.1hT=2\pi r/v=7.95\times10^4\,\text{s}=22.1\,\text{h}.

Answer: v=3.16×103m s1v=3.16\times10^3\,\text{m s}^{-1} and T=22.1hT=22.1\,\text{h}.

Common mistakes

  • Don't use altitude above the surface instead of orbital radius from the centre.
  • Don't state that every synchronous orbit is geostationary.
  • Don't make total orbital energy positive for a bound circular orbit.

Exam tip

For a derivation, start by equating gravitational force to centripetal force before substituting v=2πr/Tv=2\pi r/T.

Tier 1 · Easy

ORIGINAL

A satellite completes a circular orbit of radius 7.5×106m7.5\times10^6\,\text{m} in 6.0×103s6.0\times10^3\,\text{s}. Calculate its orbital speed.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Starting from Newton's law of gravitation, show that T2r3T^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 1.601.60.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A planet of mass 6.0×1024kg6.0\times10^{24}\,\text{kg} and radius 7.0×106m7.0\times10^6\,\text{m} rotates once every 30h30\,\text{h}. Determine the radius and altitude of a synchronous circular orbit. Calculate the total energy of a 500kg500\,\text{kg} satellite in that orbit.

[6 marks]

Total for this question: 6

3.7.3.1

Coulomb's law

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For point charges in a vacuum, Coulomb’s law gives F=Q1Q2/(4πε0r2)F=|Q_1Q_2|/(4\pi\varepsilon_0r^2), where ε0\varepsilon_0 is the permittivity of free space and rr is the separation.
  • Air may be treated as a vacuum, and a charged sphere’s external effect may be modelled as charge concentrated at its centre.
  • Like charges repel and unlike charges attract, so the equation gives magnitude while signs and geometry determine direction.
  • Electrostatic and gravitational forces between subatomic particles both follow inverse-square laws, allowing their magnitude ratio to be compared because r2r^2 cancels.
  • Examiners expect charge conversion from nanocoulombs, vector addition for several charges, and a direction justified from charge signs.
Unlike point charges separated by distance r, with equal attractive forces.
Worked example

Charges +3.0nC+3.0\,\text{nC} and 6.0nC-6.0\,\text{nC} are 0.20m0.20\,\text{m} apart in air. Calculate the force.

  1. 1.Treat air as a vacuum and convert both charges to coulombs.
  2. 2.F=(8.99×109)(3.0×109)(6.0×109)/(0.20)2F=(8.99\times10^9)(3.0\times10^{-9})(6.0\times10^{-9})/(0.20)^2.
  3. 3.Evaluate the magnitude and use opposite signs for the direction.

Answer: 4.0×106N4.0\times10^{-6}\,\text{N}, attractive.

Common mistakes

  • Don't leave nanocoulomb values unconverted before substitution.
  • Don't use the signs inside a magnitude calculation and report a negative force magnitude.
  • Don't add non-collinear force magnitudes without resolving vectors.

Exam tip

Calculate the positive magnitude first, then write a separate attraction-or-repulsion statement from the charge signs.

Tier 1 · Easy

ORIGINAL

Point charges +3.0nC+3.0\,\text{nC} and 8.0nC-8.0\,\text{nC} are 0.120m0.120\,\text{m} apart in air. Calculate the force between them.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Determine the ratio of the electrostatic force to the gravitational force between a proton and an electron. Use mp=1.67×1027kgm_p=1.67\times10^{-27}\,\text{kg} and me=9.11×1031kgm_e=9.11\times10^{-31}\,\text{kg}, and explain why their separation is not needed.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

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

[6 marks]

Total for this question: 6

3.7.3.2

Electric field strength

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Electric field strength is force per unit positive test charge, E=F/QE=F/Q, measured in N C1\text{N C}^{-1} or V m1\text{V m}^{-1}. Field lines point away from positive charge and towards negative charge.
  • A point charge produces radial magnitude E=Q/(4πε0r2)E=Q/(4\pi\varepsilon_0r^2).
  • Between parallel plates the field is approximately uniform: draw or describe straight, parallel and evenly spaced field lines, and use E=V/dE=V/d, derived from Fd=QΔVFd=Q\Delta V.
  • A charged particle entering perpendicular to a uniform field keeps constant velocity across the field but accelerates parallel or antiparallel to it, giving a parabolic trajectory.
  • Examiners expect the field direction to follow force on positive charge, so an electron accelerates opposite to the field, and may require field patterns investigated using conducting paper or an electrolytic tank.
A uniform electric field between parallel plates and the parabolic path of a positive particle entering perpendicular to it.
Worked example

Parallel plates have potential difference 1.20kV1.20\,\text{kV} and separation 30mm30\,\text{mm}. Find the field strength and force on an electron.

  1. 1.Convert V=1.20×103VV=1.20\times10^3\,\text{V} and d=0.030md=0.030\,\text{m}.
  2. 2.E=V/d=4.0×104V m1E=V/d=4.0\times10^4\,\text{V m}^{-1}.
  3. 3.F=eE=(1.60×1019)(4.0×104)=6.4×1015NF=eE=(1.60\times10^{-19})(4.0\times10^4)=6.4\times10^{-15}\,\text{N} opposite to the field.

Answer: E=4.0×104V m1E=4.0\times10^4\,\text{V m}^{-1} and F=6.4×1015NF=6.4\times10^{-15}\,\text{N} towards the positive plate.

Common mistakes

  • Don't use plate separation in millimetres in E=V/dE=V/d.
  • Don't draw the electron force in the electric-field direction.
  • Don't treat a charged particle’s path in a uniform field as circular.

Exam tip

For a uniform parallel-plate field, say the field lines are parallel and evenly spaced. Mark the plate signs and field direction before deciding the force direction for a positive or negative particle.

Tier 1 · Easy

ORIGINAL

Define electric field strength at a point.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Parallel plates have a potential difference of 1.80kV1.80\,\text{kV} and separation 45.0mm45.0\,\text{mm}. Calculate the uniform electric field strength, the force magnitude on an electron, and its acceleration magnitude.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A proton enters at 3.00×106m s13.00\times10^6\,\text{m s}^{-1} perpendicular to a uniform electric field of strength 2.50×104V m12.50\times10^4\,\text{V m}^{-1}. The field region is 80.0mm80.0\,\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 mp=1.67×1027kgm_p=1.67\times10^{-27}\,\text{kg}.

[6 marks]

Total for this question: 6

3.7.3.3

Electric potential

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Absolute electric potential is work done per unit positive charge in bringing a small test charge from infinity, where potential is defined as zero. For a point charge, V=Q/(4πε0r)V=Q/(4\pi\varepsilon_0r); potential is a scalar and carries the source charge’s sign.
  • External work in a slow transfer is ΔW=qΔV\Delta W=q\Delta V.
  • No work is done moving along an equipotential surface, which meets field lines at right angles.
  • The field points towards decreasing potential and has magnitude given by the potential gradient, while potential difference is the signed area under an EErr graph.
  • Examiners expect scalar potentials to be added algebraically, vector fields to be added separately, and recognition that zero potential need not mean zero field.
Radial electric field lines crossing circular equipotentials at right angles around a positive point charge.
Worked example

Find the electric potential 0.25m0.25\,\text{m} from a +5.0nC+5.0\,\text{nC} point charge and the external work to bring +2.0nC+2.0\,\text{nC} from infinity.

  1. 1.V=(8.99×109)(5.0×109)/0.25=180VV=(8.99\times10^9)(5.0\times10^{-9})/0.25=180\,\text{V}.
  2. 2.The initial potential at infinity is zero, so ΔV=+180V\Delta V=+180\,\text{V}.
  3. 3.ΔW=qΔV=(2.0×109)(180)=3.6×107J\Delta W=q\Delta V=(2.0\times10^{-9})(180)=3.6\times10^{-7}\,\text{J}.

Answer: V=+180VV=+180\,\text{V} and external work =+3.6×107J=+3.6\times10^{-7}\,\text{J}.

Common mistakes

  • Don't drop the sign of the source charge when calculating potential.
  • Don't add electric-field magnitudes as scalars when several charges are present.
  • Don't claim that a charge moving along an equipotential changes potential energy.

Exam tip

Calculate scalar potential first; only then use qΔVq\Delta V with the moved charge’s sign.

Tier 1 · Easy

ORIGINAL

Calculate the electric potential 0.200m0.200\,\text{m} from an isolated point charge of +4.00nC+4.00\,\text{nC}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Charges +8.0nC+8.0\,\text{nC} and 2.0nC-2.0\,\text{nC} are fixed 0.600m0.600\,\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 +3.0nC+3.0\,\text{nC} charge slowly from infinity to that point.

[6 marks]

Total for this question: 6

3.7.4.1

Capacitance

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Capacitance is charge stored per unit potential difference, C=Q/VC=Q/V, measured in farads, where 1F=1C V11\,\text{F}=1\,\text{C V}^{-1}.
  • The two plates carry equal and opposite charges, and QQ denotes the magnitude on either plate rather than a net charge on the whole capacitor.
  • For a fixed capacitor, QQ is proportional to VV, so a graph of charge against potential difference is a straight line through the origin with gradient CC.
  • Examiners expect microfarads and nanofarads to be converted to farads, the correct plate-charge interpretation, and a gradient taken from axes in the stated order.
Worked example

A capacitor stores charge 3.0mC3.0\,\text{mC} at 12V12\,\text{V}. Determine its capacitance.

  1. 1.Convert charge: Q=3.0×103CQ=3.0\times10^{-3}\,\text{C}.
  2. 2.Use C=Q/V=(3.0×103)/12C=Q/V=(3.0\times10^{-3})/12.
  3. 3.Convert the result to microfarads.

Answer: C=2.5×104F=250μFC=2.5\times10^{-4}\,\text{F}=250\,\mu\text{F}.

Common mistakes

  • Don't use the sum of the positive and negative plate-charge magnitudes as QQ.
  • Don't leave millcoulombs or microfarads unconverted in an SI calculation.
  • Don't call the gradient of a VV-against-QQ graph capacitance.

Exam tip

Write the graph axes beside C=Q/VC=Q/V before interpreting a gradient.

Tier 1 · Easy

ORIGINAL

A capacitor stores charge of 3.6mC3.6\,\text{mC} at a potential difference of 12V12\,\text{V}. Calculate its capacitance.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A charge-potential-difference graph for a capacitor passes through the point (16.0V,4.80mC)(16.0\,\text{V},4.80\,\text{mC}). Determine its capacitance and the charge stored at 27.0V27.0\,\text{V}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 470μF470\,\mu\text{F} capacitor is initially at 9.0V9.0\,\text{V}. A charge-transfer process then moves 1.2×10161.2\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.

[5 marks]

Total for this question: 5

3.7.4.2

Parallel plate capacitor

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A parallel-plate capacitor has C=ε0εrA/dC=\varepsilon_0\varepsilon_rA/d, so capacitance increases with overlap area AA and relative permittivity εr\varepsilon_r, but decreases with plate separation dd. In a dielectric, simple polar molecules rotate in the field and the resulting bound surface charges oppose the original field.
  • For fixed free charge this reduces potential difference and therefore increases capacitance.
  • If the capacitor remains connected to a fixed-voltage supply, extra charge flows onto the plates.
  • Experiments may determine relative permittivity or test the CCAA and CCdd relationships.
  • Examiners expect the overlap area, SI separation and whether charge or voltage is held constant to be identified.
Parallel capacitor plates separated by distance d with a dielectric polarised in the field.
Worked example

Plates of area 2.0×102m22.0\times10^{-2}\,\text{m}^2 are separated by 1.0mm1.0\,\text{mm} and filled by material of εr=4.0\varepsilon_r=4.0. Find CC.

  1. 1.Convert separation: d=1.0×103md=1.0\times10^{-3}\,\text{m}.
  2. 2.Use C=ε0εrA/dC=\varepsilon_0\varepsilon_rA/d.
  3. 3.C=(8.85×1012)(4.0)(2.0×102)/(1.0×103)C=(8.85\times10^{-12})(4.0)(2.0\times10^{-2})/(1.0\times10^{-3}).

Answer: C=7.1×1010FC=7.1\times10^{-10}\,\text{F}.

Common mistakes

  • Don't use total plate area rather than the overlapping area.
  • Don't say a dielectric strengthens the field for fixed free charge.
  • Don't claim free charge increases for an isolated capacitor.

Exam tip

State whether the capacitor is isolated or remains connected before describing dielectric effects.

Tier 1 · Easy

ORIGINAL

Two parallel plates in air have overlap area 2.50×102m22.50\times10^{-2}\,\text{m}^2 and separation 1.20mm1.20\,\text{mm}. Calculate their capacitance.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A parallel-plate capacitor has area 1.80×102m21.80\times10^{-2}\,\text{m}^2, separation 0.800mm0.800\,\text{mm} and is connected to a 120V120\,\text{V} supply. A dielectric of relative permittivity 3.403.40 completely fills the gap. Determine the new capacitance and charge stored.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A designer needs a 2.00nF2.00\,\text{nF} parallel-plate capacitor using a dielectric with relative permittivity 4.504.50 and thickness 0.750mm0.750\,\text{mm}. Calculate the required overlap area. Explain at the molecular level why the dielectric increases capacitance.

[6 marks]

Total for this question: 6

3.7.4.3

Energy stored by a capacitor

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Energy stored by a capacitor is the work done separating charge and is E=12QV=12CV2=Q2/(2C)E=\tfrac12QV=\tfrac12CV^2=Q^2/(2C).
  • On a graph of potential difference against charge, energy is the area under the line; for a linear capacitor the triangular area gives the factor 1/21/2.
  • The formula should be selected to match the quantities held fixed.
  • If voltage changes, energy released is the difference between initial and final stored energies, not QΔVQ\Delta V with one unchanged charge value.
  • Examiners expect the squared quantity to be retained, graph axes to be checked, and energy changes to be calculated from two complete energy values.
A potential-difference against charge graph whose triangular area is the capacitor’s stored energy.
Worked example

A 220μF220\,\mu\text{F} capacitor is charged to 12V12\,\text{V}. Calculate its stored energy.

  1. 1.Convert C=220×106FC=220\times10^{-6}\,\text{F}.
  2. 2.Use E=12CV2E=\tfrac12CV^2.
  3. 3.E=12(220×106)(12)2E=\tfrac12(220\times10^{-6})(12)^2.

Answer: E=1.58×102JE=1.58\times10^{-2}\,\text{J}.

Common mistakes

  • Don't omit the factor 1/21/2 in the capacitor-energy equation.
  • Don't forget to square voltage in E=12CV2E=\tfrac12CV^2.
  • Don't use QΔVQ\Delta V as the released energy while charge is changing.

Exam tip

For a discharge between two voltages, calculate EiEfE_i-E_f using 12CV2\tfrac12CV^2.

Tier 1 · Easy

ORIGINAL

A capacitor stores 2.0mC2.0\,\text{mC} at a potential difference of 9.0V9.0\,\text{V}. Calculate its stored energy.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

The potential difference across a 150μF150\,\mu\text{F} capacitor falls from 20.0V20.0\,\text{V} to 8.0V8.0\,\text{V}. Determine the energy transferred from the capacitor.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 330μF330\,\mu\text{F} capacitor is charged from zero to 24.0V24.0\,\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 0.800s0.800\,\text{s}.

[5 marks]

Total for this question: 5

3.7.4.4

Capacitor charge and discharge

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The time constant of a resistor–capacitor circuit is τ=RC\tau=RC. During discharge, Q=Q0et/RCQ=Q_0e^{-t/RC}, with corresponding equations for VV and current magnitude; after one time constant each is 0.3680.368 of its initial value and T1/2=0.69RCT_{1/2}=0.69RC.
  • During charging, Q=Q0(1et/RC)Q=Q_0(1-e^{-t/RC}): charge and voltage rise to limiting values while current falls.
  • Graph gradients give rates and area under an IItt graph gives transferred charge.
  • Required practical 9 includes charge and discharge graphs and a log-linear plot whose gradient is 1/RC-1/RC.
  • Examiners expect a time constant from data, not a claim that discharge completes after one RCRC.
Exponential charging and discharging curves with one time constant marked.
Worked example

A 220μF220\,\mu\text{F} capacitor discharges through 47kΩ47\,\text{k}\Omega. Find the time constant and half-life.

  1. 1.τ=RC=(47×103)(220×106)=10.34s\tau=RC=(47\times10^3)(220\times10^{-6})=10.34\,\text{s}.
  2. 2.Use T1/2=0.69RCT_{1/2}=0.69RC.
  3. 3.T1/2=0.69(10.34)=7.13sT_{1/2}=0.69(10.34)=7.13\,\text{s}.

Answer: τ=10.3s\tau=10.3\,\text{s} and T1/2=7.13sT_{1/2}=7.13\,\text{s}.

Common mistakes

  • Don't treat one time constant as complete charge or discharge.
  • Don't use et/RCe^{-t/RC} for charging without the leading 11-.
  • Don't report the log-plot gradient as RCRC instead of 1/RC-1/RC.

Exam tip

On a discharge graph, one time constant is the time to fall to 37%37\% of the initial value.

Tier 1 · Easy

ORIGINAL

A 220μF220\,\mu\text{F} capacitor discharges through a 47kΩ47\,\text{k}\Omega resistor. Calculate the time constant and state the fraction of initial charge remaining after this time.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A capacitor initially stores 6.00mC6.00\,\text{mC} and discharges through an 82.0kΩ82.0\,\text{k}\Omega resistor. Its capacitance is 100μF100\,\mu\text{F}. Determine the charge remaining after 12.0s12.0\,\text{s}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

During discharge, charge falls from 8.0mC8.0\,\text{mC} to 1.6mC1.6\,\text{mC} in 14.0s14.0\,\text{s}. The capacitance is 120μF120\,\mu\text{F}. Determine the time constant, the resistance, and the charge after one further time constant. State the gradient of a graph of lnQ\ln Q against tt.

[6 marks]

Total for this question: 6

3.7.5.1

Magnetic flux density

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A straight current-carrying wire perpendicular to a magnetic field experiences force F=BIlF=BIl.
  • Magnetic flux density BB is therefore force per unit current per unit length, and one tesla gives 1N1\,\text{N} on a 1m1\,\text{m} wire carrying 1A1\,\text{A} at right angles.
  • Fleming’s left-hand rule links field, conventional current and force directions; electron flow is opposite to conventional current.
  • Required practical 10 uses a top-pan balance to investigate how force varies with BB, II and ll, converting a mass-reading change using F=ΔmgF=\Delta mg.
  • Examiners expect the perpendicular condition, active wire length and mass conversion to be stated.
A current-carrying wire perpendicular to a magnetic field, showing the magnetic-force direction.
Worked example

A 0.080m0.080\,\text{m} wire carries 3.2A3.2\,\text{A} perpendicular to a 0.45T0.45\,\text{T} field. Calculate the force.

  1. 1.The field is perpendicular, so use F=BIlF=BIl.
  2. 2.F=(0.45)(3.2)(0.080)F=(0.45)(3.2)(0.080).
  3. 3.Round to two significant figures.

Answer: F=0.12NF=0.12\,\text{N}.

Common mistakes

  • Don't use the total wire length rather than the length inside the field.
  • Don't substitute a balance change in grams directly as force.
  • Don't use electron-flow direction in Fleming’s left-hand rule.

Exam tip

State that the field and current are perpendicular before using the full F=BIlF=BIl expression.

Tier 1 · Easy

ORIGINAL

A 0.080m0.080\,\text{m} wire carries 3.2A3.2\,\text{A} perpendicular to a magnetic field of flux density 0.45T0.45\,\text{T}. Calculate the force on the wire.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A horizontal wire of length 0.120m0.120\,\text{m} carries 4.00A4.00\,\text{A} perpendicular to a magnetic field. Switching on the current changes the balance reading by 6.50g6.50\,\text{g}. Determine the magnetic flux density.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In a balance experiment, a perpendicular wire of active length 0.0850m0.0850\,\text{m} gives a straight-line graph of balance-reading change against current with gradient 1.75g A11.75\,\text{g A}^{-1}. Determine the flux density and predict the reading change at 5.20A5.20\,\text{A}. Explain what a non-zero vertical intercept would suggest.

[5 marks]

Total for this question: 5

3.7.5.2

Moving charges in a magnetic field

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A charge moving perpendicular to a magnetic field experiences force F=BQvF=BQv, perpendicular to both velocity and field. Fleming’s left-hand rule gives the force on positive conventional current; the direction reverses for a negative particle.
  • Since magnetic force is always perpendicular to motion, it does no work and changes direction without changing speed or kinetic energy.
  • Equating magnetic and centripetal forces gives r=mv/(BQ)r=mv/(BQ) for a circular path.
  • This underlies devices such as the cyclotron, whose non-relativistic frequency is f=BQ/(2πm)f=BQ/(2\pi m) and is independent of orbit radius and speed.
  • Examiners expect correct three-dimensional directions, charge magnitude and radius rather than diameter.
A positive charged particle following a circular path in a magnetic field into the page.
Worked example

A proton travels at 4.0×106m s14.0\times10^6\,\text{m s}^{-1} perpendicular to a 0.25T0.25\,\text{T} field. Find the force and orbit radius.

  1. 1.F=BQv=(0.25)(1.60×1019)(4.0×106)=1.6×1013NF=BQv=(0.25)(1.60\times10^{-19})(4.0\times10^6)=1.6\times10^{-13}\,\text{N}.
  2. 2.Use r=mv/(BQ)r=mv/(BQ).
  3. 3.r=(1.67×1027)(4.0×106)/[(0.25)(1.60×1019)]=0.167mr=(1.67\times10^{-27})(4.0\times10^6)/[(0.25)(1.60\times10^{-19})]=0.167\,\text{m}.

Answer: F=1.6×1013NF=1.6\times10^{-13}\,\text{N} and r=0.167mr=0.167\,\text{m}.

Common mistakes

  • Don't use the same force direction for positive and negative particles.
  • Don't claim magnetic force increases the particle’s kinetic energy.
  • Don't substitute path diameter for rr.

Exam tip

Find the force direction for a positive charge first, then reverse it only if the particle is negative.

Tier 1 · Easy

ORIGINAL

A proton moves at 4.0×106m s14.0\times10^6\,\text{m s}^{-1} perpendicular to a 0.25T0.25\,\text{T} magnetic field. Calculate the magnetic force magnitude.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A proton of speed 3.20×106m s13.20\times10^6\,\text{m s}^{-1} enters a uniform 0.480T0.480\,\text{T} magnetic field perpendicular to the field. Determine the radius of its path. Use mp=1.67×1027kgm_p=1.67\times10^{-27}\,\text{kg}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An alpha particle of mass 6.64×1027kg6.64\times10^{-27}\,\text{kg} and charge +3.20×1019C+3.20\times10^{-19}\,\text{C} reaches radius 0.450m0.450\,\text{m} in a cyclotron with magnetic flux density 0.800T0.800\,\text{T}. Determine its speed, kinetic energy in MeV\text{MeV}, and cyclotron frequency.

[6 marks]

Total for this question: 6

3.7.5.3

Magnetic flux and flux linkage

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Magnetic flux is Φ=BA\Phi=BA when flux density is normal to area AA, and is measured in webers. For a coil of NN turns, flux linkage counts the flux through every turn.
  • If θ\theta is the angle between the field and the coil’s normal, NΦ=BANcosθN\Phi=BAN\cos\theta. Flux linkage is maximum when the normal is parallel to the field and zero when the coil plane is parallel to the field.
  • A negative sign indicates flux opposite to the chosen positive normal.
  • Required practical 11 uses a search coil and oscilloscope while varying angle.
  • Examiners expect the angle to the normal, not the plane, and clear distinction between flux and flux linkage.
A tilted coil showing angle theta between its normal and the magnetic field.
Worked example

A 200200-turn coil of area 1.5×102m21.5\times10^{-2}\,\text{m}^2 is in a 0.32T0.32\,\text{T} field with its normal at 6060^{\circ} to the field. Find the flux linkage.

  1. 1.Use NΦ=BANcosθN\Phi=BAN\cos\theta.
  2. 2.Substitute (0.32)(1.5×102)(200)cos60(0.32)(1.5\times10^{-2})(200)\cos60^{\circ}.
  3. 3.Evaluate in weber-turns.

Answer: NΦ=0.48Wb-turnN\Phi=0.48\,\text{Wb-turn}.

Common mistakes

  • Don't use the angle between the field and the coil plane.
  • Don't omit the number of turns when calculating flux linkage.
  • Don't state that flux is maximum when the field lies in the coil plane.

Exam tip

Draw the coil normal explicitly before choosing the angle in BANcosθBAN\cos\theta.

Tier 1 · Easy

ORIGINAL

Magnetic flux passes normally through a flat surface of area 1.5×102m21.5\times10^{-2}\,\text{m}^2. The flux density is 0.32T0.32\,\text{T}. Calculate the flux.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 240240-turn coil of area 3.50×103m23.50\times10^{-3}\,\text{m}^2 is in a 0.180T0.180\,\text{T} field. The field makes an angle of 35.035.0^\circ with the normal to the coil. Determine the flux linkage.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 500500-turn search coil of area 8.0×104m28.0\times10^{-4}\,\text{m}^2 is in a uniform 0.120T0.120\,\text{T} field. Its chosen normal is rotated from 20.020.0^\circ to 110110^\circ relative to the field. Calculate the initial and final flux linkages and their signed change.

[5 marks]

Total for this question: 5

3.7.5.4

Electromagnetic induction

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Faraday’s law states that induced emf magnitude equals the rate of change of flux linkage, ε=Δ(NΦ)/Δt|\varepsilon|=|\Delta(N\Phi)/\Delta t|. Lenz’s law gives direction: the induced effect opposes the change in flux that produced it, not necessarily the original field.
  • A straight conductor cutting field lines develops an emf, while a stationary coil in a steady field does not.
  • A current requires a complete circuit.
  • For a coil rotating uniformly, $\varepsilon=BAN\omega\sin\omega t$ and peak emf is BANωBAN\omega.
  • Examiners expect the relevant change in linkage and time interval, a direction justified by the change, and qualitative interpretation of simple induction experiments.
A magnet moving towards a coil, changing its magnetic flux linkage and inducing an emf.
Worked example

A 250250-turn coil’s flux per turn falls from 6.0×104Wb6.0\times10^{-4}\,\text{Wb} to 1.0×104Wb1.0\times10^{-4}\,\text{Wb} in 0.020s0.020\,\text{s}. Find the mean induced emf.

  1. 1.Find the flux change per turn: ΔΦ=5.0×104Wb|\Delta\Phi|=5.0\times10^{-4}\,\text{Wb}.
  2. 2.Use ε=NΔΦ/Δt|\varepsilon|=N|\Delta\Phi|/\Delta t.
  3. 3.ε=250(5.0×104)/0.020|\varepsilon|=250(5.0\times10^{-4})/0.020.

Answer: ε=6.25V|\varepsilon|=6.25\,\text{V}.

Common mistakes

  • Don't use flux change per turn but omit NN.
  • Don't say the induced field opposes the original field in every case.
  • Don't claim a steady flux through a stationary coil induces a continuous emf.

Exam tip

For Lenz’s law, name the change being opposed: increasing or decreasing flux linkage.

Tier 1 · Easy

ORIGINAL

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

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

The magnetic flux through each turn of a 250250-turn coil decreases uniformly from 3.2mWb3.2\,\text{mWb} to 0.80mWb0.80\,\text{mWb} in 40ms40\,\text{ms}. Determine the induced emf magnitude and state how Lenz's law fixes its polarity.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 180180-turn coil of area 6.50×103m26.50\times10^{-3}\,\text{m}^2 rotates at 50.0Hz50.0\,\text{Hz} in a uniform 0.240T0.240\,\text{T} field. At t=0t=0 its flux linkage is maximum and positive. Calculate the peak emf and the emf magnitude at t=2.50mst=2.50\,\text{ms}. Explain the origin and direction of the emf using Faraday's and Lenz's laws.

[5 marks]

Total for this question: 5

3.7.5.5

Alternating currents

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For sinusoidal current and voltage, Irms=I0/2I_{\mathrm{rms}}=I_0/\sqrt2 and Vrms=V0/2V_{\mathrm{rms}}=V_0/\sqrt2, where subscript 00 denotes peak value. Peak-to-peak is twice the peak.
  • An rms current has the same mean heating effect in a resistor as a direct current of that value; mains voltage is quoted as rms.
  • An oscilloscope can display ac waveforms, act as an ac or dc voltmeter, and measure period and frequency.
  • Amplitude is vertical divisions times volts per division; period is horizontal divisions times time per division.
  • Examiners expect the 2\sqrt2 relationships only for sinusoids and careful separation of peak, rms and peak-to-peak values.
A sinusoidal voltage waveform showing its peak value and period.
Worked example

A mains supply is 230V rms230\,\text{V rms}. Calculate its peak and peak-to-peak voltages.

  1. 1.V0=2Vrms=2(230)=325VV_0=\sqrt2V_{\mathrm{rms}}=\sqrt2(230)=325\,\text{V}.
  2. 2.Peak-to-peak voltage is 2V02V_0.
  3. 3.Vpp=650VV_{\mathrm{p-p}}=650\,\text{V}.

Answer: Peak =325V=325\,\text{V} and peak-to-peak =650V=650\,\text{V}.

Common mistakes

  • Don't treat the quoted mains value as a peak voltage.
  • Don't divide rms by 2\sqrt2 when finding peak.
  • Don't use the sinusoidal rms formula for a non-sinusoidal trace.

Exam tip

Label the requested value as rms, peak or peak-to-peak before applying any factor.

Tier 1 · Easy

ORIGINAL

A sinusoidal voltage has peak value 12.0V12.0\,\text{V}. Calculate its rms value.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A sinusoidal trace spans 6.46.4 horizontal divisions per cycle at 0.50ms div10.50\,\text{ms div}^{-1}. Its peak-to-peak height is 5.65.6 vertical divisions at 2.0V div12.0\,\text{V div}^{-1}. Determine the frequency, peak voltage and rms voltage.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A resistive heater rated at 1.80kW1.80\,\text{kW} operates from a sinusoidal 230V rms230\,\text{V rms} supply. Determine the peak and peak-to-peak supply voltages, the rms and peak currents, and the energy transferred in 12.0min12.0\,\text{min}.

[6 marks]

Total for this question: 6

3.7.5.6

The operation of a transformer

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For an ideal transformer, Ns/Np=Vs/VpN_s/N_p=V_s/V_p. Power is conserved, so increasing secondary voltage reduces secondary current.
  • Real efficiency is IsVs/(IpVp)I_sV_s/(I_pV_p). Losses arise from winding resistance, eddy currents, hysteresis and flux leakage.
  • Laminating and insulating the core restricts eddy-current loops, while a suitable soft magnetic core reduces hysteresis.
  • In power transmission, transformers raise voltage so the same power uses a smaller current, reducing line loss I2RI^2R, then lower voltage for consumers.
  • Examiners expect turn and voltage ratios in matching order, separate input and output power calculations, named loss mechanisms and a transmission-loss calculation using line current and resistance.
A transformer with primary and secondary coils linked by a laminated magnetic core.
Worked example

A transformer has Np=500N_p=500, Ns=100N_s=100 and Vp=230VV_p=230\,\text{V}. It supplies 10A10\,\text{A} at 92%92\% efficiency. Find VsV_s and IpI_p.

  1. 1.Vs=VpNs/Np=230(100/500)=46VV_s=V_pN_s/N_p=230(100/500)=46\,\text{V}.
  2. 2.Pout=IsVs=10(46)=460WP_{out}=I_sV_s=10(46)=460\,\text{W}, so Pin=460/0.92=500WP_{in}=460/0.92=500\,\text{W}.
  3. 3.Ip=Pin/Vp=500/230=2.17AI_p=P_{in}/V_p=500/230=2.17\,\text{A}.

Answer: Vs=46VV_s=46\,\text{V} and Ip=2.17AI_p=2.17\,\text{A}.

Common mistakes

  • Don't reverse only one of the turns or voltage ratios.
  • Don't assume input and output powers are equal for a stated non-ideal efficiency.
  • Don't use transmission voltage rather than current in I2RI^2R line loss.

Exam tip

Write primary quantities on one side and secondary quantities on the other before forming either ratio.

Tier 1 · Easy

ORIGINAL

An ideal transformer has 300300 primary turns and 12001200 secondary turns. The primary voltage is 24V24\,\text{V}. Calculate the secondary voltage.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A transformer takes 1.80A1.80\,\text{A} from a 230V230\,\text{V} supply and delivers 24.0V24.0\,\text{V} at 84.0%84.0\% efficiency. Determine the secondary current and the power dissipated in the transformer.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A station transmits 2.50MW2.50\,\text{MW} through cables of total resistance 3.20Ω3.20\,\Omega. Compare the cable power losses when transmission voltage is 25.0kV25.0\,\text{kV} and 250kV250\,\text{kV}. Explain the transformer's role and how core construction reduces eddy-current loss.

[6 marks]

Total for this question: 6

Want help turning these notes into marks?

Bring a tricky specification point or a recent answer, and we can work through the method and exam wording together.