Looking for JC H2 Physics tuition in Bukit Timah for resistance, resistivity, internal resistance, potential dividers, filament lamps or thermistors? Students often know V=IR but reverse the output across a potential divider, mistake current–voltage graph gradient for resistance irrespective of axis labels, or assume a hotter NTC thermistor behaves like a hotter metal filament. A useful tutor starts with an accurately drawn circuit and explains the physical meaning of each current and potential difference before selecting a formula.
The 2027 Singapore–Cambridge A-Level H2 Physics syllabus 9478 covers these requirements in Topic 16, Circuits, supported by Topic 15, Currents. Students should interpret circuit symbols, define resistance, use R=ρl/A, analyse I–V characteristics of an ohmic resistor, semiconductor diode, filament lamp and negative-temperature-coefficient thermistor, reason about internal resistance and source terminal p.d., combine resistors and solve potential dividers with thermistors or light-dependent resistors. RC charging and capacitor combinations also occur in Topic 16 but have their own existing BT108 guide.
eduKate Singapore’s established Bukit Timah tutorial model uses up to three pupils per group, generally 90 minutes weekly, at 8 Fourth Avenue, Singapore 268674 near Sixth Avenue MRT. This educational guide describes a way to learn the topic, not confirmation of a current specialist, lesson place, fees, laboratory equipment or timetable; families should check with the centre directly.
Official H2 Physics 9478 Topic 16 syllabus scope
The SEAB 2027 H2 Physics syllabus 9478 lists circuit diagrams, resistance and resistivity, current–voltage graphs, temperature-related effects in metals and semiconductors, internal resistance, series and parallel resistors and potential dividers. It explicitly includes NTC thermistors and light-dependent resistors. RC-circuit exponential changes are a later part of the same topic but are not the main purpose of this article.
In Topic 15, the syllabus also requires I=Q/t, I=nAvq for charge drift and P=VI, P=I²R and P=V²/R. A teacher should connect those foundations without confusing electrical power with resistance or energy per charge.
The 2027 A-Level syllabus directory confirms the examination code. A new tutoring exercise should always state whether the source is ideal and whether a potential-divider output is loaded.
Start every circuit problem with the nodes
A circuit node connects points joined by ideal conductors in the simple schematic model. Components placed between the same two nodes share the same potential difference, while components on one unbranched path carry the same current in steady state. These statements are the foundation of series and parallel relationships.
A student who identifies a parallel arrangement merely because two resistors look side-by-side on the page can make a costly modelling mistake. The correct test is whether both ends connect to the same pair of nodes.
A tutor can redraw the same circuit with different component positions while preserving its connections and ask whether the electrical relationships changed.
Conventional current is defined by positive-charge flow
Conventional current is the direction of positive charge flow. In a common metal, mobile electrons drift opposite to that convention, even though the circuit current direction is drawn as though positive charge flows. The source establishes an electric field and charge carriers respond; current is not the speed of a single electron.
The simple definition is I=ΔQ/Δt for a mean current, with current in amperes and charge in coulombs. If 6 C passes a cross-section in 3 s at a steady rate, current is 2 A.
A tutor should ask pupils to distinguish electrical current from net charge stored at one node and to mark current arrows consistently around a simple loop.
Drift velocity connects microscopic and circuit quantities
For a conductor with number density n of mobile carriers, cross-sectional area A, carrier drift-speed magnitude v and charge magnitude q, the steady current magnitude is I=nAvq. The symbol n is number per cubic metre, not the number of moles. The tiny average drift speed of electrons can coexist with rapid electrical response around a circuit.
With n=8.5×10²⁸ m⁻³, A=1.0×10⁻⁶ m², v=1.0×10⁻⁴ m s⁻¹ and q=1.60×10⁻¹⁹ C, current magnitude is about 1.36 A. A student should multiply all four factors using powers of ten carefully.
A tutor can ask which factor changes if a wire’s cross-sectional area doubles but current and carrier density stay fixed: the required drift speed halves.
Potential difference is work per unit charge
The potential difference V between two points is energy transferred per unit charge between those points in the relevant circuit model, V=W/Q. One volt equals one joule per coulomb. A source’s electromotive force, or e.m.f., describes energy supplied per unit charge in a complete circuit under its definition, rather than simply any measured voltage across any component.
If 15 J is transferred when 3 C passes a component, the p.d. is 5 V. A battery’s e.m.f. and its terminal p.d. can differ when internal resistance causes energy transfer within the source.
A tutor should insist on naming the start and end points associated with a voltage instead of writing an unexplained number next to a wire.
Ohm’s law applies under its conditions
Resistance is defined as R=V/I for a circuit component at a specified operating state with I nonzero. An ohmic conductor at constant temperature shows V proportional to I, giving constant resistance. Not every component has a constant R as voltage and temperature change.
For a fixed ohmic resistor carrying 0.50 A at 6 V, R=12 Ω. At 3 V, it would carry 0.25 A under the same constant-temperature model. A filament lamp does not generally follow the same simple proportional line because heating changes its properties.
A tutor should ask whether the component is stated to be ohmic before using one resistance value for every potential difference.
A resistance graph needs axis labels
On a graph of voltage V vertically against current I horizontally, the gradient of a straight ohmic line gives V/I=R. On an I-versus-V graph, gradient instead gives I/V=1/R for an ohmic resistor. These graph slopes have reciprocal units.
If V is on the vertical axis and the line rises from 0 to 10 V as I rises from 0 to 2 A, gradient is 5 Ω. If axes are reversed, the slope is 0.2 A V⁻¹.
A pupil who writes “gradient is resistance” without checking axes has learned a slogan rather than graphical Physics.
The I–V curve of a filament lamp is non-linear
As current through a filament lamp rises, its temperature generally rises. For a typical metal filament, lattice vibrations increase and resistivity rises, so resistance becomes greater. The I-versus-V curve is therefore non-linear, with a changing I/V ratio and changing local gradient.
A graph may appear relatively steep near the origin and flatten as larger voltages increase filament temperature. This is different from the straight I–V line of an ideal constant-temperature ohmic resistor.
A tutor should teach R=V/I at the selected point; the reciprocal of an instantaneous I–V curve gradient is not automatically equal to the point’s V/I for a non-linear device.
Why temperature usually increases metal resistance
In a typical metal, higher temperature increases atomic lattice vibrations, raising the frequency of scattering events experienced by charge carriers. The carrier drift response under a given electric field can be reduced, increasing resistivity. The syllabus’s specific explanatory focus is drift velocity for metals.
The result applies to typical metal conduction over relevant conditions rather than asserting every conceivable material has the same temperature coefficient. A special superconducting transition is a different phenomenon, not a counterexample to be inserted into basic filament-lamp questions.
A tutor can ask why the same filament offers greater resistance when hot than when cold, identifying the scattering mechanism.
NTC thermistors behave differently
An NTC, or negative temperature coefficient, thermistor is a semiconductor device whose resistance decreases as temperature rises over its intended operating range. In the syllabus model, increasing temperature can increase charge-carrier number density substantially, leading to greater conductivity. Its behaviour contrasts with a typical metal filament.
A student should not call all temperature-sensitive resistors NTC without checking their sign. For an NTC, heating causes the resistance to fall; cooling causes it to rise under the usual model.
A tutor can sketch thermistor resistance against temperature and ask for the downward trend and qualitative semiconductor explanation.
Light-dependent resistors respond to illumination
A light-dependent resistor changes electrical resistance with incident light intensity. In the standard school semiconductor device, increased illumination generally creates or frees more mobile charge carriers and lowers resistance. Its resistance is not determined merely by the lamp’s emitted heat.
A potential-divider circuit using an LDR may therefore produce a changing output as a room becomes brighter or darker. The direction of the change depends on whether the LDR is connected to the high-potential or low-potential side of the divider.
A tutor should ask the pupil to mark exactly across which component the output voltage is measured before predicting the response.
A semiconductor diode has directional conduction
A typical semiconductor diode passes substantial conventional current in its forward-biased direction once appropriate conditions are met, while conducting very little in reverse bias within its normal operating range. Its I–V characteristic is strongly non-linear and not symmetric about the origin like an ideal resistor.
A student who uses V=IR with one constant resistance for both diode directions has ignored the device’s physics. A circuit diagram’s diode orientation helps determine whether significant steady current flows.
A tutor can ask the learner to mark anode, cathode and conventional-current direction on a simple series circuit without resorting to advanced junction-physics detail.
Resistance depends on geometry and material
For a uniform conductor of resistivity ρ, length l and cross-sectional area A, resistance is R=ρl/A under the model. Resistivity describes material response under stated conditions, while resistance depends additionally on shape. A longer conductor has larger R at fixed material and area; a thicker one has smaller R.
Doubling length doubles resistance, while doubling cross-sectional area halves it. These are conditional proportionalities: changing dimensions while also changing temperature would require more information.
A tutor can ask whether cutting a uniform wire in half changes the resistivity. The material property ρ remains the same under unchanged state conditions, even though resistance changes.
Worked resistivity calculation
Suppose a conductor has ρ=1.7×10⁻⁸ Ω m, length l=2.0 m and cross-sectional area 1.0×10⁻⁶ m². Then R=(1.7×10⁻⁸)(2.0)/(1.0×10⁻⁶)=0.034 Ω. It is a small resistance because of the chosen conductive material and geometry.
The area must be in square metres, not square millimetres left unchanged. A question giving a circular wire diameter requires calculating A=π(d/2)², rather than substituting diameter as the area.
A tutor should ask pupils to predict the effect of halving diameter: area becomes one quarter, so resistance quadruples at fixed length and resistivity.
Unit of resistivity
From R=ρl/A, the resistivity unit is Ω m, since ρ=RA/l. It is not Ω/m or Ω m². The SI dimensions provide an independent check on a rearranged formula. Resistance itself has ohms, while the conductor’s resistance per unit length has yet another unit.
A common worksheet mistake writes resistivity equal to R/l, omitting area. That quantity is resistance per unit length, not intrinsic material resistivity.
A tutor can ask learners to reconstruct ρ’s unit directly from the original equation before numerical substitution.
Series resistors carry the same current
In a single unbranched steady-state path, every series resistor carries the same current. The total p.d. is the sum of individual voltage drops, so their equivalent resistance is R_eq=R₁+R₂+…. A current entering a series chain does not disappear after the first resistor.
Resistors 2 Ω and 4 Ω in series have total 6 Ω. With 12 V across the chain, current is 2 A, giving voltage drops 4 V and 8 V respectively. Their drops add to the supply.
A tutor should ask whether both components can carry 2 A while having different potential differences, highlighting what resistance actually determines.
Parallel resistors share potential difference
Two components connected across the same pair of nodes experience the same p.d. in the ideal wire model. The currents through their branches may differ, and the total current is their sum. The equivalent resistance obeys 1/R_eq=1/R₁+1/R₂+….
A 6 Ω and a 3 Ω resistor in parallel have equivalent resistance 2 Ω. Across 12 V, branch currents are 2 A and 4 A, adding to 6 A. Substituting the same 6 A into both branches would be incorrect.
A tutor should have learners identify the shared nodes before applying the parallel formula.
Equivalent parallel resistance is smaller
For positive resistor values, an equivalent parallel resistance is less than either branch resistance. Adding a parallel conductive path makes the network easier for current to flow through under the same voltage, because conductance increases. This physical check helps detect a reversed reciprocal formula.
Two equal 10 Ω resistors in parallel give 5 Ω, not 20 Ω. In series they give 20 Ω. The two outcomes represent different electrical connections, not contradictory resistor rules.
A tutor can ask pupils which arrangement draws a larger current from a fixed ideal voltage source and why.
Mixed circuits must be reduced in stages
When several resistors form nested series and parallel groups, identify sections sharing the same nodes, reduce a group to an equivalent resistance, then recompute total current and reconstruct branch values. A diagram that looks complex may contain only a few independent node relationships.
Suppose a 2 Ω resistor is in series with a parallel combination of 6 Ω and 3 Ω. The parallel equivalent is 2 Ω and the whole circuit is 4 Ω. With 12 V supply, total current is 3 A. Each 2 Ω series portion takes 6 V, so the parallel branch has 6 V.
A tutor should then calculate parallel branch currents of 1 A and 2 A and confirm their sum 3 A. This independently checks the reduction.
A potential divider splits the source voltage
Two series resistors across a fixed supply form a simple potential divider. When the output is measured across the lower resistor R₂ and no load affects it, V_out=V_in R₂/(R₁+R₂). It works because the same current passes through both resistors and V=IR holds for each.
If V_in=12 V, upper R₁=2 kΩ and lower R₂=4 kΩ, output across R₂ is 12×4/(2+4)=8 V. The output across upper R₁ would instead be 4 V.
A tutor should require pupils to circle the output measurement terminals: the ratio depends on which resistor is measured.
Potential-divider derivation makes it memorable
The series current is I=V_in/(R₁+R₂). The output voltage across R₂ is IR₂, yielding V_out=V_in R₂/(R₁+R₂). Deriving the formula avoids memorising an upper-versus-lower numerator rule with no link to physical connections.
A pupil who swaps the numerator can test the limiting case R₂→0: voltage across the lower component should approach zero, not the entire supply voltage.
A tutor can use these limits to check a result before calculating exact resistor values.
NTC thermistor below: heating can reduce Vout
Place a fixed resistor R₁ on the upper side of a divider and an NTC thermistor R_T below, with output measured across R_T. Increasing temperature reduces R_T, so V_out=V_in R_T/(R₁+R_T) decreases under the fixed-supply unloaded model.
For V_in=12 V and upper fixed 2 kΩ, if R_T falls from 4 kΩ to 2 kΩ, V_out falls from 8 V to 6 V. A student must connect the thermistor’s decreasing resistance to the component across which voltage is measured.
A tutor can swap the thermistor’s position and ask whether the observed output trend reverses.
NTC thermistor above: output can rise when hot
If the NTC thermistor occupies the upper divider position and the fixed resistor is below, output across the fixed lower resistor is V_in R_fixed/(R_T+R_fixed). Heating lowers R_T, decreases the denominator and therefore raises the output under the same unloaded ideal model.
For lower fixed resistance 2 kΩ with thermistor reducing from 4 kΩ to 2 kΩ at 12 V input, V_out rises from 4 V to 6 V. Both thermistor placements are valid but respond in opposite directions.
A tutor should ask pupils to state which resistor is temperature dependent before giving any sentence about output rising or falling.
Light-dependent resistor placement changes the signal
A typical LDR becomes less resistive as illumination increases. If it is the lower resistance in an unloaded divider, increased light generally reduces the output taken across it; if it is the upper resistance, increased light generally raises the output measured across a fixed lower resistor.
This is useful in light sensing, although actual circuits may include buffering, thresholds or other elements not captured by the two-resistor model. The syllabus expects potential-divider reasoning with LDRs, not an unsupported universal claim about every light sensor.
A tutor can ask learners to predict what happens when a torch is moved closer while retaining the same supply and wiring.
A loaded divider is not the simple ratio unchanged
A voltmeter or connected external load can affect the voltage divider if it draws appreciable current. An ideal voltmeter draws no current, while a practical load across the lower resistor introduces a parallel branch and changes the effective lower resistance. Therefore the unloaded divider equation may no longer apply unchanged.
If a load R_L is placed across R₂, replace the lower component by the equivalent R₂||R_L before calculating current and output. The result can be smaller than the original unloaded voltage.
A tutor can ask why a real sensor interface may need a high-input-resistance measurement stage, while keeping the compulsory H2 calculations within the supplied circuit.
Worked loaded-divider check
An unloaded divider has 3 kΩ upper and 3 kΩ lower across 12 V, yielding 6 V output. Add a 3 kΩ load across the lower 3 kΩ. The effective lower resistance becomes 1.5 kΩ, so V_out=12×1.5/(3+1.5)=4 V.
A pupil who still reports 6 V has treated a non-negligible load as though no current flows through it. The branch arrangement must be updated before using the simple ratio.
A tutor can ask learners to calculate total source current before and after adding the load, explaining why it rises.
Electromotive force and terminal potential difference
Electromotive force ε is energy supplied per unit charge by a source in the circuit model. If the source has internal resistance r and current I flows as the source supplies energy, part of the e.m.f. is dropped internally, so the terminal p.d. is V_terminal=ε−Ir. This can be less than the source’s e.m.f. under load.
For an ideal source r=0, terminal p.d. equals ε in the simple model. For a real source under increasing current, the internal voltage drop can grow, decreasing output terminal voltage.
A tutor should draw the internal resistance as a component in series with the ideal e.m.f. before calculating the external current.
Worked cell with internal resistance
Suppose a source has ε=1.5 V, internal resistance r=0.50 Ω and external load R=2.5 Ω. Total resistance is 3.0 Ω, so I=ε/(R+r)=0.50 A. The load voltage is IR=1.25 V, while internal drop Ir=0.25 V. Their sum is 1.5 V.
The source’s e.m.f. is not an extra 1.5 V across the external load after subtracting internal drop. A pupil who gives load voltage 1.75 V has added rather than subtracted the internal loss.
A tutor should use ε=IR+Ir as an independent audit of the answer.
Internal resistance affects output power
For a source with e.m.f. ε and internal resistance r supplying current I, electrical power delivered to its external load is P_out=I V_terminal, while the internal resistance dissipates I²r. Under the simple load model, energy conservation gives source power εI=P_out+I²r.
Using ε=1.5 V and I=0.50 A from the preceding example, source power is 0.75 W, external-load power is 0.625 W and internal loss is 0.125 W. The values balance exactly.
A tutor can ask why load output power is less than εI for a loaded real source even when the cell is functioning normally.
Short-circuit current is not a safe experiment
In the idealised internal-resistance model, connecting an extremely low resistance across a source can give a large current limited by the internal resistance and other real factors. The arithmetic is useful for understanding the model, but real short circuits can overheat sources and wiring and are not appropriate unsupervised classroom demonstrations.
For ε=1.5 V and r=0.50 Ω, the limiting ideal model with external R→0 gives I→3.0 A. An actual battery’s behaviour depends on chemistry, temperature and protection features.
A tutor should explain the physical risk and use diagrams or supplied data rather than instructing students to short real batteries.
Electrical power formulas under ohmic conditions
The general instantaneous electrical power transferred across a component is P=VI. Combining with V=IR for an ohmic resistance yields P=I²R and P=V²/R, with the chosen values referring to the same component. Using source voltage with only a subcomponent’s resistance generally creates a wrong power result.
A 6 Ω resistor carrying 2 A dissipates P=I²R=24 W, with 12 V across it. The same result follows from V²/R=144/6=24 W.
A tutor can ask why using total supply voltage across one resistor in a multi-resistor divider may overestimate its power.
Power rating and heat dissipation
Resistors and other components have practical operating limits. Electrical power delivered to a resistive component can become thermal energy, changing its temperature and sometimes its resistance. A calculated ideal circuit may therefore violate a real part’s rating if it would demand excessive power.
A 100 Ω resistor across 10 V dissipates 1 W in the constant-resistance model. Whether that is safe requires the actual device’s power rating and thermal conditions; a mathematics result is not itself an equipment safety clearance.
A tutor should separate paper-circuit calculations from real hardware operation and avoid unverified laboratory assurances.
The gradient of a V–I graph is not always constant
For a filament lamp, V/I may change with current due to heating, and the local derivative dV/dI can differ from V/I. The chord from the origin to a point represents V/I, whereas the tangent gradient describes incremental response. This distinction matters whenever a device is non-ohmic.
If a lamp has V=4 V at I=0.50 A, its point resistance is 8 Ω. A tangent gradient around that operating point could be another number; the two quantities should not be assumed equal.
A tutor can draw a curved V–I plot and ask which line is the chord and which is the tangent.
Thermistor and filament lamp graph comparison
An ohmic resistor has an approximately straight I–V relationship at constant temperature. A hot filament lamp becomes less conductive as temperature rises; an NTC thermistor typically becomes more conductive as temperature rises. Their characteristic changes follow different charge-carrier physics and must not be collapsed into one thermal rule.
A pupil can be given three current–voltage curves and asked to identify the likely ohmic resistor, filament lamp and diode based on shape, with an accompanying qualitative explanation for an NTC device under varying temperature.
The tutor should explain that device bias, temperature and axis orientation determine the precise graph, not one memorised sketch independent of operating conditions.
Resistivity versus conductivity
Resistivity ρ describes a material’s opposition to current in the uniform model, while conductivity σ is its reciprocal, 1/ρ, under the corresponding definition. These are material properties dependent on conditions; they differ from a particular component’s resistance or conductance, which include geometry.
Two copper wires at the same temperature may have the same bulk resistivity but different resistance if their lengths or cross-sectional areas differ. Saying one contains “more resistivity” solely because it is longer confuses material and component.
A tutor can ask which quantity changes when length doubles and which does not, holding material, temperature and area constant.
Current direction and the diode
A diode’s forward and reverse responses can affect whether current reaches a branch in a given circuit. The 2027 Currents syllabus also introduces a single diode for half-wave rectification of an alternating signal, but this guide’s main circuit focus is the d.c. I–V characteristic and how component orientation changes conduction.
When a question shows a diode as ideally blocking reverse current, a parallel or series resistance calculation must first account for whether that path is active. Treating it as an ordinary resistor in both orientations can change the topology.
A tutor can ask the learner to trace a permissible conventional-current path under the stated ideal diode model.
What the first tuition lesson should diagnose
A recent school test may reveal an error in identifying series versus parallel nodes, using I–V axes, computing resistivity, understanding internal resistance or predicting an LDR output. These are different intervention targets. An effective tutor asks for the original uncorrected working and finds the first incorrect model decision.
For example, a student who gets R=ρl/A correct but enters cross-sectional area in mm² needs unit discipline, while one who reverses NTC output direction needs a circuit-node and ratio explanation.
One new independent task after feedback is a stronger improvement check than repeatedly copying the corrected example.
A 90-minute three-pupil circuits lesson
Start with individual tasks identifying shared circuit nodes, two I–V graph shapes and a simple divider output. The tutor models a current-and-voltage flow through one network, then assigns personalised problems: drift velocity, internal resistance or NTC/LDR divider interpretation.
Students complete their own working before group discussion. The final new circuit changes the location of a thermistor or adds a parallel branch, requiring learners to recalculate rather than repeat a memorised trend.
Up-to-three-pupil teaching is valuable only when the actual tutor reviews each learner’s reasoning, and suitability must be confirmed directly.
Four-week H2 circuits preparation
Week one establishes current, drift speed, charge, p.d. and power. Week two covers R=ρl/A and the I–V characteristics of the syllabus components, including temperature effects. Week three develops internal resistance, series and parallel reductions, and voltage/power checks. Week four applies those foundations to potential dividers with fixed resistors, thermistors and LDRs plus unfamiliar data questions.
Retrieval of earlier errors should recur each week with different numbers and circuit layouts. A pupil with weak graph or algebra skills may require additional prerequisite teaching.
The schedule is a model learning plan, not proof of a current class timetable or a guarantee of A-Level marks.
How parents can recognise genuine progress
Ask why an NTC thermistor placed below a fixed resistor can make output voltage decrease when heated, why a lamp’s I–V graph curves, and why an external load can change a potential divider’s voltage. A good answer names the circuit nodes, physical condition and relevant ratio rather than repeating only “V=IR”.
Compare an initial school error with a new unseen circuit completed independently after a delay. Look for accurate resistor grouping, units, graph axes, power accounting and a prediction of how changing one component affects the result.
A responsible tutor should identify gains and remaining gaps without guaranteeing grades.
Frequently asked questions about 2027 H2 circuits
Are thermistors and LDR dividers part of 2027 H2 Physics? Yes, Topic 16 expressly includes them. Does an NTC thermistor gain resistance when heated? No, resistance generally decreases over its intended range. Does every I–V graph gradient equal resistance? Only with appropriate axes and, for simple equality to V/I, the relevant ohmic linear case.
Is terminal voltage always the e.m.f.? No, internal resistance may produce a voltage drop under load. Are RC circuits a separate topic? They are also within Topic 16 but have an existing dedicated guide.
Does this article confirm a specialist or equipment? No; the centre should be contacted for current details.
Related Bukit Timah H2 Physics learning routes
Start with JC H1/H2 Physics Tuition in Bukit Timah. Related but distinct pages cover Electric Fields, Capacitance and RC Circuits, Magnetic Forces and Velocity Selectors and Physics Practical Graph Analysis.
The exam source is SEAB H2 Physics syllabus 9478 (2027). The established education benchmark is the Clementi small-group tutorial reference.
Contact eduKate Singapore or WhatsApp about H2 Circuits tuition, confirming current teacher, fees, group fit and timing.